Earlier material

Tuesday, December 11 (Lecture #28)
Notes by Jinwei Yang, edited, with some comments, by the instructor.

We begin with some ideas used in the standard proof of the Riemann Mapping Theorem (RMT). We won't have time to finish the proof. But this is the now standard proof, given in many textbooks, so please read the remainder of the proof there. In the proof, a proper simply connected, connected open subset U of C is given. If there is a biholomorphic mapping from U to D1(0), then (since we know Aut(D1(0) very well now) we can require that the mapping take a selected point p in U to 0 in D1(0), and we can also ask that the derivative of this mapping at p be real. The Schwarz Lemma also suggests that we make this derivative as large as possible, so that the image will be all of the disc (this is just a suggestion!). Many proofs of the RMT create a sequence of functions {fn}, and the desired biholomorphic mapping is the limit of the sequence. We should try to see when such sequences must have a limit.

Problem Given a "collection" of functions F, when is it true that every sequence in that collection has a subsequence which converges uniformly?

A good collection of examples is important. Our function space for examples was C[0,1], the continuous functions on the unit interval. For simplicity, we will consider real-valued continuous functions.

 Example 1 The family of functions is the sequence fn(x)=n. Certainly this sequence does not converge uniformly. From this we learn that we'd better require that our family of functions is bounded (sometimes called uniformly bounded): that is, there is B>0 so that for all f's in F, we know that |f(x)|<=B. This means, geometrically, that the graphs of f(x) are between y=-B and y=B. Example 2 The family of functions is the sequence fn(x)=sin(nx). I claim that this sequence does not converge uniformly, and, indeed, that it has no subsequence which converges uniformly. Well, just look at x=0 and what happens near 0. If there is a function g which is a limit of a subsequence of the fn's, then I certainly know that g(0)=0 since all of the fn's have value 0 at 0. Notice that for some very large n, |fn(x)-g(x)|<1/2 for x's near 0. Well, since g(0)=0, I bet that we can select an interval of x's near 0 ("some [0,D] for D>0") so |g(x)|<1/2. Then for large enough n and for those x's, |fn(x)|<=|fn(x)-g(x)|+|g(x)|<1/2+1/2=1. But, golly, fn has period 2Pi/n. For large n's, this period certainly fits inside [0,D]. And the values of sine on a period are [-1,1], so there is no uniformly convergent subsequence of this sequence of functions. The sequence of functions wiggles too much.

In general, if a sequence of functions converges uniformly to a limit function, the limit function's wiggle must be close to the wiggle of almost all of the sequence. Of course, it would be helpful if the word or idea of wiggle were made more precise, and, in a real variables courses, I am sure that you will all learn about the oscillation of a function. For this course, we just need the following definition.

Definition F is equicontinuous at x0 if any e>0, there exists d>0, such that for any f in F, if |x-x0|<d, then |f(x)-f(x0)|<e.

A true fact (what's a false fact?) is that a collection of continuous functions defined on a compact metric space which is equicontinuous at every point is actually uniformly equicontinuous (e doesn't depend on x0). This follows from standard compactness arguments.

Now we can state a famous theorem which says the necessary conditions mentioned are actually sufficient. To me, this result is not obvious. Its invention needed consideration of many more examples than the few we have considered.

Theorem (Arzelà-Ascoli) If F is a subset of C[0,1] which is bounded and (uniformly) equicontinuous, then every sequence in F has a uniformly convergent subsequences.

You should try to prove this yourself, even if you've heard or read or have sat through (!) a proof. Just look at a {xn} which is dense in [0,1], and create a subsequence of any sequence in F (diagonally selected!) converge everywhere uniformly.

How can we guarantee or "force" that a collection of functions is equicontinuous at (even) one point? One simple condition follows:

Sufficient condition for equicontinuity
Suppose F is a subset of C[0,1], and suppose that all of the functons in F are differentiable. IF there is a constant K so that |f´(x)|<=K then F is equicontinuous. This follows from the Mean Value Theorem. MVT says f(b)-f(a)=f´(c)(b-a). Now if |b-a|<e/K, |f(b)-f(a)| will certainly be less than e. So d can be e/K, the same for all of the f's in F.

This is not the only way to verify equicontinuity, of course. One can ask for uniform Lipschitz conditions: there's q>0 and r>0 so that |f(b)-f(a)|<=r|b-a|q (think of sqrt(x) and companions on [0,1]: certainly not Lipschitz (a q=1 condition won't be satisfied near 0 but a q=1/2 will be) and certainly not differentiable. There was some discussion of this earlier in the diary (butterflies and boxes!).

Notice that we can put together intervals to get convergence on all of R. For example, here is a result depending A-A for closed bounded intervals:

UCC convergence on R
Suppose F is a collection of continuous functions on R and that we know that the functions in F are uniformly bounded on any compact interval in R, and also that F is uniformly equicontinuous on any compact interval in R. Then every sequence in f has a uniformly convergent subsequence.

Why is this true? Take any sequence of functions in F. We apply A-A on [-1,1] and get a uniformly convergence subsequence. Then consider this subsequence on [-2,2] and get a subsequence of that subsequence, and then ... [-n,n] ... and choose the final subsequence to be a diagonal subsequence of this sequence of subsequences. The result is a sequence which converges uniformly on any compact subset of R (not necessarily on R itself, though). So we use the sigma-compactness of R. Any subset of C is also sigma-compact, so the same approach will be successful there.

What happens when we consider holomorphic functions? For example, suppose we have a function f holomorphic in a disc, with |f´(z)|<=K. If za and zb are in the disc, then f(zb)-f(za)=pathf´(z) dz. We can take the path to be, for example, a straight line segment from za to zb. But ML then gives |f(zb)-f(za|<=K|zb-za|, a Lipschitz estimate, so a family of holomorphic functions whose derivatives have uniformly bounded modulus would also be (uniformly) equicontinuous. And since uniformly convergent sequences of holomorphic functions have holomorphic limits, we see that the limiting continuous function guaranteed by A-A is here also holomorphic.
Notice, please, we are using a sort of Mean Value inequality, which is certainly valid under these circumstances. We aren't using an equality form of MVT since that's not true.

It is amazing and wonderful and powerful that the requirements of Arzelà-Ascoli can be combined in a simpler hypothesis for holomorphic functions.

Theorem (called Montel's Theorem) Suppose U is open, F is a collection of functions which are holomorphic on U, and for any K compact in U, supf in F||f||K is finite (so the modulus of functions in F is bounded on any compact set). Then every sequence in F has a subsequence which converges uniformly on compact subsets.

The classical name for such F's is a normal family. These F's are subsets of the holomorphic functions whose closure is compact ("precompact" sets of holomorphic functions).

Proof U is the increasing union of compact sets Kn. We proved that ||f´||K<=(1/s)||f||Ks for sufficiently small positive s. Here Ks is also compact in U, and is the union of all the closed discs of radius s centered at points in K. So combine this with the previous observation, and we see that an F which is bounded on compact sets is also uniformly equicontinuous on compact sets. So A-A applies. And sigma-compactness also guarantees the existence of a (locally) uniformly convergent sequence of holomorphic functions in F. The limit function will be holomorphic.

(Not) finishing the proof of RMT: start with a well-chosen f on U. The hypothesis of simply connected insures that holomorphic square roots of non-zero holomorphic functions exist. The well-chosen f is sqrt(z-b) for b not in U. We create the initial f and then continue with a sequence of fn's. Montel's Theorem is used to guarantee a limit, and Hurwitz's Theorem (limits of 1-1 holomorphic functions are either constant or 1-1) to show that the limit is not constant. And the Schwarz Lemma essentially guarantees that the whole disc D1(0) is filled up by the image of U. Please read one of the classical proofs. The text's version is in section 6.7.

For what follows we need one more classical result called the Schwarz Reflection Principle.

Schwarz Reflection (baby version) Suppose that U is open and connected in C, and f(z)=f(x+iy)=u(x,y)+iv(x,y) is holomorphic in U. If UR is the collection of "reflected" z's in U (that is, all the zbar's for z in U), then the function g(z)=[f(zbar)bar is holomorphic in UR.

Why is this true? One way to think about this is to realize that f(z) near z0 in U is the sum of a convergent power series. The mapping z-->zbar is certainly continuous (although it equally certainly is not holomorphic). But then the series n=0an(z-z0)n becomes the series n=0{an}bar(z-{z0}bar)n upon the stated reflection, and this is a convergent power series in z, indeed.

Another way to verify holomorphicity is just to realize that g(z)=u(x,-y)-iv(x,-y). The several variable Chain Rule from calculus will show that g satisfies the Cauchy-Riemann equations.

Schwarz Reflection (adolescent version) Suppose that U is open and connected in H, the open upper halfplane (z's so that Im z>0). Also suppose that the boundary of U contains an interval I of the real line, R. Suppose that f is holomorphic in U and has real continuous boundary values on I: that is, limz-->xf(z)=Q(x) exists (x is in I and z is in U) and Q is continuous in I. Then define V to be the union of U and UR (as defined before) and I also. Let g(z) be f(z) for z in U, [f(zbar)]bar for z in UR, and Q(x) for x in I. Then g is holomorphic in V.

Proof Certainly the Baby Version of Schwarz Reflection shows that g is holomorphic in U and in UR and we need to check what happens near I. But g is continuous in all of V. If we can show that the integral of g around a rectangle with sides parallel to the coordinate axes is 0, we are done (Morera!). But it's enough to consider a rectangle with one side on the x-axis. We can move it up slightly, and the integral is 0 (Cauchy!). This integral is continuous in the slight movement, so the desired rectangular contour integral must be 0.

This sort of argument was part of an earlier homework problem and a number of students recognized it. The hypotheses imply that the original mapping of f sort of sits on one side of the image R inside the image C. The reflection principle can be used in many contexts because both the domain and the image R's can be changed into many different curves by conformal mappings. This is used, for example, to analyze the automorphism groups of complicated domains such as an annulus, and can also be used to construct the elliptic modular function, a very important classical holomorphic function with uses in number theorem and hyperbolic geometry.

The adult version of the Schwarz Lemma weakens the hypothesis in the adolescent version in one wonderful way. We ask that the limz-->xf(z)=Q(x) still exist and be real, but no additional conditions on Q are needed. Notice that one consequence of the existence of the limit, using the adult version, is that the resulting Q turns out to be real analytic: certainly this is not obvious. It is not even obvious in the adolescent version, where there is an almost magical transition from "Q is continuous" to "Q is real analytic". This result is what's important in the very last discussion of the course. The proof of the adult verions uses the solution of the Dirichlet Problem and the consequence that continuous functions which have the mean value property must be harmonic. This is discussed in the textbook. See Theorem 7.5.2 in section 7.5.

The very last, very final, wonderful conclusion of the course is the Lewy example of a simple PDE with no solutions.

Also available is a direct discussion of the Mittag-Leffler and Weierstrass Theorems, results guaranteeing the existence of holomorphic and meromorphic functions with strictly prescribed behavior. This was proofread by Mr. Castro and Ms. Naqvi, whose efforts I appreciated.

And, indeed, I appreciated the work of all the students in the course: thank you. Here is information about the final exam.

Friday, December 7 (Lecture #27)
Notes by Avital Oliver, edited, with some comments, by the instructor.

A nice fact that we will use later

Assume the following: U and V are open sets in C, f:U-->V holomorphic and h:V-->R harmonic. What do we know about hof? If V is simply connected, then h has a harmonic conjugate, thus h is the real part of a holomorphic function (h=Re(g)). In this case gof is holomorphic, so hof=Re(gof) which is a harmonic function. The image to the right was supplied by the "scribe". Appropriate thanks are certainly due to him.

If V is not simply connected, it is still locally simply connected (just choose small balls around each point), so this is all still true locally. Since holomorphicity is a local property, the result is still true:

Proposition The composition of a harmonic function with a holomorphic function is harmonic.

Another proof of this fact is given on page 213 in the textbook (using differential operators and the Chain Rule).

Why do we care about conformal (also known as biholomorphic) mappings?
For many reasons (or else the instructor would not spend so much time discussing it!). Here's an important example: The Dirichlet Problem is an important problem, which arises from many fields in Physics (electostatics, heat flow, etc.) and in probability, etc. The general question is: given boundary data on some open set in C, is there a harmonic function defined inside the open set which satisfies the boundary data?

We will consider the simplest case: The function defined on the boundary is continuous and the domain is simply connected. By the Riemann Mapping Theorem (which we will not be able to prove ... please see the textbook), this domain can be conformally mapped to the unit disc. By our "nice fact that we will use later" (later being now), solving the Dirichlet problem on our open set is equivalent to solving the Dirichlet problem on the unit disc.

Solving the Dirichlet problem on the unit disc
The instructor told us that there are many ways to solve this, some including bugs (Brownian motion), electrons ("Kelvin's Method of Images"), and other strange ideas. Also, there is the solution as given in the textbook, which is something like "Here is the solution. Just compute and see that it's right" (around page 214). Instead, we will try to find the solution ourselves.

What we know about harmonic functions:

1. h=0, h is C2, and since such an h is locally the real part of a holomorphic function, h must actually be real analytic.
2. Maximum and minimum principles: a harmonic function which extends continuously to the closure of a relatively compact set is either constant or it achieves its strict maximum or minimum on the boundary. (This actually follows from the next result!)
3. Mean-Value property: If f is a holomorphic function, then by the Cauchy Integral Formula, f(z)=, therefore, since harmonic functions are real parts of holomorphic functions, this is also true for harmonic functions.
"Abstract nonsense" discussion
Suppose b is a continuous real-valued function on the boundary of the unit disc, and we know that there is a harmonic function hb defined on the disc with boundary values equal to b. Also pick a point z inside the unit disc. The mapping Tz defined by b-->hb(z) has the following properties (all verified using what was just written):
Tz is linear: Tz(b1+b2)=Tz(b1)+Tz(b2) and Tz(rb)=rTz(b) (r is a real constant)
Tz is positive: if b>=0 everywhere, Tz(b)>=0.
Tz is bounded by 1: -||b||<=Tz(b)<=||b||.
Under such conditions, the giant machinery of measure theory tells us something important: there is always a positive measure of total mass 1 on the boundary, let's call it mz, so that Tz(b)=D1(0)b dmz. But this abstract fact doesn't describe the actual measure specifically. This can be done, and that's what the computations which follow show. The method used is fairly elementary, and is basically separation of variables, looking at the radial and angular parts of the problem.

O.k., it's time to start solving! We are trying to define a function L from the set of continuous functions on D1(0) to the set of harmonic functions on D1(0). Let b be the boundary function: b is a continuous function on D1(0), so by Fourier series (since b is also in L2), almost everywhere. The solution to the Dirichlet problem is linear in b. L also maintains limits because by the Max/Min Principles, if b1<=b2 then L(b1)<=L(b2). So we only need to solve for one of the terms.

• If n>0 then L(ein)=zn=rnein.
• If n=0 then L(ein)=1.
• If n<0 then L(ein)= zbarn (NOT 1/zn which would be singular inside the disc: notice that conjugate holomorphic functions are also harmonic!), which is r-nein
In general,
This gives us h if we know b's representation as a Fourier series and things are computationally easier if we use the complex form of the Fourier series, but we would also like a closed form. Some manipulations (!!!) bring us to the following: (* is convolution on the circle). This series can be represented more nicely.

Here is the argument, I hope without many mistakes in algebra:
Consider -r|n|ein. Split it up:
1+0r|n|ein+--1r|n|ein.
The first sum is a geometric series (converging absolutely and uniformly for r in an interval [0,A] with A is less than 1. It is rei/(1-rei). The second sum is similar, and the result is re-i/(1-re-i). Add these two and get:
rei(1-re-i)+re-i(1-rei)
----------------------------------
(1-rei)(1-re-i)
which is, on the bottom, 1-2rcos()+r2, and on the top is 2cos()-2r2. If we add the 1 which was also there (n=0) we get the form of the Poisson kernel displayed in the following result.

Solution to the Dirichlet Problem
If b is continuous on D1(0) then the following h is continuous on the closed unit disc, harmonic on its interior and agrees with b on its boundary: , where .

If you think this is hard to understand, try the textbook proof!

A proof of this is not difficult but I really think it belongs more in a real variables course. The Poisson kernel which appears in the solution quoted above is a standard example of what is called an approximate identity. Here is most of an e-mail message I sent to Mr. Pal about "approximate identities".

An approximate identity is an attempt (!) to have an identity for the convolution operation.

In basic analysis, convolution is defined usually for R or for S1 (the circle: it is easiest to think of it as [0,2Pi], or, even better, as R mod the subgroup generated by 2Pi).

Let me discuss only the situation on R. This may be easier. Also, for the purposes of this message, let me use the notation I(F) for the integral over all of R of a function F.

If f and g are functions on R, then the convolution of f*g of f and g is defined by (f*g)(y)=I(f(y-x)g(x)). This is an integrated product with a shift in one of the variables. This is defined if the integral converges, of course. If the functions are continuous with compact support or, more elaborately, are L1 (integrable functions) then f*g is defined almost everywhere. Indeed, if you know the appropriate version of the Fubini Theorem, convolution and the usual function addition make L1(R) into an algebra. Convolution is commutative and associative. The uses of convolution are many and varied. For example, Oliver Heaviside used it to compute with solutions of differential equations. Convolution is used in, say, financial math to compute such quantities as the "present value" of an income stream.

In the case of L1(R), convolution does NOT have an identity (in the sense of multiplicative identity). Maybe the easiest way of explaining this is to mention that the Fourier transform of L1 functions are continuous functions which -->0 at infinity, and that the Fourier transform of a convolution is just the pointwise product of the Fourier transforms of each of the "factors". Then a "convolution identity" would automatically Fourier transform to the function 1, which does not -->0 at infinity, so there is no identity for convolution.

In certain ways, convolution CAN have an identity. Oliver Heaviside invented and used the delta function as an identity, and formalizing this mathematically took decades. Another path is to replace the algebraic considerations of an identity by using an "approximate" identity. What is this?

It is usually either a sequence or perhaps a continuously parameterized collection of eligible smooth functions (or in L1) so that the limits behave like a convolution identity. So, for example, you could ask if there is a sequence of L1 functions, fn, so that fn*g approaches g (in any sense you might like) as n-->infinity. It may not be immediately apparent that such sequences exist.

Indeed they do. Here is one such sequence, fairly silly: take fn to be the function which is n on the interval [0,1/n] and which is 0 elsewhere. Then if g is in L1, fn*g(x)-->g(x) for almost all x pointwise and also as a sequence of functions in L1. And if g is continuous, then fn*g(x)-->g(x) for all x in R (and some weak uniformity claims can be made). But that sequence is really not so nice, because a rectangular graph is not smooth. There are other sequences of approximate identities. In the context of S1 and convolution on S1, the Dirichlet kernel, Pr(t), is an approximate identity. Here is the meaning: Pr*g-->g as r-->1- (almost always if g is L1 and more nicely if g is continuous). The proof of this fact is NOT hard (and is part of the proof that we "solved" the Dirichlet Problem: the recipe supplied gets back the boundary data as you go towards the boundary).

I decided not to give a proof principally because the proof really is a real variables fact, and is easiest to give when the correct limit theorems about integrals are known. It is relatively easy to verify that the Poisson kernel is an approximate identity. The Dirichlet kernel, which also comes up in classical Fourier series, is also an approximate identity, but this is somewhat more difficult to prove.

The qualities which make it easy to verify that Pr(t) is an approximate identity are these:
i) Pr(t)>0 for all r and all t.
ii) Pr(t) eventually decreases to 0 as r increases towards 1 for t NOT equal to 0 (and here 0 and 2Pi are the same!).
iii) The total integral over [0,2Pi] is 1.
I think this is all that's required, and then the verification that we've solved solved the Dirichlet problem works out easily.

An example Dirichlet problem on the unit disc
In fact, we can also work with boundary data in L1 and not only continuous. For this example our boundary data is: on the upper half circle the value is 1, on the lower half circle the value is 0. If we stick this in our amazing formula for L(b), we get an integral that isn't easy to understand. Instead of solving it for the unit disc, we'll solve the corresponding Dirichlet problem for the upper half plane (remember our nice fact from before!). By the "standard" biholomorphic mapping to the upper half plane, we reduce the problem to finding a harmonic function on the upper half plane such that h(x)=1 for negative reals and h(x)=0 for positive reals. We can guess what could work here: the imaginary part of the simplest branch of log will work (dividing by Pi). This is of course a harmonic function, which is equal to (1/Pi)arctan(y/x). Boom! Notice that we worked backwards. Although in general we could "solve" the Dirichlet problem in a domain by transferring it to the unit disc and solving it there, actually here we solved a disc problem by working in a different domain, and recognizing an obvious solution in the other domain! The colored lines and semicirles in the accompanying picture are intended to be the corresponding level lines of the solution function, and lines with the same colors correspond to the same values of the functions.

But this is not all! After transfering back to the unit disc, if we square root (???) we get the solution to the Dirichlet function given a quarter of the circle having value 1. By iterating this process, rotating and taking limits, we solve the Dirichlet problem for all L1 functions. Again, approximate in L1 by using a sum of step functions. This also works!

The next part of the "program" is to sketch a proof of the Riemann Mapping Theorem: that any simply connected, connected open subset of the plane which is not all of C is biholomorphic with the unit disc. In fact, Riemann's original proof (or suggested proof!) of this result relied first on solving the Dirichlet problem, and deducing the RMT from that solution. One way he considered solving the Dirichlet Problem was imagining a boundary height (the boundary data is a real-valued function) and then lowering a flexible "membrane" over the data. The membrance would naturally be tight, and would try to minimize the energy, sort of hugging the boundary data. Then (this isn't even neat enough to require a "clearly"!) the height of the membrane over the interior is harmonic, and is the solution to the Dirichlet Problem. This approach can actually be understood, and does work, but, wow, it needs (from the point of view of current mathematics) rather a lot of work. The energy is the integral over the interior of the norm squared of the gradient of the membrane height. And showing that there is actually a minimizer (something that achieves the infimum of energy) is not easy. This leads to a whole huge area of mathematics called the calculus of variations, and variational methods of solving partial differential equations.

Tuesday, December 4 (Lecture #26)
Notes by Hui Wang with the assistance of David Duncan, edited, with some comments, by the instructor.

The instructor talked about the Hopf map at the very beginning, which was described as folows. Consider S3 lying in C2 and identify CP1 with the set of one dimensional linear subspaces of C2 (which are two [R] dimensional planes!). Then each element x in S3, being a unit vector, corresponds to the unique element y in CP1 spanned by x. Let f:S3-->CP1 denote the map sending x into y. Since CP1 can also be identified with the Riemann sphere S2, f can be thought of as a map from S3 into S2. This is the Hopf map. Notice that the inverse image under f of a point is the intersection of a plane passing through the origin with S3. Thus the fibers of f are copies of S1 lying in S3. One should go to Professor Steve Ferry to ask for the details. Or see here or here or here. The phrase "hopf map" gets about 20,000 hits on Google, and these links are the first three. It turns out that the Hopf map is not homotopic to a constant map (not obvious!) and it was the first example of such a map from a higher dimensional sphere to a lower dimensional sphere.

The main goal of today is to explore the group of linear fractional transformation, LFT(C). In Conway's book, a linear fractional transformation is defined as f(z)=(az+b)/(cz+d). And Conway uses the term Möbius transformation for the case when ad-bc is nonzero. However, we will always identify LFT's with Möbius transformations.

Remark As we did in the last lecture, LFT(C) is actually the group Aut(CP1). It is also PSL2(C).

It's impossible for us to completely explore the group LFT(C), which will require all branches of mathematics. Here we only give 3 easy properties.

0. Linear fractional transformations are conformal maps of CP1.
Proof This is because it is an element in the group Aut(CP1). Mr. Pal correctly asked what the meaning of "conformal" is for a map defined at or having values at . For such maps, please look in the appropriate local "chart" (as defined in the previous lecture, the mapping from CP1 to Cw) which takes to 0. Then looking there, the correctly composed mapping will be conformal.

1. LFT(C) is exactly triply transitive on CP1.
(The "exactly" means that there is a unique element of LFT(C) taking a given ordered triple to another given ordered triple.)
Proof We only need to prove that, given 3 distinct points z2, z3, and z4, there is an unique LFT T taking z2 to 0, z3 to 1 and z4 to . Actually, given two ordered triplies of 3 distinct points {z2,z3,z4} and {w2,w3,w4}, if the above assertion holds, there are LFT's T1 and T2 with T1 taking z2 to 0, z3 to 1, and z4 to and T2 taking w2 to 0, w3 to 1 and w4 to . Then (T2)-1oT1 takes z2 to w2, z3 to w3 and z4 to w4, which is what we want. For the existence of such a T, we only need to check that T(z)=

```[(z3-z4)(z-z2)]
---------------
[(z3-z2)(z-z4)]```
is such a map.
For the uniqueness, we just must show that only one LFT takes 0 to 0, 1 to 1 and to . If an LFT T with T(z)=[az+b]/[cz+d] takes to , then c must be zero. Hence, T(z)=za/d+b/d. Since it takes 0 to 0 and 1 to 1, a/d=1 and b/d=0. Therefore T is the identity map, which complete the proof of uniqueness.
As mentioned in class, the "triply transitive" feature is quite remarkable. Very few "interesting" (means simple!) finite groups have this property.

Definition The cross ratio (z1,z2,z3,z4) is T(z1), where T is the LFT taking z2 to 0, z3 to 1 and z4 to defined above.

The student scribe bravely wrote: "Why do we need to introduce this definition? I don't know." My answer follows. And, by the way, as mentioned in class, the notation, etc. is all taken from Conway's book. The instructor's confusion during the lecture, especially in this discussion, is taken from inside his own head.

Proposition For S in LFT(C), let w_j = S(z_j), j=2,3,4. Then (z,z2,z3,z4)=(S(z),w2,w3,w4).
Proof Let T(z) be (z,z2,z3,z4). Then ToS-1 takes w2 to 0, w3 to 1 and w4 to . Thus, ToS-1(z)=(z,w2,w3,w4). i.e. T(z)=(S(z),w2,w3,w4).
I still don't understand. But let me try to answer the question, "Why cross ratio?" LFT(C) becomes more complicated (really!) as you study it more. It is a very strange group, associated with lots of geometry and number theory. Therefore any tool to understand it is valuable. The cross ratio tries to help us: it is an invariant of LFT(C)'s action on CP1. For example, here is an easy first proposition: (this should be read after the next section, maybe!)
Proposition Four distinct points z1, z2, z3, and z4 of CP1 are on a straight line or circle if and only if the cross ratio (z1,z2,z3,z4) is real.

Proof and discussion Certainly this seems, at first sight, very strange. This weird number, (z1,z2,z3,z4), has enough information to tell whether there is a nice geometric subset of CP1 containing all four points. I do know that through every three distinct points in CP1 there is exactly one straight line or circle (follows, if no other way, from the next property of linear fractional transformations). O.k., here is the very easy proof: consider the element of LFT(C) given by T(z)=(z1,z2,z3,z4). The image triple of z2, z3, and z4 is on the real line extended (this is R union ). Where is T(z1)? If it is on the real line, the cross ratio, which is T(z1) must be real (it is computed with real numbers). If it is not on the real line extended, then it has some i's in it, and the cross ratio is not real. (Easy algebra). So we are done with the proof, since the cross ratio is an invariant of ordered four tuples, and the linear fractional transformations preserve straight lines and circles.

2. Any element of LFT(C) permutes the set {STRAIGHT LINES, CIRCLES}.

To prove this property, we need the following lemma.
Lemma A linear fractional transformation is the composition of dilations Ma(z)=az, a complex and nonzero, translations Tb(z)=z+b, and the inversion V(z)=1/z.
Proof Every LFT is of the form TeoMfoVoTg(z) for appropriate e,f,g in C.

Then we only need to prove that dilations, translations, and the inversion permute the set {STRAIGHT LINES, CIRCLES}. For dilations z-->az, the image of the circle |z-c|2=r2 is the circle |z-c/a|2=(r/|a|)2. A line can be written as sx+ty+r=0, or is(z+zbar)+t(z-zbar)+2ri=0, or (is+t)z+(is-t)(zbar)+2ri=0, or wz-(wz)bar+2ri=0, w=t+is, The image of this line under the dilation is the line (wa)z-((wa)z)bar+2ri=0. For translations z-->z+b, the image of the circle |z-c|2=r2 is the circle |z-(c-b)|2=r2, and the image of the line wz-(wz)bar+2ri=0 is the line wz-(wz)bar+wb-(wb)bar+2ri=0 (wb-(wb)bar is purely imaginary so this can be recognized as a straight line). For the inversion z-->1/z, the image of the circle |z-c|2=r2 is |(1/z)-c|2=r2, which is 1-cz-(cz)bar+|cz|2=|rz|2. If |c|=|r|, the image is a line. If |c| isn't |r|, it is a circle. The image of the line wz-(wz)bar+2ri=0 is the line w(zbar)-(wbar)z+(|z|2)2ri=0, which completes the proof.

I have found that what is listed above is sufficient for almost all "casual" users of complex analysis. Again, certainly if you are interested in number theory or geometry, you will learn and use much more, but what's above should be enough for most people.

We then considered examples of conformal maps. The goal was to try to map open, connected, simply connected proper subsets of C to the unit disc. The Riemann Mapping Theorem states that this is possible. The reasons for getting explicit mappings will be explained in the next lecture.

Example 1
We will find the conformal map from the left half plane {z | Re z>0} to the unit disk. (Common linguistic usage is that "conformal map" is the same as "biholomorphic".) That map should take the boundary to the boundary. Assume that T takes 0 to -1, i to i, and to 1. Then T(z)=(z-1)/(z+1). One more thing we should check is that T maps the inside of the domain to the inside of the co-domain (I call this the range.). Checking only one point is enough. T(1)=0. Notice that while it takes circles to circles, T doesn't necessarily take centers to centers. For example, the image of the circle |z-1|=1 is the circle |z+1/3|=2/3. The center 1 was sent to 0. Learning that the image circle is correct involves also realizing that the y-axis in the domain is a circle through , and it and the circle |z-1|=1 are tangent. Also the circle |z-1|=1 meets the x-axis (another circle through ) perpendicularly. Since elements of LFT(C) are conformal the image circle must have the resulting image geoemtry with respect to the images of the two axes. This identifies the image circle precisely.

Conformal Remark #1: An easy way to avoid an error
Check that the image under an element of LFT sends the "inside" of your domain object to the "inside" of your range object.

For example, the mapping T(z)=1/z sends the unordered subset {0,1,} of CP1 to itself, but the image of the inside of the unit disc gets sent to the outside of the unit disc. In this case the error of assuming "inside gets sent to inside" is apparent, but in more complicated computations this can be difficult to see.

Conformal Remark #2: An easy way to avoid an error
Do not assume that the center of a circle gets sent to the center of an image circle. This is almost always false!

Look at what happens to circles centered at 0 in the unit disc when we move the origin to another point with an automorphism of the disc, say, z-->(z+a)/(1+abarz). This moves 0 to a. The concentric circles all must intersect one fixed diameter of the unit circle perpendicularly (the image of the diameter through 0 and -a, actually). But these circles are "packed" in a strange way, if you are not acquainted with hyperbolic geometry.

Example 2
We map the strip K={z | 2<Re z<3 } conformally to the unit disk. Take K={z | 2<Re z<3 } to H={z | 2<Im z<3 } by h(z)=iz, take H={z | 2<Imz<3 } to G={z | 0<Im z<1 } by g(z)=z-2i, take G={z | 0<Im z<1 } to J={z | 0<Im z<Pi } by j(z)=zPi, take J={z | 0<Im z<Pi } to upper half plane by exp(z), and take the upper half plane to the unit disk by f(z)=(z-i)/(z+i). Since every step is conformal, the composition of them is conformal, and takes the strip K={z | 2<Re z<3 } to the unit disk.

Diary entry in progress!

The pictures below show the first few steps graphically. These are the routine steps: rotating and translation and scaling of the initial strip.

The pictures below this are an attempt to show the true magical step: somehow exp holds (?) the strip at - on the x-axis (!!) and opens the top border until it gets back to the negative x-axis, while the bottom border gets squeezed (??!) into the positive x-axis. I can not ... well, Pi i goes to -1 and 0 goes to 1 on the border.

Conformal Remark #3: write as compositions!
Many examples are best understood and written as compositions of familiar, simpler mappings. Try to avoid the ambition to "solve" the problem by writing things using one monstrous formula. That's usually difficult, at least for poor mortal human beings. Think simple, please.

Example 3
We map the 3/4 plane {z | Im z>0 or Re z<0} conformally to the unit disk. Take the 3/4 plane to the upper half plane by f(z)=z2/3, and take the upper half plane to the unit disk by g(z)=(z-i)/(z+i). The composition is what we want.

Conformal Remark #4: boundary properties
Mappings do not need to be conformal or nice in any way on the boundary. Don't be distressed about this. We only "care" about the interior.
In fact, there are powerful theorems which address how conformal mappings extend to the boundary, and some of them are stated and proved later in the textbook. But a beginner (?) shouldn't be too concerned: specific examples usually have easily discernible boundary properties.

Example 4
We map U=D(2,2)\D(1,1) conformally to the unit disk. Notice that U is simply connected, since 0 is not in U, so no closed curve can go "around" what could carelessly be considered the inside hole. Consider the inversion f(z)=1/z. it sends 0 to , 4 to 1/4, 2+2i to (1-i)/4, 2 to 1/2, and 1+i to (1-i)/2. Therefore it sends the circle |z-2|=2 to the line Rez=1/4, and the circle |z-1|=1 to the line Re =1/2. Since it sends 3 to 1/3, the connected open subset U is taken to the strip {z | 1/4 Example 5
We will find a conformal map from U=D(1,1) union D(i,1) to the unit disk. Consider the inversion f(z)=1/z. Since it takes 0 to , 2i to -i/2, and 1+i to (1-i)/2, the image of the circle |z-i|=1 is the line Im z=-1/2, which is parallel to the x-axis. Also, since it takes 2 to 1/2, the image of the circle |z-1|=1 is the line Re z=1/2, which is parallel to the y-axis. Finally, it takes 1 to 1 and i to -i. Since the LFT preserves the "inside", the image of U=D(1,1) union D(i,1) under the inversion is a 3/4 plane. What we did in class was wrong. Brave Mr. Nakamura commented after class that he was trying to signal the correct answer to the instructor, who was apparently then touring Mars. Look carefully at the picture, and at how the boundary and interior points are mapping, and at the right angle indicated which is preserved. The image is a 3/4 plane. Then translate by z-->z-(1-i)/2, and then rotate by -Pi/2 (this is z-->-iz). The result is the starting domain for Example 3, so we're done.

Conformal Remark #5: Your "toolkit" for conformal mappings
should at least include LFT(C), z-->zQ (powers and roots), exp, and ln.

I gave out a very brief picture dictionary in class (copied from the third edition of Marsden and Hoffman's text, Basic Complex Analysis). That has a few more examples (a mapping using sin z, for example) which are obtained from careful study of the compositions of the toolkit mappings already mentioned. You should look at them. I don't memorize them, but it is comforting that they exist. Sigh. Almost all of the examples I can compute explicitly have domains which are diangles (like triangles, but one less side). Even an average polygon (if # sides>2) is inaccessible if explicit mappings are requested (such mappings are studied using the Schwarz-Christoffel formula, which generally can't be written in terms of standard elementary functions).

A remark inspired by the solution of a homework problem
All students claimed to know the simplest version of the Weierstrass Approximation Theorem:
Suppose [a,b] is a closed interval in R and f is a continuous function on [a,b]. Given any e>0, there is a polynomial P such that supx in [a,b]|f(x)-P(x)| Therefore, if f is any continuous function on R, there is a sequence of polynomials {Pn} such that Pn-->f uniformly on any compact subset of R as n-->. (Just take Pn to be the approximation to f on [-n,n] with e=1/n). Therefore every continuous function on R is a locally uniform limit of real analytic functions on R (polynomials are quite real analytic!). So uniform convergence alone on R is not enough to guarentee smoothness of the limit. You need uniform convergence on open subsets of C for that! Mr. Duncan and I computed formulas for the semidirect product in Aut(D1(0)). This is what I mean: if we're given two elements, ei1[(z-a1)/(1-a1barz)] and ei2[(z-a2)/(1-a2barz)], then their composition (which is multiplication in Aut(D1(0))) should have the same form. After about half an hour of mistakes, we got a formula which gave the 3 and a3 which describe this composition. The formulas were complicated. He has the results; I don't.

Wednesday=Friday, November 30 (Lecture #25)
Notes by Camelia Pop, edited, with some comments, by the instructor.

Ms. Pop actually sent me a LaTeX file (here it is). Probably the various versions of TeX are the most effective way to typeset math, and you should learn one of them if you haven't already. I am using html because it is a simple way to show stuff on the web, but I do realize how imperfect it is.

### Projective space

The most pretentious approach
Let V be a vector space over a field F. Then we let P(V) be the set of all 1 dimensional subspaces of V (the "lines"). In particular, V can be chosen to be L2(R) (important in Quantum Mechanics, for example), or a finite dimensional vector space Fn (important in Combinatorics).

More pedestrian and low-class
That approach is maybe too difficult and too abstract for this course. Here I just want to study of CP1: one-dimensional complex projective space. So, we will define it first using an equivalence relation, in the following way:
C2 is the collection of (a,b) with a and b in C. C2* will be all pairs except for (0,0). On C2*, we define ~ by
(a1,b1)~(a2,b2) if there is a non-zero complex number so that a1=a2 and b1=b2.
The non-zero complex numbers will be denoted C* here.

CP1, as a set, is just C2*/~. The equivalence classes in CP1 are the one dimensional subspaces of C2* over C.

How to define a topology on CP1
Suppose [(a,b)] is a "point" in CP1. Actually, people mostly just write [a,b] and this is called the homogeneous coordinates of the point. They are, of course, defined only up to a non-zero common multiple of the pair. We can try to define a unique representative of [a,b]. So:

[a,b]=[a/b,1] if b is not 0 (we will call such "points" elements of Cz). Otherwise, certainly [a,b]=[1,0].
In the first case, we will relabel a/b as z, and z can be any point in C. So CP1 seems to be, as a set, a copy of C together with [1,0], which most people call (and so will we!).

Notice that if we have a sequence {zn} in C and if limn-->|zn|=, then for n large, [zn,1]=[1,1/zn], and this seems to suggest that, in CP1, a natural "limit" of the sequence {[zn,1]} is the point [1,0]=.

So one way to define a topology on CP1 would be to take the usual topology on C and put it on Cz, and then define neighborhoods of to be together with any complement of bounded discs in C. So an open neighborhood of is: take A>0, then make the set together with all the [1,1/z]'s with |z|>A. This makes our use of the symbol more believable.

Of course, we made an arbitrary choice of what is to be . Dually, we could look at [a,b]=[1/b/a} when a is not 0, and call this [1,w]. If we call 0 in CP1 the point whose homogeneous coordinates are [0,1], then we have described CP1 as a union of 0 and Cw where w is a/b.

It is a remarkable fact that the topologies gotten from both of these descriptions match up very well. The reason is that the overlap mapping from C* (as a subset of Cz) to C* (as a subset of Cw) is so neat and simple:
z-->[z,1]=[1,1/z]-->1/z.

This mapping essentially "sews together" two copies of C continuously, making CP1 into a topological space. This topological space is, of course, the one-point compactification of C.

Another way
to put a topology on CP1 is to consider the natural quotient map P from C2* to CP1 defined by taking (a,b) to [a,b]. C2* has a topology since it is a subset of the metric space C2 (which is R4). Put the strongest topology on CP1 which makes P continuous (the most open sets, I guess: a set is open in CP1 if and only if P-1 of the set in C2* is open). This topology is equivalent to the one just defined. (To me this is not totally obvious!)

Yet another equivalent way
is to consider the stereographic projection of the Riemann sphere. This should make the identification of CP1 with S2 perhaps clearer.
The picture shows some of the standard realizations of stereographic projection (that is, the mappings from parts of CP1 to Cz=C and Cw=C. Other ways of projecting put the plane tangent to or tangent to 0. This method has the interior of the unit disc, which is lightly tinted in magenta, correspond to the upper hemisphere. Notice, please, that the same point in CP1 corresponds to two different points in C under the two different projections. One is outside the unit circle, and the other is inside the unit circle. Hey: z and 1/z are usually like that!

Higher dimensions?
The elementary aspects are more or less the same. The replacements are:

• C2 gets replaced by Cn+1.
• C2* gets replaced by C(n+1)*.
• CP1 gets replaced by CPn (a compact topological space, with more complicated topology).
• Cz and Cw get replaced by n+1 copies of Cn, but "sewed together" by similar simple mappings (in components, zi/zj).
This is where classical algebraic geometry "lives" and there are still many interesting unsolved problems in analysis and algebra.

Mappings from CP1 to CP1
What would be a natural (?!) collection of mappings from CP1 to itself? Since CP1 is a collection of equivalence classes, we could look for mappings from C2* to itself which "respect" the equivalence relation. By that I mean specifically an F:C2*-->C2* so that (understanding that F(a,b)=(f(a,b),g(a,b)))
if (a1,b1)~(a2,b2) then (f(a1,b1),g(a1,b1)~(f(a2,b2),g(a2,b2).

More specifically, this means:
If there is in C* with a1=a2and a1=a2, then there is another w in C* so that f(a1,b1)=wf(a2,b2) and g(a1,b1)=wg(a2,b2)

Some though makes one consider homogeneous polynomials in two variables: elements P of C[z,w] of degree n so that P(z,w)=nP(z,w). Every polynomial can be written as a unique sum of homogeneous polynomials of distinct degrees. (Yes, yes, there is a direct sum around.)

Example, n=3 P(z,w)=Az3+Bz2w+Czw2+Dw3.

So let's get an F by taking two such polynomials, Pj(z,w)=Ajz3+Bjz2w+Cjzw2+Djw3, j=1,2, and then, using homogeneous coordinates, map [a,b] to [P1(a,b),P2(a,b)]. If we take C and push it into CP1 by using the C to Cz identification, and then, on the other side (the range copy of CP1), we get back to C, then this specific mapping will look like:
z-->P1(z,1)/P2(z,1).
This is a rational function! In fact, every rational function can be identified with one of these mappings induced from the equivalence relation on C2*. There is a nice correspondance between C(z), rational functions in one variable, and homogeneous polynomials in two variables. Then considerations about and dividing by 0 are not so problematic when looking at the C[z,w] picture.

How can we make such mappings holomorphic in some sense? They are already continuous (this isn't hard to check, since they come from continuous mappings on C2*).

Riemann surfaces
We'll define a >Riemann surface. Briefly, this will be a 1-dimensional (over C!) complex manifold. This is distinct from the idea of a Riemannian manifold, which is a manifold having an inner product nicely specified on its tangent spaces, making it possible to define a distance. So here: Riemann surfaces. Now to the definition. A real discussion of such a definition could easily take several lectures. Here only what is needed to consider CP1 will be mentioned, plus a bit more to put things into some context.

We say that a topological space X is a n-dimensional topological manifold if X is a topological space locally homeomorhic to Rn. These local homeomorphisms are frequently called coordinate charts. In the case of CP1, the coordinate charts we consider are, of course, the Cz and Cw identifications (or, equivalently, the stereographic projects from 0 and ).

Usually we want to make our life easier so we will require that the manifold satisfies some additional properties as: it is Hausdorff, connected, sigma-compact (X can be written as an ascending union of compact sets, which will allow us to consider only a countable family of charts), or paracompact (whatever that means, but of course WIKI!!! has more than enough).

We made several observartions as for instance that a connected space is not necessarily Hausdorff. For this we considered the real line from which we deleted 0 and replaced it by 2 points. The topology changes in that the neighborhoods of the 2 additional points become the neighborhoods of 0 from which we delete zero and add the appropriate point. Such a space is connected and is still locally homeomorpic with R1. It is not Hausdorff because the 2 additional points cannot be separated by disjoint neighborhoods.

The sigma-compactness is also important, but I did not state it very well. There are topological spaces which are connected and Hausdorff such that every neighborhood of such a space is homeomorphic with a neighborhood of 0 in R1, but these spaces are not sigma-compact. They are too big. One example is the long line, and its construction needs some simple facts about uncountable ordinal numbers. Please see here if you are interested, or any good point-set topology textbook.

Suppose that we have two coordinate charts whose domains overlap (that means: "have non-empty intersection") on a two-(real!)-dimensional manifold. We could ask that the composition of the appropriate mappings from an open subset of R2 (that's C!) to another open subset of R2 (C again) be holomorphic. Of course, this is exactly what happens with the Riemann sphere, with the z-->1/z overlap. Then such a manifold is called a Riemann surface.

We say that a continuous mapping f between two Riemann surfaces X and Y is holomorphic if no matter how we choose a point p in X and a chart around p and a corresponding chart around f(p), then the composition of the appropriate subset of C to X, then, transported by f to Y, then backwards to C again, is holomorphic.
This is a certainly very abbreviated presentation. Basically I wanted to reassure you that the mappings we have created using homogeneous polynomials were as nice as possible: holomorphic. In fact, just using what we already know, it is possible to prove that any holomorphic map from CP1 to CP1 is of the form given: a quotient of homogeneous polynomials. (Hint: CP1 is compact, so how many preimages of can a non-constant holomorphic function have? Then use properness.)

An important and amazing result which is easy to state but difficult to prove is the
Uniformization theorem If X is a simply connected Riemann surface then X is biholomorphic to exactly one of the unit disc, the complex plane, of CP1.

This generalizes the Riemann Mapping Theorem, which asserts that simply connected open proper subsets of C are biholomorphic with the unit disc.

Now, we come back to CP1 viewed as S2 and cover it by 2 charts - the projection from the North pole (0,0,1) and from the South pole (0,0,-1). This is a way to define CP1 as a Riemann surface.

We want to determine Aut(CP1), the holomorphic automorphisms of CP1: f is 1-1, onto, and holomorphic (just as before, such maps are biholomorphic, since the inverse is locally o.k. by earlier discussion [if it is 1-1 then then derivative isn't 0, so the inverse is differentiable and satisfies the CR equations, so is holomorphic]).

We look at the f's which stabilize (so f()=. But then f restricted to Cz is a proper holomorphic mapping. Because f is also bijective, from what we have proved before (remember Aut(C}!), such an f has the form f(z)=az+b, with both a and b complex and a not equal to 0. Now we consider the transitive part of Aut(CP1). That is the f's with the property f()=z0 C. If we compose such an f with g(z)=1/(z-z0) , we obtain a mapping which fixes (a mapping in the stabilizer of Aut(CP1)). Then we can see that such an f has the form f(z)=(az+b)/(cz+d), where the determinant of the matrix is not zero. This is exactly a mapping of the type we analyzed earlier: it comes from two homogeneous polynomials of degree 1, and the determinant condition is needed so that the resulting map on CP1 is a bijection.

Theorem Aut(CP1) is the collection of mappings z-->(az+b)/cz+d), where the determinant of the matrix is not zero (equivalently, where (a,b) and (c,d) are linearly independent).

Then Aut(CP1) is a group and its elements are called linear fractional transformations (you can see why, I hope!), Möbius transformations, etc. (many names!). These transformations compose as if the coefficients were entries in matrices (because they come from linear automorphisms of C2). This group is also denoted PGL2(C) and contains SO(3) (rotations of the unit sphere), SU(1,1) (automorphisms of the unit disc), Aff(C)={az+b: a not 0} (automorphisms of the complex plane). Please look at the the earlier discussion about geometry if you haven't yet.

Wednesday=Friday, November 27 (Lecture #24)
Notes by Jeffrey Amos, edited, with some comments, by the instructor.

Humph: cos(sqrt(z)) and sqrt(cos(z)) are interesting.

Schwarz's Lemma Let f be a holomorphic function from D1(0) to D1(0) with f(0)=0. Then |f'(0)|<=1 and |f(z)|<=|z| for all z in D1(0).
Furthermore, if either |f'(0)|=1 or |f(z)|=|z| for some non-zero z, then f is a rotation: there is a real number, t, so that f(z)=eitz for some real t.

Proof Let g(z)=f(z)/z for z not 0, and define g(0) to be f´(0). Then g is certainly holomorphic away from 0, and since f is differentiable at 0, g is continuous in the unit disc (the limit of g(z) as z-->0 is g(0)!). Therefore the singularity of g at 0 is removable: g is holomorphic on D1(0). Let 0r(0) (closed) its maximum modulus is on the boundary. But when |z|=r, |g(z)|<=1/r (the top has modulus at most 1 since f's values are in the unit disc, and the bottom has modulus r). But for a fixed z, this inequality holds for |z|<=r<1. Thus |g(z)|<=1 proving the first half of the lemma.

Now suppose either |f´(0)|=1 or |f(z)|=|z| for some non-zero z. In either case, g has an interior maximum and is hence a constant function of modulus one, say eit. Thus f(z)=eitz.

Now for a silly example. Suppose someone walks up to you (this is supposing you live in a weird neighborhood), and says "I'm thinking of a strip S in the complex plane consisting of numbers with imaginary part between -Pi/2 and Pi/2. A holomorphic function F from this strip S to the unit disc fixes 0. How big can |F(1)| be?"

First, realize that S and D1(0) are biholomorphic. Of course, the Riemann Mapping Theorem declares that this is true, but the utility of Schwarz's Lemma lies in the fact that frequently the assertion can be verified with a computable (or approximately computable!) mapping. ez takes S to the half-plane with positive real part, and changes 0 to 1 and 1 to e. Next, (z-1)/(z+1) takes this half plane to the unit disc (verification of this fact can be done with "elementary" considerations, such as manipulation of inequalities, but we will see shortly that there are ways to "guess" and then verify such formulas easily); we will call its inverse B. This mapping takes 1 to 0 and e to (e-1)/(e+1). ez and (z-1)/(z+1) are holomorphic bijections (over the appropriate domains), so they have holomorphic inverses. Consider f(z)=F(log(B(z))). This is a holomorphic function from D1(0) to D1(0). By Schwarz's Lemma, (e-1)/(e+1)>=|f((e-1)/(e+1))| = |F(1)|. So (e-1)/(e+1) is the desired bound (and we can even find out which functions f attain this bound if we wanted!).

Now, consider the same problem except with S replaced with St={z in C such that |Im z|<(Pi/2)t}. The function mt(z)=z/t sends St to S. Let f(z)=F(m-1t(log(B(z)))), so that (e1/t-1)/(e1/t+1)>=|f((e1/t-1)/(e1/t+1))|=|F(1)|.

When t is very large, then |F(1)| is very small. Intuitively, this means that wide strips have to be "shrunk" to fit in D1(0). As t-->, F(1) goes to zero. The limiting picture follows from Liouville's Theorem.

Comment O.k., I will admit it. The previous example is actually not totally silly and not totally random. There are various kinds of (differential) geometry currently being studied (examples are contact geometry and CR geometry) where mapping and distortion properties similar to what is analyzed here (a very simple case!) play a very large role in understanding the many notions of geometric size. So ... if you learn of such things, you saw it here first!

If U is open in C, then Aut(U) will denote the holomorphic automorphisms of U. A function f is in Aut(U) if f maps U to U bijectively (1-to-1 and onto) and if f is holomorphic. Schwarz's Lemma is the principal way to analyze the structure of many examples of Aut(U).

Please note that this already implies that f-1 is holomorphic as well. Why? We know that a holomorphic mapping always has a nice local description. So near w, say, f(z)=(z-w)Nh(z) where N is a non-negative integer, and h is holomorphic with h(w) not 0. If N>1, then f is not 1-to-1 near w. So N=0 or N=1, and f-1 is not holomorphic. Thus Aut(U) is a group under composition.
This contrasts with what happens in calculus: there are C functions which are bijective on R whose inverses are not even C1: x3, for example.

Theorem The stabilizer of 0 in Aut(D1(0)) is the rotations around 0.

Proof Let f be in Aut(D1(0)) with f(0)=0. By Schwarz, |f´(0)|<= 1. Also, 1<=|f-1´(0)|=|1/f´(0)|. Thus |f´(0)|=1 and f must be a rotation around 0. (Of course, all such rotations fix 0.)

What about the "transitive part" of Aut(D1(0))? For any w in the unit disc, we will find f in Aut(D1(0)) with f(0)=w. Let H be the half plane with positive imaginary part. We saw on day 1 that C(z)=i(1-z)/(1+z) is a biholomorphic mapping from D1(0) to H. For any a+bi in H, the mapping z-->bz+a is in Aut(H) and sends i to a+bi. Motivated by this, let f(z)=Re C-1(w)+(Im C-1(w))z. Now C-1((f(C(z))) is in Aut(D1(0)) and C-1((f(C(0)))=C-1(f(i))=w.

Theorem All f in Aut(D1(0)) can be written as f(z)=eit((z-w)/(1-{w conjugate}z)) for some real t and complex w in D1(0).

Some details need to be checked. If I have time, I will. But one "detail" is that the mapping C-1((f(C(z))) described before the statement of the theorem is actually (z-w)/(1-{w conjugate}z) (it is actually the inverse of that map). Once that is checked, then the description follows from this idea:
Take f in Aut(D1(0)). Then f maps 0 to some w in D1(0). Compose f with the mapping (z-w)/(1-{w conjugate}z). This is in Aut(D1(0)) and maps 0 to 0. Therefore it must be a rotation around 0 by the result on the stabilizer subgroup. Then f must be equal to the inverse of (z-w)/(1-{w conjugate}z) followed by a rotation, as above.
I don't want to do any of these computations until we have some better knowledge about the whole group of linear fractional transformations, and we will get that very soon.

The theorem is not a "group theoretic" description of the set because compositions of automorphisms do nasty things to t and w.

Theorem All f in Aut(C) are of the form f(z)=az+b for some a,b in C.

Proof f is proper, so the inverse image of Dr(0) (closed) is bounded for all r. Thus by 4.7.5, f is a polynomial, say of degree n. f has n zeros counting multiplicities and is 1-to-1, so f(z)=v(z-w)n for some complex v and w. Now v(z-w)n-v has zeros that are w plus an nth root of unity. Thus n different z's satisfy f(z)=v. Since f is 1-to-1, n must be 1 and f is linear.

Is the group Aut(C) isomorphic to the group Aut(D1(0))? Learning about the automorphism groups of domains is really learning about different kinds of geometry. The information below is useful background for all sorts of things ranging from teaching geometry in high school to doing research in geometry. Take a look. Not everything there is precisely correct, but the intention is to show you the beginning of the truth. It is descriptive and not deductive (some of the claims take a good amount of effort to verify).

### Some Euclidean and non-Eucliean geometries

Geometry #1

The set and some name R2 or C; Euclidean geometry.
"Triangles" Triangles whose interior angles' sum is Pi.
Parallel "axiom" A point external to a line always has exactly one parallel line (doesn't meet the original line.
Length? From the L2 norm of R2.
Description of the automorphism group z-->eitz+b
Also use z-->z-conjugate if orientation reversal is allowed.

Geometry #2

The set and some name D1(0) (the Poincaré disc) or H, the upper halfplane (the Poincaré halfplane); hyperbolic geometry.
"Lines" Straight line segments through 0 or circular arcs intersecting the boundary perpendicularly.
"Triangles" Triangles whose interior angles' sum is less than Pi.
The triangle's (hyperbolic!) area is related to this "defect".
Parallel "axiom" A point external to a line always has infinitely many lines through it which don't meet the original line: infinitely many distinct parallel lines exist.
Length? The hyperbolic distance from 0 in the D1(0) model goes to as z-->the boundary, so the diameter is infinite. (I think the specific function is arctanh(|z|)!)
Description of the automorphism group z-->eit((z-w)/(1-{w conjugate}z))
Also use z-->z-conjugate if orientation reversal is allowed.

Geometry #3

The set and some name S2 (best seen as CP1!); elliptic geometry.
"Lines" Arcs of great circles
"Triangles" Triangles whose interior angles' sum is greater than Pi.
Exercise: find a triangle with three right angles!
Parallel "axiom" Two great circles must meet, so a point external to a line has no parallel lines.
Length? The great circle distance on the sphere.
Description of the automorphism group The rotations of the sphere. This is O(3) or SO(3) if orientation preservation is needed. The group can also be realized as a group of linear fractional transformations: more later.

Wednesday=Friday, November 21 (Lecture #23)
Notes by Gene Kim, edited, with some comments, by the instructor.

A really terrific UC Berkeley qualifying exam problem
How many roots does the polynomial P(z) = z5+z3+5z2+2 have inside the annulus 1<|z|<2?

Answer Let's try to use Rouché to satisfy the following hypotheses:

s is a simple, closed curve in U open and connection, and s is homotopic to a point in U. Also, f and g are holomorphic functions in U, and |f(z)|>|g(z)| for all z on s.
If the hypotheses are satisfied, then we can conclude that #{f-1(0) inside s}=#{(f+g)-1(0) inside s}.
Take s to be |z|=1, f(z)=5z2, and g(z)=z5+z3+2. If |z|=1, then |f(z)|=5, and |g(z)|<=|z|5+|z|3+2<=4. Since 4 < 5 we can apply Rouché: hence #{f-1(0) inside s}=2.

Now, take s to be |z|=2, f(z)=z5, and g(z)=z3+5z2+2. Then |f(z)|=32, whereas |g(z)|<=|z|3+5|z|2+2=30<32. So P has 5 roots inside s, and there must be three roots in the annulus.

Mr. Amos noted that if |z|>=2, then P(z) never vanishes, since then |z|5>|z|3+5|z|2+2.

Maybe a problem not on an exam
Suppose we have a sequence of functions, {fn} which are defined on a connected open set, U, of C. If {fn} converges uniformly on compact subsets of U, and each fn is 1-to-1, what can be said about the pointwise limit, f?

Such questions come up in iterative methods to find roots, etc., and certainly occur in proofs of the Riemann Mapping Theorem.

There are some useful examples to consider. One is fn(z)=(1/n)z. This is 1-to-1, but if n-->, the limit is the constant 0, which certainly is not 1-to-1.

Suppose the limit function is not 1-to-1. Then design a closed curve (a "contour") so that a contradiction will be apparent. This is:

Hurwitz's Theorem Suppose U is open and connected and {fn} is a sequence of 1-1 holomorphic functions in U. Suppose that fn-->f uniformly on compact subsets. Then either f is 1-1 or f is constant.

If w is in f(U), then f-1(w) is closed in U. Since f is holomorphic:
(1) If f is constant then f-1(w)=U.
(2) If f not constant then f-1(w) is closed and discrete.

What if there are z1 not equal to z2 such that f(z1)=f(z2)=w? Define F(z)=f(z)-w, and Fn(z)=fn(z)-w. Draw a simple closed curve s containing z1 and z2 (like a barbell). Such a curve can be drawn since U is connected and open, hence arcwise connected. Then the inf over s of |F(z)| is m>0 since f-1(w)is a discrete subset of U, and we can wiggle around it if necessary (as in the exam problem). Choose n so that that sup(over sigma) of |fn(z)-f(z)|<m/37. By Rouché (considering F and fn), fn is NOT 1-1. Contradiction! Hence, f is 1-1.

One definition of "barbell": [Sport] an iron bar with a series of graded discs at each end, used for weightlifting exercises.

Speaking of the exam ...
On the midterm, we looked at: f is holomorphic on C\N, and f(z)=n=1[(1/n!)(1/(z-n))]. We saw that the residue of f at n was 1/n!.

What if we want a function f such that f is holomorphic on C\N and meromorphic in C with simple poles at all n in N with residue at each n equal to 1? We'll create such a function by "cheating".

We suppose that f(z)=n=1[1/(z-n)+STUFFn(z)].
Look at Dn/2(0). There we want |1/(z-n)+STUFFn(z)|<1/2n for all z in Dn/2(0). Let's create STUFFn(z).

Approximating in |z|<n/2: note that 1/(z-n) is holomorphic in this disc. So (geometric series):
1/(z-n)=(-1/n)[1/(1-z/n)]=(-1/n)k=0[zk/nk]. This is valid since |z|<n. This is equal to
(-1/n)k=0m[zk/nk]+(-1/n)k=m+1[zk/nk]=(-1/n)k=0m[zk/nk]+(-1/n)(z/n)m+1/(1-z/n) and finally we can convert this back, reversing the geometric series, and get
(-1/n)k=0m[zk/nk]+(z/n)m+1/(z-n).

Choose m so that |(z/n)m+1/(z-n)|<1/2n if |z|<n/2. Since |z|<n/2, we know that |z/n|<1/2. Hence |(z/n)m+1/(z-n)|<=(1/2)m+1/|z-n|<=[(1/2)m+12]/n=1/(n2m)<=1/2m. So if we set m=n, then we know that |(z/n)m+1/(z-n)|<1/2n.

Therefore a formula for a suitable f is f(z)=n=1[1/(z-n)+(1/n)k=0n[zk/nk]. f(z) converges due to the Weierstrass M-test: write f as n=12[z]+n=2[z]+1, where [z] denotes the ceiling function. Then each term of the latter series (the "infinite tail") is less than 1/2n.

Written Thanksgiving morning:

YOU TOO CAN MAKE \$10,000!!! easy money ...
So let's see how to create, fairly explicitly, a meromorphic function with singularities at sqrt(n) which are each simple poles with residue n56. That is, we're trying to find the appropriate convergence terms ("STUFFn") so that
N=1((n56/{z-sqrt(n)})-STUFFn(z)) converges.

So let's start by imitating what we did before.

• (n56/{z-sqrt(n)})=-(n57/sqrt(n))(1/{1-[z/sqrt(n)]}) and do geometric series.
• (1/{1-[z/sqrt(n)]})=j=0(z/sqrt(n))j=j=0N(z/sqrt(n))j+j=N+1(z/sqrt(n))j and N will be selected later.
• -{n57/sqrt(n)}j=N+1(z/sqrt(n))j=-{n57/sqrt(n)}([z/sqrt(n)]N+1/(1-{z/sqrt(n)})=n57(z/sqrt(n))N+1/(z-sqrt(n)) by doing geometric series backwards.
• Take z's satisfying |z/sqrt(n)|<1/2. Then the expression found above for the infinite tail of the series can be estimated (not pretty, but it works!):
|n57(z/sqrt(n))N+1/(z-sqrt(n))|<n57(1/2)N+1(2/sqrt(n)).
The only weird part of that is estimating |1/(z-sqrt(n)| which is like many things we've done before, but I still mess it up. Here I know |z/sqrt(n)|<(1/2) implying |z|<sqrt(n)/2 so that (reverse triangle) |z-sqrt(n)|>=sqrt(n)-|z|>sqrt(n)-sqrt(n)/2=sqrt(n)/2. Then |1/(z-sqrt(n))| must be overestimated by 2/sqrt(n) which is part of what is above. Whew!
• Now we need to select N. Well, we would like this infinite tail to be less than 1/2n in modulus, say. So we want
n57(1/2)N+1(2/sqrt(n))<1/2n
And if we take log2 and work diligently, we need
n+56.5log2(n)<N.
• Let's temporarily call kn the smallest integer larger than n+56.5log2(n) (here n is a positive integer, so everything is o.k.). Actually, since log's grow very slowly, I bet that I could take kn to be (Some constant)n so that would appear to be simpler but would actually (sigh!) be a bit "wasteful".

With this choice of kn, then an f satisfying the requirements would be
f(z)=n=1( (n57/{z-sqrt(n)})-n57j=0kn(z/sqrt(n))j); kn=an integer larger than n+56.5log2(n).

Ain't that purty?! Well, not really, but the "job" is done.

Here is something much more serious, with a bit of novelty also.

Mittag-Leffler Theorem Suppose W is a closed, discrete subset of C, and suppose that for each w in W, a polynomial Pw in C[z] with no constant term is selected. Then there is a meromorphic function f defined in C with poleset W such that the principal part of f at each w in W (this is the sum of the terms of negative degree in the Laurent series for f at w) is Pw(1/(z-w)).

Proof We first write W as a disjoint union of sets Wn. Here n is a non-negative integer. So W0 will be those w's in W with |w|<=1, while, more generally, if n is at least 1, Wn is the collection of w's in W with n<|w|<=n+1. Note that since W is discrete and closed, each of the Wn has at most finitely many elements. Of course, some of them may be empty. Then define Qn (for n>=0) to be Qn(z)=w in WnPw(1/(z-w)) (this is a finite sum!). If Wn is empty, then Qn(z) should be 0 These are the sum of the principle parts, the pieces of the singularities, in each of the annular regions between an integer and its successor (integer+1).

Now consider Wn for n>=1. All w's in this Wn must have |w|>n so that the sum defining Qn is holomorphic in some disc of radius r, where r>n. This is because Wn is finite, and each w of Wn has modulus greater than n, and a minimum of a finite set is one of the set's elements (a specific |w|=r>n). Since Qn is holomorphic in Dr(0), it is equal to a power series centered at 0 valid in all of the disc. The series will converge uniformly on compact subsets. Therefore there is a partial sum Vn of this power series (just a polynomial) so that if |z|<=n then |Qn(z)-Vn(z)|<{1/2n}.

Now we can write a "recipe" for f to prove the theorem! It will be defined by f(z)=Q0(z)+n=1(Qn(z)-Vn(z)).

Some verification is necessary.

1. We will show that if z is not in W, the sum defining f converges absolutely. Well, we know there is an integer N with |z|<=N. Break up the sum of f(z) in two pieces, fin(z)+fout(z), where
fin(z)=Q0(z)+n=1N(Qn(z)-Vn(z)) and fout(z)=n=N+1(Qn(z)-Vn(z)).
The series defining fout (the infinite series, and the other is a finite sum) converges absolutely and uniformly for all z with |z|<=N. This follows because if n>N>=|z|, |Qn(z)-Vn(z)|<{1/2n}, and then we use the Weierstrass M-test.
2. We also now observe that given any disc centered at 0 in C, the series defining f can be written as a sum of a rational function plus a function which is holomorphic in that disc. Additionally the rational function has the desired principal parts for all w's in W which happen to be in the disc. This of course also follows from the fin+fout decomposition described previously. Surely fout is holomorphic, since it is the result of a uniformly convergent series of functions holomorphic in the disc, and fin is a finite linear combination of inverse powers of z-w (for those w's in W which are in the disc) and of polynomials in z. That's certainly a rational function! And, of course, the rational function does have the desired principal parts.
Review of the proof I think this proves the Mittag-Leffler Theorem for C. Please look over the proof. The first paragraph "dissects" or breaks up W into parts which are each finite and nicely chosen geometrically so consequences of the organization lead to simple estimates. The second paragraph defines the polynomials which will control the principal parts in the appropriate discs. Then f's recipe is given, and the next two paragraphs verify the properties of the recipe.

First comment My comments in class on Friday/Wednesday were maybe a bit deceptive (no: wrong!) when I asserted that, oh my, the "bookkeeping" involved in such a proof would be forbidding. I was thinking more algorithmically. I wanted a more definite "formula" or "procedure" for the function f, that is, for selecting the polynomials which are involved with its description. Well, certainly if W is a sequence of w's whose growth (that is, rate of increase of |w|) is known, then we can use geometric series arguments effectively to get the polynomials we need to balance the principal parts. But the proof above avoids that consideration. There is no "effective formula" given, but the phrases "will converge uniformly" (applied to the power series for a holomorphic function) and "we can choose a partial sum ... so that" applied to the same series are the existential (?) version of effectively estimating the geometric series remainders.

Second comment Be very aware that the theorem does not assert that w in WPw(z) converges or that n=1Qn(z) converges! That might be true (someone nice might have thrown in some 1/n! factors, after all). The convergence of the series defining f is a bit delicate. We have "tweaked" the terms with somewhat subtle adjustments so that the series for f does converge in the way we would like.

Third comment A version of Mittag-Leffler is true for any open subset of C, as I declared in class. But a proof needs some topological "dissection" of the set as a nice increasing union of compact sets. Some further knowledge of analysis is very useful also, since we need the analog of partial sums of Taylor series. A result called Runge's Theorem (also classical) provides such approximations. But I won't prove Runge's Theorem in this course.

Friday, November 16 (Lecture #22)
Notes by Yusra Naqvi, edited, with some comments, by the instructor.

Lemma 1 If f has an isolated singularity at w and f is meromorphic at w, then f(z) = h(z)(z-w)n, where n is an integer, h is holomorphic near w with h(w) not 0.
So h is a unit in the ring of functions holomorphic in some neighborhood of w. We saw also by a direct computation that the residue of f´/f at w is equal to n.

Lemma 2 If f is holomorphic on U, s is a closed curve in U with f not 0 on s, then 1/(2Pi i)s[f´(z)/f(z)]dz is an integer.

We then interpret this integral using our BIG Theorem (the Argument Principle).
BIG Theorem Suppose f is meromorphic on U and s is a simple closed curve so that f has no zeroes or poles on s. Then 1/(2Pi i)s[f´(z)/f(z)]dz can be interpreted in the following ways:

• The change in the argument of f as z goes "around" s.
• The total number of zeroes and poles inside s, counted with multiplicity (so the function f in Lemma 1 has a zero/pole at w with multiplicity |n|: n is positive if f has a zero, and n is negative if f has a pole).
• The winding number of the curve f(s) around 0.
"Proof" The first statement follows from the observation that f´/f is (locally) the derivative of log(f). The second statement follows from the fact that the zeros and poles of f are poles of f´/f, with residue equal to their multiplicity. The third statement follows from the observation that if u=f(z), then du=f´(z) dz, and so the integral about s of f´/f is equal to the integral about the curve f(s) of 1/(u-0), which gives us the winding number of f(s) around 0.

This led to the development of the concept of degree in topology. If f is a continuous map of Sn (the unit sphere in Rn+1) to itself, then the degree of f is an integer with the following properties:

1. deg(fg)=deg(f)deg(g) (here fg is f composed with g)
2. deg(identity)=1
3. If f is not onto, then deg(f)=0
4. If f is homotopic to g, then deg(f)=deg(g).
The study of degree arose from looking at simple closed curves s (which are homotopic to the circle) in the complex plane, and seeing how f maps this curve into C. The local degree of f at a point w is the order of f at w, i.e., the multiplicity of the zero or pole w.

There are many references to the idea of degree and it is prominent in any study of algebraic topology, or, indeed, in algebraic geometry. One lovely presentation, very non-traditional when it was written, is in Milnor's Topology from the Differentiable Point of View.

If we omit the word "simple" from our big theorem, then we get that 1/(2PI i)s[f´(z)/f(z)]dz is equal to the sum over the zeroes and poles of f of the order of f at each of those points multiplied by the index of s at that point.

Another change we can make to theorem is to keep the word "simple", but then multiply f´(z)/f(z) in the integral by another function g which is holomorphic on U. Then
s[f´(z)/f(z)]g(z)dz=w zero or pole of fg(w)·(Order of f at w).
This is because locally, near w, f´(z)/f(z)=n/(z-w)+Higher Order Terms and g(z)=g(w)+g´(w)(z-w)+Higher Order Terms. Thus: g(z)[f´(z)/f(z)]=ng(w)/(z-w)+Higher Order Terms and the Residue Theorem gives the result.

This version of the theorem has several applications in algebra and combinatorics. For instance, we can generate the symmetric functions of zeroes of polynomials by the sums of nth powers of these zeroes. This integral allows us to calculate this by letting f be the polynomial we are considering, and by letting g(z)=zn. Then we get the sum of the nth powers of the roots of f if we take any simple closed curve which is large enough to contain all the roots of f. (These sums generate the ideal of all symmetric functions of the roots.)

Another place this version is used is in the proof of the Lagrange inversion theorem, which gives an explicit way of finding the power series for the inverse of a holomorphic function which is given by a power series.

Here is an exercise (taken from Complex Variables by George Polya and Gordon Latta) which uses our big theorem: Consider the polynomial P(z)=z3+z2+6i-8. By considering a large quarter circle in each quadrant and using the Argument Principle, show that there is a zero in each of the second, third and fourth quadrants.

 To verify this, we will consider what happens to the image of a certain closed curve under mapping by P. The closed curve is shown to the right. It is a line segment up the imaginary axis, then a quarter circle in the third axis (R is supposed to be large), and finally a closing line segment along the negative real axis to 0.
 Certainly P(0)=6i-8. As we move up along the imaginary axis, P(iy)=-i·y3-y2+6i-8 so as y grows, we move from P(0) to a point in the third quadrant far away from either axis, but very far "down" so the argument is quite close to 3Pi/2. The curve shown is the image of the line segment from A to B (with B precisely 5i in this case). P(A) is in the second quadrant, and P(B) is in the third quadrant, appearing "close" to the negative imaginary axis.
 We then observe that on points on the curved part of s, P(z) is dominated by the z3 part, because R is very LARGE. Thus, P takes the quarter circle almost to a three-quarter circle, taking us all the way around back to the second quadrant. What's shown is the image under P of the quarter circle when R=5. One might not think that very large, but the image curve looks very much like three quarters of a circle (that's the cubing effect, the highest order term). So this is the curve going from P(B) to P(C).
 Finally, the result of mapping the line segment from C back to A is shown. The static image is not too interesting and is shown below: But the "kinetic" aspect is actually a bit more amusing. To the right is a picture of the real part of the position of the curve. As you can see, due to the cubic, as the curve comes "back" to 6i-8, it actually overshoots the point a bit. But it still stays in the second quadrant.
Here is a more symbolic picture of the image of the curve s under the mapping by P. You can see clearly that the winding number of f(s) about 0 is 1, and hence there is one root contained inside the curve s. If you look closely, the diagram even shows the overshoot near 6i-8.

Out of cowardice, we didn't do the third and fourth quadrants in detail. There are people who understand the Argument Principle very, very well, and they can "see" the curves easily, and detect where roots are. These ideas are very useful in many areas of pure and applied math. For example, in applied math, "stability" of the solutions of certain systems of differential equations is assured if the location of some polynomial roots can be established, and this is frequently done with the Argument Principle.

Another important consequence of the argument principle is Rouché's Theorem.

Rouché's Theorem Suppose U is a connected open set in C, f and g be holomorphic on U, and s is a simple closed curve homotopic to a point in U. Also suppose that if z is on sigma, |f(z)|>|g(z)|. Then the number of zeroes of f(z) inside s equals the number of zeroes of f(z)+g(z) inside s.

Proof We know that the number of zeros of f(z) inside sigma equals 1/(2Pi i)s[f´(z)/f(z)]dz. (Note that f is non-zero on s since |f(z)| is strictly greater than the non-negative number |g(z)| for all points z on s.) We now consider 1/(2Pi i)s[{f´(z)+tg´(z)}/{f(z)+tg(z)}]dz where t goes from 0 to 1. We call this second integral F(t), and note that it is a continuous function of t (t in [0,1]) that is defined everywhere, since |f(z)|>t|g(z)| on s, so |f(z)+tg(z)|>|f(z)|-t|g(z)|>0, and because the function (t,z)-->[{f´(z)+tg´(z)}/{f(z)+tg(z)}] is jointly continuous on [0,1]xs. So its integral is continuous.
However, we know that F(t) is always an integer (Lemma 2 above!), and so it must be the same integer for all t since it is a continuous integer-valued function. Hence, F(1)= F(0), and so the number of zeroes of f+g inside s is the same as the number of zeroes of f inside s.

The classic application of Rouché's Theorem is apparently in solving qualifying exam problems. What unbecoming cynicism! Here is a standard such problem: If |a|>e, show that ez=azn has n zeroes in |z|<1.

To solve this, we use Rouché's Theorem by taking s to be the unit circle, f to be equal to -azn and g to be ez. When |z|=1, |f(z)|=|a|, and |g(z)|=e|Re(z)|1<|a|. So |f(z)|> |g(z)| on |z|=1, and since f(z) has n zeros in |z|<1, f+g must have n zeros also.

Note that here we have used the fact that inf |f(z)|>sup |g(z)|, where the inf and sup are taken over all z in s. This certainly implies that |f(z)|>|g(z)|. However, the converse is not true, and it is enough to show that the inequality |f(z)|>|g(z)| holds pointwise for z in s for us to apply Rouché's Theorem.

How can the zeros of a limit function and the zeros of the approximating functions "match up"? I honestly believe that almost everyone working in mathematics (or an area using mathematics) will at some time try to find roots. Here is a very simple example to show you what can go wrong or, rather, why the complex numbers are nice.

Suppose we consider the sequence of real, calculus functions, fn(x)=(1-{1/n})x2+{1/n} on the interval [-1,1]. This is arranged so that the values of fn are all positive (no roots!), and for all n, fn(1)=1 and fn(-1)=1. Of course, and n-->, these functions uniformly go to x2 on [-1,1], which does have (a?) root: f(0)=0. It would be nice to consider situations where, if the limit function has a root, then closely approximating functions also have roots, and the number (!) of roots (counted correctly) are the same.

Or we could consider, say, fn(x)=x(x-{1/n}) in [0,1]. Then counting the roots: fn certainly has two roots, but as n-->, the result is x2 which (as a set!) has only one root.

Well, we should consider complex roots, of course, and also count roots with multiplicity. Then the seeming irritation of 0>1 (first example) and 2>1 (second example) as n--> will no longer occur.

 I tried to explore how roots of a polynomial would change when the polynomial is perturbed. If p(z) is a polynomial of degree n (n>0) then we know (Fundamental Theorem of Algebra) that p(z) has n roots, counted with multiplicity. Now look at the picture. The smallest dots represent roots of multiplicity 1, the next size up, multiplicity 2, and the largest dot, multiplicity 3. (I guess, computing rapidly, that n=11 to generate this picture.) Now suppose that q(z) is another polynomial of degree at most n, and q(z) is very small in some sense (say, |q(z)| is less than some small positive number in some big ). The only sense I won't allow is to have q(z) uniformly small in modulus in all of C (because then [Liouville] q(z) would be constant). Where are the roots of p(z)+q(z)? In fact, what seems to happen is that when q(z) is very small, the roots don't wander too far away from the roots of p(z). Things can be complicated. The roots of multiplicity 2 could possibly "split" (one does, in this diagram) or the root of multiplicity 3 could split completely. How can we prove that this sort of thing happens? Well, suppose we surround the roots of p(z) by "small" circles. Since all the roots of p(z) are inside the circles, |p(z)| is non-zero on these circles, and (finite number of circles, compactness!) inf|p(z)| when z is on any of these circles is some positive number, m. Now let q(z) be small. I mean explicitly let's adjust the coefficients of q(z) so that |q(z)| has sup on the set of circles less than m. Then the critical hypothesis of Rouché's Theorem is satisfied, and indeed, the roots of p(z)+q(z) are contained still inside each circle, exactly as drawn.

Certainly one can make this more precise. But I won't because I am in such a hurry. But I will remark on this: if we consider the mapping from Cn+1 to polynomials of degree<=n (an (n+1)-tuple is mapped to the coefficients), then for a dense open subset of Cn+1, the polynomial has n simple roots (consider the resultant of p and its derivative). On this dense open set, it can be proved that the roots are complex analytic functions of the coefficients. An appropriate version of the Inverse Function Theorem proves this result and it is not too difficult. But when the roots are not simple, then complicated things can occur. For example, consider the following smooth family of quadratic polynomials: fs(z)=z2-s2. The roots are, of course, +/-s. Notice that (absurdly!) fs(|s|)=0. So the non-smooth curve which maps s to |s| forms a collection of roots of the (surely!) smooth family of polynomials {fs}. The problem is that the curve is non-smooth exactly where the family of polynomials has multiplicity>1.

Tuesday, November 13 (Lecture #21)
An exam was given.

Friday, November 9 (Lecture #20)
The integral festival continues ...
Mr. Cantillo presented -[sin(x)/x]2dx
A tricky computation. A "simple" integration by parts changes this to -[sin(x)/x]dx, and then analysis of the integrals goes on. The residue (even though outside!) contributes half its value through a limiting argument on the small semicircle. The discussion on the large semicircle is more intricate, and some subtlety is called for (the integral over that arc does approach 0, but some care is needed.

The functions involved are important in mathematical physics. For example, sin(x)/x is (essentially, except for scaling) the Fourier transform of a "square wave".

Mr. Nanda presented a proof that n=11/n2 equals Pi2/6.
This is a standard computation to those who know the style! I made the pole at 0 larger and red, because it has order 3 (Pi cot(Pi z)/z2) while the other poles have order 1.

Mr. Nanda followed proof #8 in this reference which has fourteen (that's 14) proofs of the equality. Proof #7 is Euler's proof, which I vaguely suggested in class.

The lecturer's residue integral
The lecturer returned after writhing with anxiety. What if f is meromorphic at w? This means f(z)=(z-w)Nh(z) where N is some integer (positive, negative, or zero) and h is holomorphic in a neighborhood of w (N>0 means f has a zero of order N, and N<0 means f has a pole of order N). Then we "computed" f´(z)/f(z).

Well, f´(z)=N(z-w)N-1h(z)+(z-w)Nh´(z), so f´(z)/f(z) turns out to be, locally, N/(z-w)+h´(z)/h(z). Since h is holomorphic and non-zero near w, g=f´/f has an isolated singularity at w, and the residue of g at w is N. This is really neat.

How to look at rational functions
Suppose R is a rational function. Then we can write R(z)=P(z)/Q(z), where P and Q are polynomials. Since we know that C is algebraically closed, we can (in theory, at least!) completely factor P and Q. We can strike out common factors, so P and Q will be relatively prime. Then we would have a "description" of R which would be multiplicative and unique: R(z) is a product of a constant multiplied by (z-zj)Nj, where the Nj's are integers.

Example?
Here is a rational function:

```  5        4      3      4        3      2        2
z  - 2 I z  + 2 z  + 3 z  - 6 I z  - 2 z  - 6 I z  - 3 z - 2 I z - 1
----------------------------------------------------------------------
4      3       2
z  + 8 z  + 18 z  - 27```
This is (but not obviously) the same as
``` [(z+1)3(z-i)2]
----------------
[(z+3)3(z-1)1]```
Then applying the "construction" above -- the residues of the derivative divided by the function -- we seem to get a geometric picture of 4 points, 1 and -i and -3 and -1, together with integer weights at each of the points. The integers represent zeros with their order or multiplicity and poles with their order or multiplicity.

The picture I would like you to see is to the right. It has all (or almost all the information about the rational function, because two rational functions which have the same zeros and poles can differ only by a non-zero multiplicative constant!) of the information about the rational function, but displayed in a rather pleasing and possibly informative geometric manner. The poles of the function are "at" the points which have negative integer weights. The zeros of the function are "at" the points which have positive integer weights. This assemblage (?) is called the divisor of the function.

Now notice that the Residue Theorem will provide a way of adding up some of the divisor's information. So if s is a simple closed curve (this could be done for any closed curve, but simple, where winding numbers are +1 or -1, make things easier) which does not go through any of the poles or zeros of f, the integral of f´/f around s will count the total (+ for zeros, - for poles) number of poles and zeros inside s with their multiplicity, and, yes, somewhere there is a 2Pi i.

So suppose I have a meromorphic function f, and suppose I have a simple closed curve s, oriented positively (so all the winding numbers will be +1 for simplicity) and suppose that the curve does not go through any of the poles and zeros of s. Let's have s defined on the interval [a,b}. What is (1/{2Pi i})s[f´(z)/f(z)]dz? One interpretation has something to do with logs. In a simply connected open set where f is not 0, f´(z)/f(z) is the derivative of log(f(z)) or, rather, of some branch or determination of log(f(z)). Let's see:

We cover the image of s by a finite collection of open discs, disjoint from the poles and zeros of f. In each disc, we get lj(z), which is a version of log(f(z)). We get the antiderivative as usual by taking values of lj (j going from 0 to M, say) at two points in each disc. So the value of the integral is j=1Mlj(s(tj)-lj(s(tj-1). Notice, please that the sum "telescopes" in the sense that s(tM)=s(b)=s(a)=s(t0). Notice further that lj(tj)-lj-1(tj) must be 2Pi i multiplied by an integer since we are subtracting two "determinations" of logarithm. So all this tells me that the integral will be 2Pi i multiplied by an integer. This integer, if you ponder it carefully, is the winding number of the curve f(s) around 0. That is the value of the sum! Therefore the integral also evaluates the change in the argument of f around the curve s.

The Argument Principle
If f is meromorphic and if s is a simple closed curve which avoids the poles and zeros of f, then (1/{2Pi i})s[f´(z)/f(z)]dz is the net number of poles and zeros (counted with multiplicity) contained inside s. It is also the winding number of f(s) with respect to 0. Why? Because if you look at the integral, and make the "substitution" u=f(z), then the integrand becomes {1/u}du, and the curve integrated over is f(s), the image of s under f. Therefore the integral measures the change in the argument of f as we "travel" around s, s(arg f).

An example (!!)
O.k., this is a bit risky, but let's try to see an example at work. We've got good technology, and things should be visible. So let's use the function f(z) discussed earlier:

``` [(z+1)3(z-i)2]
----------------
[(z+3)3(z-1)1]```
And I would like to use the following simple closed curve: s(t)=2eiPi t with t in the interval [0,2Pi]. This is, of course, a circle with radius 2 and center 0. So the circle encloses three of the four points of the divisor of f. The count should be -1+2+3=4. The image curve in the complex plane is shown in an animation created using Maple. If you are not using a fast connection, then I apologize. The file size is half a megabyte! The picture on the right is a still picture of the image of the circle. That may be easier for people to see and study. I know as I tried to count and get the winding number from the moving picture, I almost got dizzy!

Do you believe the theorem? I will admit that getting a nice-looking example took some experimentation.

Tuesday, November 6 (Lecture #19)
The integral festival!!!
Mr. Oliver presented 0[sqrt(x)/(1+x2)]dx
A fairly standard computation. Maybe some care is needed to compute the residue.
Mr. McRae presented 0ln(x)/(1+x2)2
Here we needed an indented contour to take care of problems at 0. The integrand had to be chosen with some care in order to get a good definition of log.
Mr. H. Wang presented 0dx/(1+x4)
Two simple poles. With luck skill, the result is real and even positive.
Mr. Castro presented 02Pid/(5+3sin()).
Clearly (NO!) change the integral into an integral around the unit circle.
Ms. Naqvi presented 0dx/(1+x3)
Many coincidences. Choosing the correct contour works, but it is emphatically not obvious!

Friday, November 2 (Lecture #18)
Notes were to be by Mr. Nanda, but he has been relieved in order to give him time to prepare a somewhat intricate application of the Residue Theorem. Brave Mr. Nanda! Thus this report is written by the instructor.

Teaching Rule #4 Avoid clever proofs!
This is totally a personal preference, but I feel that clever proofs, which are almost never the first proofs of a result, may discourage or intimidate students (hey, and some faculty!). So maybe just compute, compute ... "stupidly" until there is insight built in the minds of students. This also is how most mathematical results I've been involved with have been discovered.

In honor of the Motion Picture Association of America rating system, I warn the young scholars of the class that what's presented today is rated VC, for very clever, and does not obey the rule stated above. Thus you should not assume that almost anything shown here is clear or obvious or ...

 "A child of five could understand this. Fetch me a child of five...."                 Groucho Marx (1890-1977)

Suppose U is an open subset of C, and f is holomorphic in U with isolated singularities. So there is a discrete set W in U with f holomorphic in U\W. Suppose also that s:[a,b]-->U\W is a closed curve which is homotopic in U to a point. Then
sf(z)dz=w in W2Pi i Resf(w)Inds(w).

This is undoubtedly one of the great theorems of mathematics, and special cases include versions of Cauchy's Theorem and the Cauchy Integral Formula. Those of you who are obsessively precise (or merely trying to understand!) may desire some definitions (residue and index, or as I prefer, winding number). The text has these on p.123 and p.124, respectively. But let me try. Please note, though, that the sum in the theorem is over a set, W, which may be infinite, so it is not even clear it converges.

Definition of the winding number (or index)
Suppose s:[a,b]-->C is a closed curve (so s(a)=s(b)), and w is not in s([a,b]). Then the index of s with respect to w (also called the winding number of s with respect to w is [1/{2Pi i}]s{1/[z-w]}dz.

Properties of the winding number

1. It is an integer.
Let's see: consider one of the methods we used to evaluate integrals. We need to cover s([a,b]) be a collection of open discs, Drj(zj), 1<=j<=N, so that w is not in any of the discs because then the integrand, 1/(z-w) (a function of z!), will be holomorphic in each disc. There should be an accompanying partition, a=t0<t1<...<tN=b so that s([tj-1,tj]) is contained in Drj(zj) for all j's between 1 and N. We can choose an antiderivative of 1/(z-w) in each disc. Considering the function, we will choose a "branch" of log(z-w) in each disc. That is, a function lj(z) so that elj(z)=1/{z-w} in the jth disc. Notice, please, that on the overlap values, s(tj), lj and lj+1 need not be the same, but they will differ by roots of esomething=1, which are 2Pi i integer. So the integral in the jth disc is just lj(s(tj))-lj(s(tj-1)). This certainly seems formidable! We won't "compute" it but we will consider the nature of this as it is summed from j=1 to j=N. Notice that what we have are just a collection of values of the form lj+1(s(tj))-lj(s(tj))). We have just "shifted" the summing (this is subtle -- send for that "child of five"!). We also note that the endpoints are the same sort of thing, since the curve is closed. That is, l1(s(t0)), which appears with a minus sign, and lN(s(tN)), which appears with a plus sign, are both log's of the same complex number. So this alternating sum, as observed above, is a sum of 2Pi iinteger. Therefore the whole integral has than nature, and when we "normalize" (divide by 2Pi i) the result will be an integer.
2. Which integers are possible?
Suppose n is an integer (positive, negative, or 0). Consider the closed curve defined by the formula s(t)=e2Pi i t defined on the domain [0,n]. We will compute the winding number of this curve around 0. So s[1/z]dz=0ne-2Pi i (2Pi t)e2Pi i tdt=(2Pi i)t]0n=(2Pi i)n. Dividing by 2Pi i gives the result n, so all integers occur as winding numbers.
3. It is a holomorphic function, so ...
Consider f(w)=[1/{2Pi i}]s{1/[z-w]}dz. Now f, as a function of w, for w in the open subset of C which is C\s([a,b]) is a holomorphic function of w (remember the early homework assignment, please: you can differentiate f by analysis of the difference quotient, etc.). But f is an integer-valued holomorphic function. Since f is continuous and the integers are a discrete subset of C, f must be constant in connected components of C\s([a,b]). So in the previous example, we showed that Inds(0)=n for the closed curve defined there. We've computed Inds(w) for all w with |w|<1: it must be n, also.
4. Way out, winding number is 0.
The logic of the homework assignment persists. The f(z) must satisfy |f(w)|=O(1/|w|) for |w| large. Now consider that s:[a,b]-->C is continuous, and by compactness s([a,b]) is contained in DR(0) for some R>0. f is defined in C\DR(0) and the estimate and the fact that f is integer-valued all unite to declare that f must be 0 in C\DR(0).
 Example (Even a VC lecture can have an example!) To the right is an example of a closed curve with the connected open subsets of its complement in C labeled with integers: the winding numbers in each piece. I certainly hope that I have not made any errors! There are some things to note about this. There are points "inside" the curve which have winding number equal to 0. The "net number of times" that the curve wraps around those points is 0, even though there is no curve in the complement of our given curve which would connect the points to . And we can have regions with negative winding number. There are points "inside" the curve which have winding number equal to -1, also. The word "inside" seems to be complicated. This diagram may actually convince you (it does me!) of the utility of a different way of algebraizing (!) topology, which is homology. I won't discuss that method in this course, but be aware that maybe another approach may have some validity.
5. A homotopy property
Suppose s1 and s2 are closed curves, and w is in the complement of both of their images. If s1~s2 in C\{w} (that is, s1 and s2 are homotopic as closed curves in the complement of w in C) then Inds1(w)=Inds2(w). This is a consequence of our homotopy version of Cauchy's Theorem applied to f(z)=1/(z-w).

Official definition of residue
If f is holomorphic in De(w)\{w}, then the residue of f at w, abbreviated as Resf(w) is a-1 if f(z)=n=-an(z-w)n is the Laurent series for f in De(w)\{w}.

If you do analytic number theory or mathematical physics (or many other fields!) you will compute many, many residues. There are a plethora (!!) of methods, and, indeed, most computer algebra systems (such as Maple and Mathematica) have powerful algorithms for computing residues. Professor Iwaniec told me to assure you that most of analytic number theory is a consequence of the Residue Theorem.

Proof of the Residue Theorem
Since s~a point in U, there exists a homotopy H which changes s to a point map continuous in U. The image of this homotopy is a compact subset, K, of U. If w is not in K, then (property 5 above) Inds(w)=0. Therefore in the sum, w W2Pi i Resf(w)Inds(w), which appears in the statement of the Residue Theorem, we need only consider those w's which are in K. Since W is discrete and K is compact, there are only a finite number of such w's. Let's call them w1, w2, ..., and wM. Since f has an isolated singularity as each wj, we know that f can be written as a sum of fin+fout (the non-negative and negative terms in the Laurent series at wj). The fout is actually holomorphic in C\{wj}. Let me call this fout, Sj, the singular part of f at wj. Now g=f-j=1MSj is holomorphic in U\W, and g has removable singularities at each wj for j from 1 to M, because the local Laurent series doesn't have any negative terms. Therefore when we consider the integral of g over s, we can use the Cauchy integral theorem. g (extended suitably over the M removable singularities) has integral 0 over s because the integral homotopies to an integral over a point. So sg(z)dz=0. This means that sf(z)-j=1MSj(z) dz=0. Thus (it must be a serious proof, "thus" is used!) sf(z)dz=j=1MsSj(z)dz.

Now consider each of the integrals sSj(z)dz. Sj is holomorphic in C\{wj}, and so Oh, baby, baby we can apply the "baby" version of the Residue Theorem. But this is exactly (2Pi )Inds(wj)ResSj(wj), which equals (2Pi i)Inds(wj)Resf(wj). And we have obtained the sum needed to complete the proof of the Residue Theorem.

Application: a finer form of the CIF than you'll ever need
Suppose f is holomorphic in U, w is in U, and s is a closed curve in U\{w} which is homotopic to a point ("homotopic to 0") in U. Consider the holomorphic function g defined by g(z)=[f(z)-f(w)]/[z-w]. This is certainly holomorphic in U\{w}, but limz-->wg(z) exists (it is f´(z) as several students reminded me!) and therefore g has a removable singularity at w, which we will consider "removed": g is holomorphic in U. Then by our previous version of Cauchy's Theorem, we know that sg(z)dz=0. This means s[f(z)]/[z-w] dz=s[f(w)]/[z-w] dz=(s[1/[z-w] dz)f(w)=(2Pi i)Inds(w)f(w), which reduces to the Cauchy Integral Formula we've been using constantly (circles, with the index computation as in 2 above).

I don't think that this rather elaborate version of the CIF will be used frequently by many people in the class, but I think such a formula is really neat.

The traditional application: computing an improper integral
Let's compute the Fourier transform of 1/(1+x2). This is an L1 function (it is essentially the probability distribution function of the Cauchy random variable, used in statistics). The Fourier transform is -[eimx/(1+x2)]dx where m is a real number. Fourier transforms are used in a large number of parts of mathematics.

Notice that since eimx=cos(mx)+i sin(mx), and sin(mx) is odd and 1/(1+x2) is even, and the interval (-,) is "balance" around 0, the sine part of the improper integral is 0.

We need to compute -[cos(mx)/(1+x2)]dx. I can't compute this using the Fundamental Theorem of Calculus directly because I don't know any antiderivative of the integrand (there is no simple formula for that, actually). The integral is absolutely convergent, however (that's the L1 comment). This means that although the actual definition of the improper integral is the limit as, say, Q1 and Q2 both -->+ in -Q1Q2[cos(mx)/(1+x2)]dx, we actually only need to analyze -QQ[cos(mx)/(1+x2)]dx.

How to do this
The traditional way to compute -QQSOMETHING(x) dx as Q--> is to relate the integral to CQSOMETHING ELSE(z) dz, where the most frequent choice of CQ is the simple closed curve shown to the right. CQ is the "sum" of IQ (the integral over [-Q,Q] on the real line) and, say, a semicircle UQ as shown. The SOMETHING ELSE(z) is a holomorphic function with "nice" isolated singularities in the upper half plane, and which, further, is nicely related to the original SOMETHING(x) and which, somehow, on the contour UQ, goes to 0 fast enough as Q--> that the integral on UQ will go to 0. Then the original integral will be computed by looking at the residues of SOMETHING ELSE(z) in the upper half plane and using the Residue Theorem.

Technique, you bet!!!
I hope that you see there is a great deal of choice involved in all this, and very many techniques have been developed to get contours and functions for which the idea above works. For example, we had cos(mx)/(1+x2). One SOMETHING ELSE(z) which is a candidate is cos(mz)/(1+z2) but cos(mz) has enormous (exponential!) growth in imaginary directions. So this choice won't work in the template discussed above.

A good selection
Go back to the Fourier transform (as Ms. Naqvi said) and use f(z)=eimz/(1+z2). Also let's restrict ourselves to m>=0 here when we use UQ. We do this because of the following:
IF Im(z)>=0, |eimz|=|eim(x+iy)|=e-y<=1.
So the "top" of the suggested SOMETHING ELSE(z) is certainly bounded in modulus in the upper halfplane. Also please note that |1/(1+z2)|<=1(|z|2-1) if |z|>1 by our "reverse triangle inequality".

In fact, we now know that for m>=0 and for Q>1, |UQf(z)dz|<=ML[1/(Q2-1|](Pi Q). This will -->0 as Q-->.

Also, IQf(z)dz-->-f(z)dz, and f(z)=eimz/(1+z2=[{cos(mx)+i sin(mx)}]/[1+x2] when z=x is real. The sine integral again is 0, so we will get, as the limiting value when Q-->, the desired integral.

But now we need to compute the integral of f(z)=eimz/(1+z2) over CQ when Q is large positive (we just need Q>1). Now eimz/(1+z2)=eimz/[(z+i)(z-i)]=[1/(z-i)][eimz/(z+i)]. The factor [eimz/(z+i)] is holomorphic in a neighborhood of i and non-zero at i. Therefore f(z) has a simple pole (a pole of order 1) at i. The Laurent series of f centered at i begins a-1/(z-i)+a0+a1(z-i)+... and suppose we multiply this by (z-i). The result is a convergent power series which begins a-1+a0(z-i)+a1(z-i)2+... so we can get the residue just by "plugging in" z=i (only a-1 will remain then). So let's do that with f(z)=eimz/[(z+i)(z-i)]. We multiply by z-i and get eimz/(z+i) and plug in z=i to get e-m/(2i). Are we in trouble? Do we get an "imaginary" answer for a real integral? No! We must remember to multiply by 2Pi i (the Residue Theorem), so the result is (2Pi i)e-m/(2i), or Pi e-m.

After we take the limit as Q-->, we know that, for m>=0, -[eimx/(1+x2)]dx=Pi e-m. You can check this because when m=0, the integrand does have an antiderivative (arctan!) and its value is Pi by elementary means.
For m<0, the integral could be computed with a semicircular "completion" of IQ in the lower halfplane. The result is that the Fourier transform is Pie-|m|. You likely will later see in real variables that the non-smoothness of this function (the absolute value) is a consequence of the original function being in L1 but x times the function is not in L1.

Next time, a great festival of integrals!!

Tuesday, October 30 (Lecture #17)
Notes by Camelia Pop, edited, with some comments, by the instructor.

Laurent series results, specialized for this lecture
Suppose f is holomorphic in the punctured disc centered at 0 with radius R (that is, the collection of z's with 0<|z|<R. Then

• f(z)=n=-anzn for all z with 0<|z|<R. This series converges absolutely for all such z, and uniformly for any compact subsets of this annulus.
• an=(1/[2Pi i])Dt(0)(f(s)/sn+1)ds, for any t between 0 and R (in fact, for all such t: the integral does not depend on these "eligible" t's!).
• f=fin+fout, where fin is holomorphic in the whole disc DR(0) and fout is holomorphic in C\{0}.

For me, it is important to note that we "developed" the material in the following table sequentially. We did not present it as all done at once, and obvious by some sort of divine inspiration. It certainly is not that. It was done by human beings, fallibly, with many examples and attempts with varying success.

Classification of isolated singularities
Name of singularityLaurent seriesTheorem or behavior
(if sing. is at 0)
Examples
(the sing. is at 0)
Removable singularityThe an=0 if n<0, so that f=fin. Riemann Removable Singularity Theorem If f is bounded in DR(0)\{0}, then there is F holomorphic in DR(0) with F=f if |z|>0.
The same is true under other circumstances: for example, see the later discussion, or prove the result is true if f is L2 in the disc.
z3 (silly, but an example!); [sin(z)]/z, [cos(z)-1]/z2.
PoleThere is N<0 so that aN is not 0, and an=0 when n<N f(z)=zNh(z) for some negative integer N, where h is holomorphic in a neighborhood of 0 with h(0) not 0. This is a pole of order N.
limz-->0f(z)=.
Suppose P and Q are in C[z} (polynomials) and (P,Q)=1 (that is, P and Q have no common factor -- remember by Fund.Thm of Alg., we "can" factor all polys). Then P/Q has a pole at every singularity (where Q(z)=0).
1/[ez-1]: this has a pole at 0 if you consider the power series of exp, factor out a z, and then recognize (?) h(z). The behavior is the same at 2Pi n i, for all integer n.
Take the mth (integer) power of the preceding example. The result certainly has poles of order m at all 2PI n i.
Mysterious black hole!!!
Essential singularity
There are infinitely many an's with n<0 which are not 0. Casorati-Weierstrass Theorem For all e>0, f(De(0)\{0}) is dense in C. e1/z and many others.

Riemann's Theorem If f is bounded in DR(0)\{0} then f can be extended holomorphically to DR(0).

Proof (There is an alternate proof at the beginning of chapter 4). The boundedness condition satisfied by f implies that there is M >0 so that |f(z)|<=M for all z with 0<|z|<R. Then |an|=|(1/[2Pi i])Dt(f(s)/sn+1)ds|<=ML[(2Pi t)/(2Pi)]M/tn+1=M/tn.
Now if n<0, the final expression-->0 as t-->0. So all of those an's are 0, and f=fin, and fin is the desired extension.
This is very much like the Liouville's Theorem proof. As are other proofs ... as you will see.

Counterexample to show that this theorem does not work on R
Consider sin(1/x). This is bounded. It is differentiable (indeed, real analytic!) in R\{0} but it cannot be extended by continuity at 0. (Look at what happens in "imaginary" directions!)

The limit means the following statement:
For all M>0, there is e>0 so that |f(z)|>M when )<|z|<e.
But please notice that if |f(z)|-->, it must "go to " in a very precise fashion: essentially proportional (constant of proportionality: h(0)) to an inverse integer power of |z|. This is a very "stiff" requirement.

The astonishing Casorati-Weierstrass Theorem Suppose f has an essential singularity at 0. Then for all e>0, f(De(0)\{0}) is dense in C.

Proof Suppose the conclusion is false. Then there is e>0, a in C, and b>0 so that for all z in De(0)\{0}, |f(z)-a|>=b. If you can contradict a complicated quantified statement, then you too can prove the Casorati-Weierstrass Theorem!

Consider g(z)=1/(f(z)-a). Then |g(z)|<1/b, so that g is bounded in De(0)\{0}, and we can apply Riemann's Removable Singularity Theorem. We know that there is G holomorphic in De(0) which extends g. Consider the behavior of f(z)=(1/g(z))+a.
If g(z) is not 0, then f has a removable singularity at 0, and the Laurent coefficients an for n<0 are all 0. This is false.
If g(z)=0, then (since g is not identically 0) f has a pole of some order at 0, and only finitely many of the an's for n<0 are non-zero. This is also false.
This contradiction shows that the original assumption must be false.

There is a much more "powerful" result called The Great Picard Theorem which asserts that if f(z) has an essential singularity at 0, then on any open neighborhood of 0, f(z) takes all possible values, with at most one possible exception. (Think of e1/z, where the exceptional value is 0.)

Now let's discuss some applications.

A simple application
Suppose f is holomorphic in C\{0}, and we know that |f(z)|<=[1/sqrt(|z|)]+sqrt(|z|). Then f is constant (and we can even say a bit about the constant!).
This problem is from a previous final examination in Math 503.

Proof
Let's call FROG the term [1/sqrt(|z|)]+sqrt(|z|).

First claim: z=0 is a removable singularity. Just write |an|=|(1/[2Pi i])Dt(f(s)/sn+1)ds|<=ML[(2Pi t)/(2Pi)]n+1=FROG/tn.
Now consider the pieces of FROG for n<0. One piece gives the upper bound t-n-1/2 and the other piece gives the upper bound t-n+1/2. If n is a negative integer, then both of the exponents are positive so that as t-->0, the upper bound-->0. The negative Laurent coefficients are all 0.

Second claim: The positive Laurent coeffients are all 0. Do the same (more or less!) thing. Write |an|=|(1/[2Pi i]) Dt(f(s)/sn+1)ds|<ML[(2Pi t)/(2Pi)]n+1=FROG/tn.
And the pieces of FROG give you t-n-1/2+t-n+1/2. If n is a positive integer and if t-->, then each of the exponents is negative, so the sum-->0. These coefficients are all 0 also.

So f(z)=a0, a constant. Wow. Indeed, |a0|<=[1/sqrt(|z|)]+sqrt(|z|) for all non-zero z. Calculus tells me that the minimum value of g(x)=[1/sqrt(x)]+sqrt(x) is 2, so actually |a0| must be less than or equal to 2.

Here is perhaps a more interesting application. We begin with a definition.
Suppose X and Y are topological spaces and f:X-->Y. Then f is said to be proper if f-1(K) is compact in X for all K which is compact in Y.

Examples and counterexamples
We will consider continuous f's going from R to R.
f is NOT proper
f=constant (take K={that constant}).
f(x)=arctan(x). This is 1-1, but f-1([-Pi/2,Pi/2]) is not compact.
f is proper
f(x)=x ("clearly").
f(x)=x2 (maybe not quite as "clearly").
It is possible to "construct" a continuous proper map from R to R so that the cardinality of f-1(y) is infinite! Just take f(x)=x for x<0, f(x)=0 if x is in [0,1], and f(x)=x-1 if x>0. (You can get a smooth version of this fairly easily using the functions we described earlier.)

There is a sequential way to characterize proper functions.
f is proper if (and only if) for all sequence {xn} with |xn|-->, then |f(xn)|-->.
Another way of characterizing proper maps, as Professor Ferry reminds me, is that a continuous map is proper if and only if it can be extended continuously to the one-point compactifications. (It will take the domain's to the range's .) Lovers of point set topology may contemplate this.)

Suppose an entire function is proper. So if f:C-->C is holomorphic and proper, consider g(z)=f(1/z) (this is a standard way of taking things "at " and bringing them to 0). What sort of singularity can g have at 0?

If g's singularity is removable, then when z is close to 0, g(z) is close to the "value" g's extension would have at 0. Then g is bounded in a neighborhood of 0, so f is bounded for |z| large. But such an f would be bounded for all C (it is [nearly] trivially bounded for |z| inside a disc). So (Liouville) f must be constant, and hence not proper.

If g's singularity is essential, we can find a sequence {zn} with 0<|zn| with limn-->zn=0 so that limn-->g(zn)=1, say. Then f(1/zn)-->1 but limn-->1/zn=. This contradicts properness.

So g must have a pole at 0. But then g's Laurent series centered at 0 is the result of substituting 1/z for z in f's Taylor series centered at 0. Thus g's Laurent series has no positive terms, and only finitely many negative terms. So f must be a polynomial. We have proved:

Proposition If an entire function is proper, then it must be a non-constant polynomial.

Corollary The collection of holomorphic automorphisms of is the maps z-->az+b, with b in C and a not zero in C.

Proof Each holomorphic automorphism must be a homeomorphism, therefore a proper function and we have proved that all entire proper functions must be polynomials. The only polynomials which are one-to-one are the linear ones WHY? with non-zero dominant coefficient.

Here the group "operation" is composition. So the result of a1z+b1 followed by a2z+b2 is a1a2z+a2b1+b2. Doesn't this make you think a little bit about matrix multiplication? z-->az+b resembles, just a little bit,
(a b)
(0 1)
and the group composition operation seems like matrix multiplication. This is not a coincidence.

A baby (?) version of the Residue Theorem
Suppose f is holomorphic in D1(0)\{0} and s is a closed curve in D1(0)\{0}. We describe how to "compute" sf(z)dz.

sf(z)dz=sn=-anzndz=n=-anszndz. The interchange is justified since the series converges uniformly on s. But szndz is 0 for all n not equal to -1, since these functions all have holomorphic antiderivatives in D1(0)\{0} (the antiderivatives are zn+1/(n+1) if you have forgotten). So the only term left is a-1s(1/z)dz.

Now we define our way out of the difficulties of the theorem (!!!).

a-1 will be called the residue of f at 0 and s(1/z)dz will be called (2Pi i multiplied by) the winding number of s around 0 (the text calls this the index of s around 0).

Friday, October 26 (Lecture #16)
The instructor went over an important global/local idea which he had previously described inadequately.

Ms. Pop discussed the Phragmen-Lindelöf Theorem (assisted by Mr. Bate, Mr. Cantillo, and Mr. Pal. She followed notes by Professor Paul Garrett, a number theorist at the University of Minnesota.

The most important single technique for analyzing holomorphic functions in circularly symmetric domains is the Laurent series, a generalization of power series. A biography of Pierre Alphonse Laurent (1813-1854) declares that "his memoir was not published" (referring to the paper in which the Laurent series was first described). This is not so nice. Laurent series will be especially important in our discussion of the isolated singularities of holomorphic functions, to be done next time.

We suppose that 0<=r<R<=, and define an annular region A by A={z in C with r<|z|<R}. (Here we will have our annulus and our Laurent series "centered" at 0. If you wish to center it at p, then |z-p| and (z-p) replace |z| and (z) in the following discussion.) We suppose that f is holomorphic in A. If z is in A and e is sufficiently small, then the Cauchy Integral Formula implies that f(z)=(1/2Pi i)De(z)(f(s)/(s-z))ds. We now deduce the extremely useful Laurent expansion for f.

Deforming the integration contour continuously
The preliminary step is to distort the integration contour nicely. The function g(s)=(f(s)/(s-z)) is holomorphic in A\{z}. We "distort" De(z) as shown in the picture.

The integrals over the line segments connecting the two circles cancel. We are left with integrals over two circles, one oriented positively (in the usual, counterclockwise, direction) and the other oriented negatively. Then Cauchy's Theorem implies that the integral of g over De(z) is equal to the difference of two integrals, one over Db(0) and the other over Da(0) with r<a<|z|<b<R. Now we handle each of the integrals. I'm just going to outline the process -- it appears in virtually every complex analysis text. The opportunity for making sign errors and summation index errors arises frequently.

The outer integral
Well, consider (1/2Pi i)Db(0)(f(s)/(s-z))ds. We will "expand the Cauchy kernel". So:
We know |s|=b>|z|, so 1/(s-z)=((1/s))(1/(1-[z/s])). Now we use the geometric series, since |z/s|<1. The series will consider absolutely for each such s, and uniformly for s on Db(0). Then we interchange integration and summation.

The inner integral
Well, consider (1/2Pi i)Da(0)(f(s)/(s-z))ds. We will "expand the Cauchy kernel". So:
We know |s|=a<|z|, so 1/(s-z)=-((1/z))(1/(1-[s/z])). Now we use the geometric series, since |s/z|<1. The series will consider absolutely for each such s, and uniformly for s on Da(0). Then we interchange integration and summation.

A difficulty: convergence of doubly infinite sums
Before we state the result, think a bit about how n=-qn should be defined. One possibility is to look at n=-NNqn and require convergence as N-->. This has some defects. For example, then the double series with qn=-1 for n<0 and +1 for n>0 would converge (I guess to q0). This series certainly would not converge absolutely, and the implication "absolute convergence implies convergence" is useful and we should preserve it. Also, if we used that definition, then relabeling n-->n+1 might change the sum of the series. So the more accepted definition is as follows:

The series n=-qn converges and its sum is S if, given any real w>0, there is some positive integer N so that if m1 and m2 are both greater than N, then |n=-m1m2 |qn-S|<w.

Then the double series will converge if and only if each of the half series (?), n=0qn and n=-0qn, converges, and all of the expected statements will be correct.

Laurent Series Theorem
Suppose that 0<=r<R<=, and define an annular region A by A={z in C with r<|z|<R}. Further, suppose that f is holomorphic in A. Define the doubly infinite sequence of complex numbers {an} by an= Dt(0)(f(s)/sn+1)ds for any t with r<t<R. Then

• f(z)=n=-anzn
• This series converges absolutely for every z in A, and converges uniformly on compact subsets of A.
• There is exactly one such series representing f in A.
• f can be written as a sum of fin, holomorphic in {|z|<R}, and fout, holomorphic in {|z|>r}. This decomposition is unique if also limz-->fout(z)=0.
Discussion of proof
Notice that in the formula for an, the integral Dt(0)(f(s)/sn+1)ds, there seems to be a dependence on t. But Cauchy's Theorem declares that the result is the same for all t with r<t<R.

The existence of the series is obtained by following the path of expanding the Cauchy kernel twice, as mentioned before the statement of the theorem. If there is such a series, then the coefficients can be obtained by integrating f(s)/sn+1 around a circle, and interchanging sum and integral (possible because of the uniform convergence).

fin is the sum of the non-negative z powers, and fout is the sum of the negative z powers.

Examples
I tried diligently to find various Laurent series expansions of a rational function, using the partial fraction expansion (that the partial fraction expansion always exists is a consequence of the Fundamental Theorem of Algebra, by the way!).

I "found" the (-1)st coefficient of ez+(1/z) in its Laurent expansion in C\{0}. This is a value of a certain Bessel function, and is k=01/(k!(k+1)!). Finding every coefficient of a Laurent series explicitly for a random function, even one defined by a classical formula, can be very difficult or even impossible.

Since Laurent series converge absolutely, and they converge uniformly on compact sets, then any of the standard manipulations of algebra (multiplication, addition, division, etc.) and calculus (differentiation, integration) are justifiable, and any sort of trickery which gets the series is justified.

Tuesday, October 23 (Lecture #15)
Notes by Brian Nakamura, edited, with some comments, by the instructor.

Recall the
Cauchy Integral Formula (CIF) Suppose f is holomorphic in an open set, U, and that the closed disc of radius r centered at p is contained in U. Then f(z)=(1/(2Pi i))Dr(p)(f(s)/(s-z))ds.

We differentiated this formula and obtained:

(CIF)(n) (Assume the same hypotheses as the previous result.) Then f(n)(z)=(n!/(2Pi i))Dr(p)(f(s)/(s - z)n+1)ds

Then, a consequence of this is

Theorem (Cauchy Estimates) Suppose f is holomorphic in an open set which contains the closed disc of radius r centered at z. Then |f(n)(z)|<=(n!/rn)Mr(f), where Mr(f) is sup|s-z|<=r|f(z)|.

Recalling the past homework assignment, we had ||f´||K<=(1/s)||f||Ks. The theorem is essentially this inequality with K={z} and Ks=the appropriate closed disc centered at z. The Cauchy estimates can be proved using the (CIF)(n) and the ML inequality or by iterating the n=1 case.

At this point, a digression was made as the instructor attempted to explain a dubious pun. Mr. Bates explained to the class that a Louisville Slugger was a type of baseball bat. In addition, he expressed regret that a Theorem was named after a bat. The instructor also clarified that this "bat" is not a flying mammal. The Little Brown Myotis ("little brown bat", Myotis lucifugus) is commonly observed in central New Jersey (most easily seen during the evening in spring and summer). A picture is to the right.

Liouville's Theorem If f is entire (f holomorphic in all of C) and f is bounded (so supz in C|f(z)| is finite), then f is constant.

Proof The idea is to show that the derivative must be 0. We used the Cauchy estimate with n=1. Notice that, for any r>0, Mr(f)<=B (the value of supz in C|f(z)|). Then |f´(z)|<=(1/r)B for all r>0. If r-->, we see that f´(z)=0 for all z. Therefore f is constant.

There are many applications. A startling one is that the spectrum of a linear operator is non-empty. The setting is the following:
Suppose V is a vector space and T:V-->V is a linear transformation. The spectrum of T (usually denoted with the Greek letter "small sigma") is the collection of w in C so that T-wI is not invertible. If V is finite dimensional, a test for non-invertibility is easy using the determinant. That the spectrum (which then consists of eigenvalues) is non-empty reduces to the Fundamental Theorem of Algebra, which we are about to deduce. In infinite dimensions, more varied phenomena occur, and a direct use of Liouville's Theorem is needed.

Observation D1(0), the unit disk, and C are homeomorphic. The mapping defined by rei-->(r/(1-r))ei is such a homeomorphism. Note, please, that this mapping is actually a diffeomorphism. So these open sets are differentiably "the same".

Theorem There does not exist f:D1(0)-->C which is biholomorphic.

Here is the verification: if such an f were to exist, consider f-1. Certainly f-1 is a bounded entire function, which by Liouville must be constant. This result therefore declares that, somehow, the geometry (in this case, conformal geometry, since biholomorphic mappings preserve angles) has more information about the domain than just topological or even differentiable mappings.

Corollary There does not exist f:H-->C which is biholomorphic. (Here H is upper halfplane, those z's in C with Im(z)>0.)

Proof We verified on the first day of class that H and D1(0) are the "same" (they are biholomorphic), using the mapping by z-->(z-i)/(z+i). This verification was a direct computation.

Corollary (one of perhaps infinitely many the instructor could recite) W=C\{z with Im(z)<=0 and Re(z)>=0} is not biholomorphic with C.

Proof We just need to show that this domain, three-quarters of the plane (!), is biholomorphic with half the plane (!!). The mapping z-->z2/3 is defined in W (it is exp(2/3{log(z)}) and since W is a simply connected open set not containing 0, W has "a branch of log"). The mapping just defined does not preserve angles through 0: it multiplies such angles by three halves, so, indeed, it changes 3 right angles (in W) to 2 right angles (in H). The lack of conformality doesn't matter because it happens on the boundary of the domains, not inside! So W is biholomorphic with H, and therefore in view of the previous fact, cannot be biholomorphic with C.

One of the major results of complex analysis is the Riemann Mapping Theorem, which I hope we'll get to later. The statement of the theorem is quite simple: any simply connected open subset of C which is not all of C is biholomorphic with the unit disc, D1(0). So there are two distinct "equivalence classes" of angle geometry, all of the complex numbers, and the unit disc.

A consequence of Liouville's Theorem is the Fundamental Theorem of Algebra:
If P an element of C[z] and deg(P)>0, then P has a root.

Before proving this, let's see what a polynomial mapping looks like asymptotically. First, an example:

Suppose P(z)=z5-(3+i)z3-4z2+7i does have 5 roots as predicted by FTA. Where are they? How about where are they not? Can we find R so that if |z|>=R, then |P(z)|>0?

Define the "small part" of P be S(z)=-(3+i)z3-4z2+7i. Then let us (over)estimate S(z) if |z|>= R.
|S(z)|<=|3+i| |z|3+4|z|2+|7i|<=(|3+i|+4+|7i|)|z|3 <= 100|z|3, assuming that R>=1 in order to insure that the second inequality is correct.

Now let the "big part" of P be B(z)=z5. Then
|B(z)|=|z|5, and P(z)=B(z)+S(z), which implies that |P(z)|>=|B(z)|-|S(z)| and |P(z)|>=|z|5-100|z|3=(|z|3)(|z|2-100). ).
By taking R>=50, we know |z|>=R>=50, which implies that |z|2-100>=(1/2)|z|2 (Check this!).

Therefore, |P(z)|>=(1/2)|z|5 if |z|>=R>=50. So roots of P must be inside a disc of radius 50 centered at 0 in C.

Proposition If P is an element of C[z] with n=deg P>00, then there exists R>0 and A,B>0 so that if |z|>=R, then A|z|n<=|P(z)|<=B|z|n.

The proof of this is similar to what we just saw in the example. Take P(z)=j=0najzj. We know that an is not 0. Let's consider Q(z)=P(z)/an=zn+j=0n-1ajzj which the first term will be B(z), the "big" term, and the second term, the sum, will be S(z), the "small" term. Then we overestimate the small term.

|S(z)|<=|j=0n-1ajzj|<=j=0n-1|aj| |z|j<= (j=0n-1|aj|)|z|n-1=W|z|n-1. The last inequality is true if |z| is at least 1. And W is just a "label" for j=0n-1|aj|.

Just as before we have |Q(z)|=|B(z)+S(z)|>=|B(z)|-|S(z)|>|z|n-W|z|n-1=|z|n-1(|z|-W). Now I want |z|-W>=(1/2)|z| and this is equivalent to (1/2)|z|>=W which is the same as |z|>=2W. So if |z|>max(1,2W), we know that |Q(z)|>=(1/2)|z|n. Wow! Now we have |P(z)|>=(|an|/2)|z|n.

The other side of the desired bound, the overestimate, on |P(z)| is much easier. Look:
|P(z)|<=|j=0najzj|<=j=0n|aj| |z|j<=V|z|n which is valid if |z|>=1 and if V=j=0n|aj|.

So you can put it all together now, and see that the R and A and B desired and predicted do exist. They can even computed from the coefficients of the polynomial.

Now returning to the Fundamental Theorem of Algebra:

Proof (of FTA): If P has no root, let f(z) = 1/P(z) and observe that this f is an entire function, holomorphic in all of C. Also, using the previous corollary, we get an R>0 and A>0 such that if |z|>=R, then |f(z)|<=1/(ARn) (actually bounded by 1/(A|z|n but what's written is enough). Thus f is bounded outside of the disc of radius R centered at 0. Also, if |z|<=R, |f(z)| is bounded (by compactness). Then Liouville's Theorem implies that f is constant, a contradiction!

One consequence of FTA (together with polynomial manipulation, as done in high school) is that polynomials with complex coefficients "split completely". That is, any non-constant polynomial can be written as a product of linear polynomials.

As I mentioned in class, there are now probably hundreds of proofs of FTA, some quite clever. The original proof of Gauss is not the one given above, I believe. The proof given above "appeared" after a hundred years of cleverness, and it is very efficient. Some of the other proofs are worth looking at. Remmert's book discusses some of the other proofs.

Notice that this result does not address the problem of finding or approximating the roots of a polynomial. Root finding is not easy. I hope that you know that Newton's Method, unassisted, doesn't always converge to a root of a polynomial. The literature on this (numerical, symbolic, etc.) is enormous.

Some Liouville-like results are:

Theorem If there exists a natural number k so that |f(z)|<=A|z|k for |z| large (and f entire), then f is a polynomial of degree at most k.

Proof We want to show that (k+1)st derivative is 0. The Cauchy estimates give: |f(k+1)(z)|<=(constant/rk+1)Mr(f)<=(constant/rk+1)Ark=(another constant/r). Then let r-->.

Silly observation If f is entire, with f(0)=37 and |f(z)|<=sqrt(6000+|z|), then f is constant. So the modulus of a non-constant holomorphic function has certain minimal rates of growth -- things can be rather subtle.

Now if R is commutative ring with multiplicative identity 1 and no zero divisors, we can "construct" a quotient field.

The canonical example is Z, the integers, where the quotient field can be realized as Q, the rational numbers.

Here's (approximately) how the quotient field is constructed.
Look at S=Zx(Z\{0}) and define an equivalence relation:
(a1,b1)~(a2, b2) if and only if a1b2=a2*b1.

Comment Since I can't think algebraically too well, the only way I keep track of this is by almost constantly thinking about rational numbers etc. This is very abstract to me.

• ~ is an equivalence relation (verification needs "no zero divisors"). Then S/~ has a ring structure defined as follows.
• Addition [(a1,b1)]+[(a2,b2)]=[(a1*b2+a2*b1,b1*b2)]. (Many details to be checked in order to see that the definition does not depend on representatives of the equivalence classes, etc.)
• Multiplication [(a1,b1)]·[(a2,b2)]=[(a1·a2,b1·b2)].
• Further, if a is not 0, [(a, b)]·[(b, a)]=[(1, 1)] (muliplicative inverses and multiplicative identity).
• Also there is a ring monomorphism from R to S/~ defined by a-->[(a, 1)]). There's also a universal mapping property, which can also "define" the quotient field as the smallest (?) field containing a copy of the ring, R.

Some rings with no zero divisors are C[z], C{z}, C[[z]].

What "is" the quotient field of C{z} in some perhaps more "realistic" fashion than the description above? Note that a "realistic" description of C{z} is the collection of functions, f, which are holomorphic in some neighborhood of 0.

If f, g are elements of C{z} and g is NOT the 0 series, we know that g(z)=(zn)h(z) for some positive integer n and for some holomorphic h with h(0) not equal to 0. Then [(f, g)]=[(f, znh)]= [(f/h,zn)], so we could think of the element [(f,g)} as corresponding to a function (1/zn)(f(z)/g(z)) defined in some deleted neighborhood of 0 (a deleted neighborhood of 0 is a neighborhood from which 0 has been removed). So if [(f, g)] is a non-zero element of the quotient field, there exists a unique integer n so if z is nonzero, f(z)/g(z)=znh(z) with h holomorphic and h(0) not equal to 0.

We seem to be led to the following definition, of a holomorphic functions with isolated singularities. What is it?

Let U be open subset of C and W be a discrete subset of U. W discrete means here: for all w in W, there exists ew > 0 so that intersection of D(w, ew) and W is {w}. This is equivalent to requiring that relative topology which the subset W inherits from U is the discrete topology.

We say that f is a holomorphic function with isolated singularities in U if there is a discrete subset W of U so that f is holomorphic in U\W.

Since the quotient field construction above seems to imply that certain local representations of holomorphic functions may be more interesting, we have another definition: f is meromorphic if f can be described as (z-w)nh(z) with h holomorphic and h(w) not equal to 0 near every point of W.

Description of an "algebraic" result
Here is a weird, important, and non-trivial result. Consider the ring R of holomorphic functions on U. If U is connected, then this ring has no zero divisors. (This fact itself needed some effort, so you should think about it!) What is the quotient field of R? It turns out that the quotient field of the holomorphic functions is the meromorphic functions. This is not obvious. Let me give an example to explain the statement perhaps a bit better.

Suppose U=C, the whole complex plane. Then the sum n=11/(z-n) does not converge (just insert z=0 and get minus the harmonic series). So, much like we did previously when we constructed a function that blows up on magnifications of all of the roots of unity, we can make it converge: the sum n=1(1/n!)(1/(z-n)) certainly does converge on all of C\N. It is a function with isolated singularities in C. Since the isolated singularities are rather apparent, it is also meromorphic in C. The result mentioned previously then implies that this function can be written as f(z)/g(z), where f and g are holomorphic in C. What are f and g? Why are there such f and g? This is emphatically not clear to me. Further comment, added 11/9/2007
Let me persuade you that the series above does represent a holomorphic function away from the positive integers. Fix a positive integer N, and consider the disc DN(0), an open disc of radius N and center 0. Now look at the sum n=1(1/n!)(1/(z-n)) and break it up: (the alert reader is allowed to criticize, but I am not trying for optimality!)

n=1(1/n!)(1/(z-n))=n=12N(1/n!)(1/(z-n))+n=2N+1(1/n!)(1/(z-n)). The first, finite sum is clearly holomorphic away from the positive integers. We'll only consider the second sum, therefore. But for \$|z|<N\$ and \$n>2N\$ we know that |z-n|>=n-|z|>2N-N=N. So for those z's and n's, the modulus of (1/n!)(1/(z-n)) is bounded above by (1/n!)(1/2N). Thus, inside of the disc of radius N centered at 0, the "infinite tail" is bounded termwise by n=2N+1(1/n!)(1/(2N)), an absolutely convergent series of real numbers. Therefore (Weierstrass M-test) that infinite series converges absolutely and uniformly to a nice (continuous and even holomorphic!) function in DN(0). So we are done.