### Thursday, March 10 (Lecture #14)

Cauchy's Theorem (one version)
Suppose that f(z) is analytic in a simply connected domain and f´(z) is continuous. Then the integral around any closed curve in the domain of f(z) is 0.

For this of course I needed to discuss simply connected: no holes, or the inside of any closed curve in the domain is also entirely in the domain. See pp.107-8 in the text.

I used the complex version of Green's Theorem for this result. See p.107 of the text. I remarked that the assumption of continuity for the derivative is not necessary. This is verified by Goursat's proof (see section 2.3.1) which is witty but will not be discussed by me, because I can't really see another use of the proof technique.

Then I went through the proofs of important theorems (see pp.107-111), including the Cauchy integral formula. I will continue with some of section 2.4 next time.

On Thursday
I will lecture from 6:10 to 7:10, take a 10 minute break, and then lecture for another hour. I would like to do this for the next few Thursdays and catch up with the other section.

Darn snow!

### Thursday, March 3 (Lecture #13)

I gave
an exam but withheld the answers until I returned the graded exams. The grading procedure and some statistics for the exam are discussed here.

### Tuesday, March 1 (Lecture #12)

We lost a day due to some snow. This won't happen again.

The object of today's class is to again look at power series: SUMn=0infinityanzn. Such series have a radius of convergence, R. Inside the radius of convergence, the series will converge to a sum, S(z), depending on z. This sum is a continuous function inside the radius of convergence. But more is true.

• If SUMn=0infinityanzn has radius of convergence R, then SUMn=0infinityannzn-1 (the "term-wise" differentiated series) also has radius of convergence R.
• The function S(z) is complex differentiable, and inside the radius of convergence, S´(z)=SUMn=0infinityannzn-1.
This looks so routine and pleasant, and it certainly shouldn't surprise you. In fact, you should be a bit suspicious. About a century went by in mathematical history until people realized that power series were really special.

 THIS  IS  NOT  COMPLEX  ANALYSIS  SO  YOU  CAN  DISREGARD  IT! A Fourier series example Fourier series were basically introduced and used by Fourier in the 1820's to solve heat conduction problems. I handed out some pictures of Fourier series. I did not prove anything about these pictures, but I strongly suggested that they encouraged the belief that you can not always differentiate series of functions, term-by-term. Proving that the original Fourier series in these examples does converge to a continuous function is not difficult. Proving that the function is not differentiable and that the differentiated series does not generally converge would be much more difficut, and is something done in more advanced courses. But, please: the pictures should encourage you to be a bit careful.

I gave a good proof of the second fact above, that S(z) is differentiable, and that its derivative is the term-wise differentiated series. This is a technical proof, probably the most complicated to be done in the whole course. I did want to convince people that the result was true. The proof in the textbook actually is more-or-less the same, perhaps reordered in ways that I don't find so appealing. But basically you constantly compare to geometric series, you use the fact that [{S(z+h)-S(z)}/h]-S´(z) can be written nicely using the Binomial Theorem, and that the binomial coefficients are positive, so that things work well with the triangle inequality and rearrangement of series.

I did not give a proof or much of a discussion of the first claim, that the differentiated series had the same radius of convergence. The textbook's version is good, I think. Basically the assertion is that putting n in front of the terms of a power series doesn't much change how it converges (think of the ratio test!)

We can now apply our results again and again, to see that the sum of a power series can be differentiated as often as we want. And we can also see that if S(z)=SUMn=0infinityanzn, then by repeatedly differentiating and letting z be 0, the coefficients an must be Sn(0)/n!. So the power series is actually the Taylor series of its sum inside the radius of convergence.

Much more surprising things will occur!

### Tuesday, February 22 (Lecture #11)

So here is what I hope to do this evening:
• A chain rule computation from section 1.6 which leads to a neat result about evaluation of line integrals. I should have done this last time but I forgot.
• An example of pathology from the real calculus.
• Verification of the initial power series results, and discussion of some examples.
Suppose that C(t) is a parameterized smooth curve in the plane. That means C(t)=x(t)+iy(t) for t in some interval [a,b] of R, and both x(t) and y(t) are differentiable functions of t. Also suppose that f(z) is analytic in some domain containing the curve. Therefore the composition f(C(t)) is defined. What is the derivative of f(C(t))? Since this is a composition, any result is called the Chain Rule but let's see: f(C(t+a))=f(x(t+a)+iy(t+a))=f(x(t)+x´(t)a+HOT+i[y(t)+y´(t)a+HOT])=f(x(t)+iy(t)+h) where h=x´(t)a+HOT+i[y´(t)a+HOT]) so that, since f(z) is analytic, this is f(x(t)+iy(t))+f´(x(t)+iy(t))h+HOT=f(x(t)+iy(tt)) +f´(x(t)+iy(t))[x´(t)+iy´(t)]a+HOT. "Therefore" the derivative of f(C(t)) is f´(z)[x´(t)+iy´(t)]. The reason for the quotes around the word derivative is that there are lots of details which need to be verified. But let me show you how this can be used to evaluate some line integrals. It is basically the Fundamental Theorem of Calculus.

If I know that F(z) is analytic, and that F´(z)=f(z), and I have a curve C(t), for t in [a,b], then Cf(z) dz=F(C(b))-F(C(a)). That is, the line integral of the derivative is the value at the END minus the value at the start.

So now we have another method to evaluate line integrals. Start with C f(z) dz. IFwe know

• f(z) is the derivative of an analytic function, F(z).
• The domains of f(z) and F(z) include the curve, C.
THEN Cf(z) dz=F(END of C)-F(START of C).

For example, suppose f(z)=z5. This is an entire function, and I know it is the derivative of the entire function F(z)=[1/6]z6. Suppose C is any curve that starts from 2-i and ends at -1-i. Then Cz5 dz=[1/6](2-i)6-[1/6](-1-i)6.

You should not suppose, though, that the restrictions about domain of f(z) and F(z) including C can be disregarded. We know that unit circle(1/z) dz=2Pi i. Since if we did have an f(z)/F(z) pair where f(z)=1/z, the integral over a closed curve would be 0, we can conclude that there is no function analytic in a domain which includes the unit circle whose derivative in that domain is 1/z. "Locally" on pieces of the circle (in fact, any piece of the circle which discards one point) we can in fact find such an antiderivative. The function Log(z) is an antiderivative of 1/z, where we discard -1 from the circle.

THIS  IS  NOT  COMPLEX  ANALYSIS  SO  YOU  CAN  DISREGARD  IT!
 An example having little to do directly with this course, but it is an example which you should keep in mind because it will help you appreciate the special and rare beauty of the results of this course.
Consider the real function of a real variable defined in a piecewise fashion by

```      { exp(-1/x) for x>0
f(x)= {
{ 0 for x<=0.```
To the right is a Maple graph of this function. On the left, of course we have a half line. Since the real exponential function of negative reals is between 0 and 1, the graph on the right must be between y=0 and y=1 (if x>0 it can't equal 0 since the exponential function is never 0, and it can't equal 1 because 1/x can never be 0). Of course, on the left all of the derivatives are 0. What about on the right, where x>0? There f´(x)=e-1/x[1/x2]. Since this is always positive, we know that f increases. In fact, the right-hand limit at 0 of e-1/x "is" enegative infinity so it is 0. The function is continuous at 0, and therefore continuous everywhere. As x-->infinity, e-1/x-->e0=1. The Maple graph is totally justified.

I would like to investigate finer properties of the function f(x). In particular, I would like to look at its differentiability at 0. The left-hand derivative of f(x) at 0 certainly exists and is 0. What about the right-hand derivative? For x>0, [{f(x)-f(0)}/{x-0}] is the same as (1/x)e-1/x. What happens as x-->0+ to this? Well, this is a Calc 2 exercise in L'Hopital's rule. I guess I would do it this way:
As x-->0+, then w=1/x-->+infinity. And (1/x)e-1/x becomes w e-w=[w/ew]. This satisfies the criteria for L'H, so its limit is the same as the limit (if it exists) of [1/ew] as w-->+infinity. That's certainly 0. In fact, as I hope most of you remember, exponential growth is much faster than any polynomial growth.

I now know that f(x) is differentiable, and that its derivative is

```       { (1/x2)exp(-1/x) for x>0
f´(x)= {
{ 0 for x<=0.```

Let me do one more step. f´´(x) must exist away from 0 (I can use the product and Chain Rules on the right). What about f´´(0)? From the left again we see that the appropriate limit is 0. From the right, we officially need to consider [{f´(x)-f´(0)}/{x-0}] as x-->0+. Let's insert the formula for f´(x) and the value (just 0) of f´(0), so that we see we need to consider (1/x3)exp(-1/x) as x-->0+. But (w=1/x) this is w3/ew as w-->infinity, and the limit is 0.

An official proof needs induction, etc., but I don't need no steenkin' official proof ... In fact, this function f(x) is infinitely flat at the origin. It and all of its derivatives exist at 0, and the values of the function and all of its derivatives at 0 are 0.

Suppose we wanted a power series which would equal f(x), a power series centered at 0. Well, if the series work the way they should (and they do) then the power series will be a Taylor series for f(x) at 0. But all of the coefficients will be 0. So here's a function which can't be equal to a power series in any interval centered at 0, even though we can differentiate the function as much as we want.

Nothing that bad occurs in complex analysis.

Elementary theory of power series
A power series centered at z0 is a sum of the form SUMn=0infinityan(z-z0n where all of the an's are complex numbers. Most of the time I will write power series centered at z0=0 because I am lazy. We say that the power series converges if the sequence of partial sums has a limit. If the power series converges absolutely (that is, the series SUMn=0infinity|an| |z|n, a sequence of non-negative numbers) then the series will converge. I discussed this in lecture #4.

Suppose SUMn=0infinityanzn converges for some non-zero z (any power series will always converge at its center, here when z=0). Then the pieces of the series can't get large: that is, the max of |an| |z|n is finite (or else the successive partial sums, which should converge, will take bigger and bigger "steps" from each other. Suppose this max is some number V. Look at what happens if we take a complex number w with |w|<|z|. Then |SUMn=0infinityanwn|<=SUMn=0infinity|an| |w|n<SUMn=0infinity|an| |z|n[|w|/|z|]n. Now look closely.
|w|/|z| is some number r with r<1. We concluded that the other "stuff", |an| |z|n, is bounded by some big number, V. So therefore we are left with the overestimate, SUMn=0infinityV rn with r a positive real number which is less than 1. But this series converges. So the original series, SUMn=0infinityanwn, converges absolutely, and therefore converges.

Now we need to think about the situation a bit. As soon as you know the power series converges at a complex number, the series will converge absolutely at any complex number closer to its center. So the situation is like this: either the series converges for all complex numbers (perfectly possible, consider SUMn=0infinity[zn/n!]) or it converges for no complex number except for the center (perfectly possible, consider SUMn=0infinityn! zn]) or there are some complex numbers for which it diverges and some for which it converges. But then the computation we just made shows that if we consider the modulus of every complex number for which the series converges, then a circle (centered at the center of the power series) of such modulus must have all convergent numbers in it. Take the union of all those circles. The result will be a circle, and its radius is the radius of convergence.

Continuity of the sum
Look at SUMn=0infinityanzn and suppose it has radius of convergence R>0. Here "infinity" is a possibility and infinity will be >0. Let's call the sum of this series, S(z). If |z|<R, the series converges absolutely. Fix some z inside the circle of convergence. Take some number r>0 so that |z|+r is still less than R (possible -- you could take r=(1/2)[R-|z|], for example). Then SUMn=0infinityan(z+w)n will converge absolutely for all w's with |w|<r. Not only that, but if we compare "infinite tails" of series, we also know that SUMn=Ninfinity|an| |z+w|n<SUMn=Ninfinity|an|(|z|+r|)n. Since the series SUMn=0infinity|an|(|z|+r|)n converges, I can choose N so that the tail, SUMn=Ninfinity|an|(|z|+r|)n is less than epsilon. But then look at what we have left out: SUMn=0N-1an(z+w)n. This is a polynomial. If w-->0 this polynomial will eventually differ by epsilon from its value at z, because polynomials are continuous.

Hey, I've just shown you that by neglecting an appropriate infinite tail, you can make the terms of a power series vary continuously. That means that the sum of the series, S(z), will be a continuous function inside its radius of convergence. This is dandy.

### Thursday, February 17 (Lecture #10)

Motto: I won't bother you with PROOFS unless both the method and the content will be very useful, even needed, in the remainder of the courss!

We reviewed and tried to understand what differentiability means
What's complex differentiability? Well, we want the limit

```      f(z+h)-f(z)
lim -------------
h-->0      h```
to exist, and this limit will be called f´(z). Actually, the division sign makes this definition a bit difficult to work with. It is easier to try to "unroll" the definition in the following fashion:
``` f(z+h)-f(z)
------------- = f´(z)+Err(f,z,h)
h```
where Err(f,z,h) is an error term which depends on f and z and h, and all I know about it is that Err(f,z,h)-->0 as h-->0. We can then rewrite the equation and eliminate the division and the minus sign:
`f(z+h)=f(z)+f´(z)h+Err(f,z,h)h`
Here the term on the left-hand side is a perturbed value of f, and the terms on the right-hand side describe a way of looking at the effects of the perturbation. The first piece on the right is the "old" value of f, then comes a first-order part, and then comes a part which is higher than first-order. And to me that's what complex differentiability means. (This way of looking at the derivative also makes other things, like verifying the Chain Rule) much easier.

Compatibility conditions
I remarked that linear algebra tries to study collections of linear equations systematically (a little math joke in that phrase, sorry). So maybe one studies M equations in N unknowns. There are only a few rather rough things one can learn in general. For example,
a homogeneous system of M equations in N unknowns with more unknowns than equations (N>M) must always have non-trivial (non-zero) solutions.
And here's another:
If you have more equations than unknowns (M>N), in general the equations can't be solved: they don't have solutions unless certain additional conditions ("compatibility conditions") are satisfied.

For example,

```2x+3y=A
x+2y=B
5x+4y=C```
has no solution unless C+7B-6A=0 (I think this is correct but I make lots of mistakes with linear algebra.)

And now back to complex analysis ...
Last time we tried to solve

```vx=-2y
vy=4x3```
and couldn't get a solution. Well, we (sort of) have two linear equations in v, and maybe there should be a compatibility condition. There should be, and here it is: the derivative of v with respect to x and then with respect to y should equal the derivative of v with respect to y and then with respect to x. Certainly there are some very weird functions where this doesn't happen, but polynomials are nice. I quote:

Fact from calc 3 If a function has is continuous second partial derivatives, then its mixed second partial derivatives must be equal. The proof uses the Mean Value Theorem.

So look at the Cauchy-Riemann equations:
ux=vy
uy=-vx
Differentiate the first with respect to x and the second with respect to y. We get
uxx=vyx
uyy=-vxy
Since vyx=vxy I know that uxx=-uyy, usually written
uxx+uyy=0
This is Laplace's equation, and the stuff on the left-hand side is called the Laplacian of u. Solutions of Laplace's equation are called harmonic functions, and harmonic functions come up all over the place. Notice, please, that x4+y2 is not harmonic, so it cannot be the real part of a complex differentiable function.

Harmonic; conjugate harmonic
In fact, if f(z) is complex differentiable, with f(z)=u(x,y)+iv(x,y), then both u(x,y) and v(x,y) are harmonic functions. v(x,y) is called a harmonic conjugate of u(x,y). It is interesting to realize that a "random" polynomial in x and y is very unlikely to be harmonic. In fact, a goal of the course is to justify interest in these rather rare (?) objects.

The converse to CR
Suppose that u(x,y) and v(x,y) have continuous first partial derivatives. Further suppose that the Cauchy-Riemann equations are correct at a point (x,y):
ux=vy
uy=-vx
Then f(z)=u(x,y)+iv(x,y) is complex differentiable at z=x+iy.

I actually attempted to verify this. I know another fact from calc 3. I h=a+ib (think of a and b as very small):
u(x+a,y+b)=u(x,y)+ux(x,y)a+uy(x,y)b+Err1(u,x,y,a,b)(|a|+|b|)
v(x+a,y+b)=v(x,y)+vx(x,y)a+vy(x,y)b+Err2(v,x,y,a,b)(|a|+|b|)
and Err1(u,x,y,a,b) and Err2(v,x,y,a,b) both -->0 as a and b both -->0. Now let's look at f(z+h)=u(x+a,y+b)+iv(x+a,y+b). This is
[u(x,y)+ux(x,y)a+uy(x,y)b+Err1(u,x,y,a,b)(|a|+|b|)]+i[v(x,y)+vx(x,y)a+vy(x,y)b+Err2(v,x,y,a,b)(|a|+|b|)]
Let us reorganize this stuff:
u(x,y)+iv(x,y) (this is f(z))
{ux(x,y)a+uy(x,y)b}+i{vx(x,y)a+vy(x,y)b}
Now if A=ux=vy and B=uy=-vx, this becomes
{Aa+Bb}+i{-Ba+Ab} and we can recognize this as (A-iB)·(a+ib). If we think about this, then f´(z) should be A-iB and a+ib is h, so we have f´(z)h.

What's left over? This:
{Err1(u,x,y,a,b)(|a|+|b|)}+i{Err2(v,x,y,a,b)(|a|+|b|)}
I would like this to be Err(f,z,h)h. Remember that all the Err terms go to 0 as a+ib-->0. What's left to consider? Well, |h|=sqrt(a2+b2) compared to |a|+|b|. So I'll compare them.

Inequalities
Surely sqrt(a2+b2)<=|a|+|b| because if you square both sides, you get a2+b2<=|a|2+2|a| |b|+|b|2. Now cancel a2 with |a|2 and b2 with |b|2 and we get 0<=2|a| |b| which is certainly true.

Since sqrt(a2+b2)<=|a|+|b|, I know that if |a|+|b|-->0, then |a+ib|-->0. What about the converse?

Look at |a|+|b|<=2sqrt(a2+b2). I think I tried 2 in class. Now square and see if we can clarify things:
|a|2+2|a| |b|+|b|2<=4(a2+b2). We can cancel some terms, and get
2|a| |b|<=3(a2+b2).
But 2|a| |b|<=a2+b2 is true, because 0<=a2-2|a| |b|+b2=(|a|-|b|)2 is certainly true.

Since |a|+|b|<=2sqrt(a2+b2) I now know that if |h|=|a+ib|-->0, then |a|+|b|-->0. Wow! We are done.

If f(z)=|z|2=x2+y2, at which points is f(z) complex differentiable? I just need to check the Cauchy-Riemann equations when u(x,y)=x2+y2 and v(x,y)=0. Then:
ux=vy becomes 2x=0.
uy=-vx becomes 2y=0.
So |z|2 is complex differentiable only at the point z=0.

More vocabulary: analyticity (holomorphicity?)
It turns out that complex differentiability is not quite the correct definition. Here's the correct definition:
A function f(z) defined on a connected open set is (complex) analytic if f(z) is complex differentiable at every point of the set.
This definition allows a neat theory. I remark that the word holomorphic is used instead of analytic sometimes.

f´=0
Here's a first fact: if f(z) is analytic and if f´(z)=0 for all z, then f(z) is constant.

Notice that if the domain of f(z) is not a connected open set, then this result is false, because you could have f(z) equal to one constant in one "blob" and another constant in another "blob".

Well, if f´(z)=0 for all z, then then Cauchy-Riemann equations, combined with the observation that f´(z)=ux+ivx show that ux=0 and uy=0 and vx=0 and vy=0 for all x and y. This means that u and v must be constant on horizontal lines (because of the x derivatives) and on vertical lines (because of the y derivatives). But in a connected open set, any two points can be connected, not only by line segments, but also by line segments which are made up entirely of horizontal and vertical line segments. So the values at any two points must be equal because of the previous observations about vertical and horizontal lines.

Other conditions
If f(z) is analytic and has only real values, then f(z) is constant.
Why? Because then v is 0, and so vx=0 always and vy=0 always, then the Cauchy-Riemann equations tell you that f´(z) is 0 always. So f(z) must be constant.

If f(z) is analytic and has only imaginary values, then f(z) is constant.
Why? Interchange u and v in what we just did.

If f(z)=u(x,y)+iv(x,y) where u and v are real, and if u2+v2 is constant, then f(z) is constant.
Look at the clever proof on p.82 of the text. Here is another proof, not so clever: if u2+v2=C, then d/dx this equation. The result is
2u ux+2v vx=0.
Of course we can do the same thing with d/dy:
2u uy+2v vy=0.
Now use the Cauchy-Riemann equations. This will get us:
2u ux-2v uy=0.
2u uy+2v ux=0.
Look closely at this system of equations:

```(2u -2v)  ( ux )   (0)
(      )  (    ) = ( )
(2v  2u)   (0)```
This is a system of two homogeneous linear equations in the two "unknowns", ux and uy. From linear algebra we know that such a system always has the trivial solution (where the two unknowns are 0). This would be fine, because then we could conclude that u is constant, and then (CR) v is constant. But does the system ever have non-trivial solutions? Yes, when the determinant of the coefficient matrix is non-zero. Here that determinant is 4u2+4b2. This quantity we are assuming is constant. Hey, if it is the zero constant, then both u and v must be 0, so f is constant. If it is not the zero constant, then linear algebra tells us that ux=0 and uy=0. Therefore, as before, u is constant and so must v be.

Why? Geometry for the future
The previous results are actually reflections of the following truth: if an analytic function is not a constant function, then the image of the domain of the analytic function will always be an open subset of C. This is not obvious, and nothing like it is true in elementary real analysis.

Exam warning!
I'll give an exam a week from Tuesday. The exam day is therefore Tuesday, March 1.

### Tuesday, February 15 (Lecture #9)

This is very sketchy, since I am pressed for time. Chapters 2 and 3 are the heart of the subject. I hope we can get through as much of them as possible. I promised to follow the text closely. At least for a while, just a while.

The definition of complex differentiability
Here we take advantage of the fact that C is a field and just copy the definition from R. The limit of the difference quotient, if it exists, is again called the derivative.

Examples
zn is complex differentiable everywhere (it is entire). Constants are entire. Sums, products, and quotients (this where defined) of appropriate functions are complex differentiable. The chain rule holds. (I recommend that you try to verify the chain rule.)

Exp
I attempted to, and finally succeeded in, with students' valiant help, verifying that the exponential function was complex differentiable. This was conceived as a lesson, not just of the subject matter, but also of the difficulty of reading a math textbook. There is a need for intensity and engagement, a need for active participation. That is what I hope students got from this episode.

Counterexample
The function f(z)=Conjugate(z)=x-iy is not complex differentiable at any point! This we showed by first taking h=a+ib and making a=0 and b-->0, and then making b=0 and a-->0 in the definition. The results should be the same, since these difference quotients should approach the same limit. But the results are two distinct constants, and never equal.

A question
I didn't raise this is class, but now a "sophisticate" might ask a question: is the definition correct? Here I use the word in a uncommon way which I will explain. If nice functions (and Conjugate(z) is fairly nice: it is linear) don't satisfy the definition, maybe the definition is too strict for use. We'd like to have a rich theory, covering many examples with many applications. Well, it turns out that there will be many examples and applications, although that is not clear now. And Conjugate(z) is real linear, but it is not complex linear, so we shouldn't worry too much about it.

The Cauchy-Riemann equations
Suppose f(z)=f(x+iy)=u(x,y)+iv(x,y), where u(x,y) and v(x,y) are real differentiable functions. If f(z) is complex differentiable, then both first partial derivatives of u(x,y) and v(x,y) exist, and
ux=vy
uy=-vx

This pair of equations is called the Cauchy-Riemann equations. We proved this by imitating our two limits which we used to investigate the function Conjugate(z).

From the real part
I suggested that we try the following exercise: if u(x,y)=3y+7xy, then find a corresponding v(x,y) and see if we can write a corresponding f(z) in terms of z=x+iy. I suggested that we try the following line of investigation:

1. Compute ux and uy.
2. Use the Cauchy-Riemann equations to get vx and vy.
3. Reconstruct v from its first partial derivatives.
4. Try to use algebra to recognize f(z).
We did exactly this, and we were successful. Let me try to show you:
1. ux=7y and uy=3+7x.
2. vx=-3-7x and vy=7y.
3. Integrating the first equation:
v(x,y)=-3x-(7/2)x2+STUFF1(y). Integrating the second equation:
v(x,y)=(7/2)y2+STUFF1(x).
Now we compare the two descriptions of v(x,y), and learn that v(x,y)=-3x-(7/2)x2+(7/2)y2+possible constant. The various STUFF's of course conceal parts of each description. This process was studied in calculus and also in some aspects of differential equations. It reconstructs a function from its gradient. Here I'll set the possible constant to 0, but warn that at some times this is not appropriate.
4. Now u(x,y)+iv(x,y) is 3y+7xy+i[3x-(7/2)x2+(7/2)y2]. After scrambling (?) with some algebra, we saw that this corresponded to f(z)=f(x+iy)=-3iz+(7/2)iz2.
Another example?
This was vicious. I suggested we try another polynmial for u(x,y): just x4+y2. We followed the same outline, but ran into some difficulty in step 3. What's happening?

HOMEWORK
the previously assigned problems. Some of them are difficult.

### Thursday, February 10 (Lecture #8)

We have so many things to celebrate:
• Mardi Gras!
• The Chinese New Year, the year of the Rooster!!
• Valentine's Day!!!

 Cu(z) dz

Here are some facts we need to know about line integrals.

The definition

In this course, line integrals will be defined for piecewise differentiable parameterized curves, C, and for u(z)'s which are continuous on the curve's image. I noted that it is possible to define the integrals using a sort of Riemann sum procedure, by sampling the function u(z) along the curve and multiplying by the lengths of the edges of a corresponding polygonal approximation to the curve. Then we would take the limit as the mesh (the length of the longest side in the polygon) goes to 0. Then, using the Mean Value Theorem, we'd get the same result for our collection of curves. Also, line integrals frequently are associated with actual physical quantities such as work, flux, etc. These quantities are most easily recognized using the Riemann sum definition.

Independence of parameterization

One consequence of our definition of line integral is that we should prove or verify that value of the line integral does not depend upon parameterization. I gave an example last time which showed this. That is very different from proving that equality holds in all cases, as I hope you realize.

Linearity in the integrand, u(z).

If k is a constant, and u1 and u2 are eligible integrands, then Cku1(z)+u2(z) dz=kCu1(z) dz+Cu2(z) dz.
This certainly looks o.k.: it looks like something we've been accustomed to for years.

"Linearity" (as much as you'll let me!) in the curve, C.

What the heck does this mean? Well, for one thing, when we can add curves, the corresponding integrals add. So if C1 and C2 are curves with the correct HEAD/TAIL relationship, then C1+C2u(z) dz=C1u(z) dz+C2u(z) dz.
Reversals work nicely, also: -Cu(z) dz=-Cu(z) dz.
These things are very convenient in practice, and I will probably use them (I did later in class!) without even a proper citation. They are just so familar and neat -- after they have been learned!
People have extended "linearity in C" in various ways whenever it has been useful. We may need some of this later.

Estimating the integral, with weak and strong justification.

Let me begin with an example, which I should have done in class. I need to choose the darn example very carefully, to have a hope of computing it "by hand".

Let's compute 01t2+2ti dt. (Why do I need to be so careful? Wait, wait.) I can antidifferentiate polynomials. This works out to (1/3)+i. The modulus of this complex number answer is sqrt(10)/3, which is about 1.054092553. Therefore the modulus of the integral is about 1.054092553.

Much of mathematics is not very subtle. A great deal consists of trying to find out if changing the order of doing things changes the answer. Here the "things" are integration and taking modulus. Onward!

The integrand is t2+2ti, and its modulus is sqrt(t4+4t2) which is t sqrt(t2+4). Hah! The alert Calc 1 student will now see that this is a function whose antiderivative can be easily written: it is (1/3)[t2+4]3/2. So I can compute the integral of the modulus of this function exactly, and the answer is (5/3)sqrt(5)-(8/3), A numerical approximation of this is 1.060113297.

These computations have been facilitated by my friend, Maple.

What have I done? I have computed 01|g(t)| dt and |01g(t) dt|. The first is approximately 1.060113297 and the second is approximately 1.054092553.

Now let me give you a heuristic reason for a more general assertion. Of course, heuristic means

```
1. allowing or assisting to discover.
2. [Computing] proceeding to a solution by trial and error.```

Suppose g(t) is a complex-valued function defined for t's in the interval [a,b]. Now abg(t) dt is, to me, really a BIG (Riemann) sum: SUMj=1Ng(tj)(tj-tj-1). How large can this be? Since the complex numbers can't be ordered, I use modulus instead, and ask how large can |SUMj=1Ng(tj)(tj-tj-1)| be? The Triangle Inequality applies here, and an overestimate is SUMj=1N|g(tj)|(tj-tj-1). I don't need the | | around (tj-tj-1) because (tj-tj-1) is a positive real number. But SUMj=1N|g(tj)|(tj-tj-1) is a Riemann sum for another integral: ab|g(t)| dt. All of this leads me to believe that
|abg(t) dt|<=ab|g(t)| dt.

This inequality is correct. We have seen one numerical agreement. The discussion using Riemann sums is quite difficult to convert into a proof. There is a very nice, rather tricky, device to prove this result on p.61 of the text, which you might want to look at.

The ML inequality

The previous result will be used a great deal in this course but it will be applied to line integrals. If z(t) is the parameterization for the curve C, when t is in the interval [a,b], and if z(t) is differentiable (both x(t) and y(t) are differentiable), then Cu(z) dz=t=at=bu(z(t))z´(t)dt. The inequality we just developed then states:
|t=at=bu(z(t))z´(t)dt|<=|u(z(t))| |z´(t)| dt. The next step depends on clever recognition of a formula and knowing what the aim of the manipulation is. As I described in class, the aim is sort of a decoupling (as things are called in physics) or a disconnection between the curve, C, and the integrand, u(z). We will necessarily lose some precise information but in applications it will turn out that the estimates we can make are the important consequences. So consider: suppose that we know, from somewhere, that the function u(z) has some overestimate in modulus on the curve, C. That is, there is a non-negative real number M so that |u(z)|<=M for all z on C. Then |u(z(t))| |z´(t)| dt<=M |z´(t)| dt. But,hey!, |z´(t)| is sqrt[{x´(t)}2+{y´(t)}2 and maybe you recognize the integral: it defines (or computes, depending on your choice of textbook!) the length of the curve. Whenever possible in this course, I will label the length of C with the letter L. So:
|Cu(z) dz|<=ML.

This inequality will be used about 370 times by the instructor and by students in the course. Let me show you some examples, which will be silly because they are out of context.

ML example #1
I considered the curve SR, which I drew on the board. It was a semicircle of radius R in the upper halfplane, centered at the origin. The integrand was 1/z2. I wanted to investigate the asymptotics of SRdz as R-->infinity. In this case the integral can be evaluated exactly, but I wanted ot use the ML inequality. So:
|SRdz|<=ML. Here L is half the circumference of a circle of radius R, and that's (Pi)R. What about M? Well, on the curve, |z|=R. so |1/z2|=1/R2. Therefore the modulus of the integral is at most (Pi)R(1/R2). The R2 on the "bottom" dominates as R-->infinity, so certainly this integral must-->0.

ML example #2
I used the same SR, but now I wanted a more interesting u(z). I think I tried something like u(z)=1/[z4+5z+4]. The exact numbers don't matter: what does matter is the estimation. Again, to understand the asymptotic behavior as R-->infinity, I used ML, and L was PiR as before. Since u(z) is a bit more complicated, I had to use the techniques introduced in Lecture #2 to estimate it. We did some examples then which you may review.

Suppose |z|>10 (mostly picked at random). Then |z4+5z+4|>=|z|4-|5z+4|. But how large can |5z+4| be? Well, on SR where |z|=R>10, I know that |5z+4|<=5|z|+4<=2|z|2. But 2<=(1/2)|z|2 for these z's, and therefore |5z+4|<=(1/2)|z|4, and |z4+5z+4|>=|z|4-(1/2)|z|4=(1/2)|z|4. There is a great deal of sloppiness and waste in here: the inequalities are likely to be very far from equalities but I just want to get an answer.

Now I know for z on SR, |u(z)|<=1/[(1/2)|z|4] so that I can take 2/R4 as my M. The modulus of the integral is therefore bounded by (Pi)R[2/R4] and this surely -->0 as R-->infinity. In this case, I don't it is at all possible to compute exact values of the integral, but the estimation is very easy.

Green's Theorem

So here is what it's about: line integrals over closed curves. I will compute a line integral over a closed curve. Indeed, the closed curve I will use will be the boundary of the unit square (both x and y in the interval [0,1]), and the integrand will be the innocent function, x. So let me compute Bdry squarex dz. I will compute it using four parameterizations and the whole integral will be the sum of four pieces. There are several reasonable ways to parameterize some of these pieces and some of these ways are different from those I choose.
 t is in [0,1] z=t x=Re(z)=t dz=dt 01t dt=1/2 t is in [0,1] z=it+1 dz=i dt x=Re(z)=1 01i dt=i/2 t is in [0,1] z=-t+1+i dz=-dt x=Re(z)=-t+1 01(-t+1) -dt=-1/2 t is in [0,1] z=i-it dz=-i dt x=Re(z)=0 010 i dt=0

Therefore the integral is 1/2+i/2-1/2+0=i/2. This isn't 0: what a disappointment. Sometimes people misremember Green's Theorem and it might sound like the line integral over almost any closed curve of almost anything is 0. That is far from true.

A sort of statement of Green's Theorem
Well, let me quote the result in the text:
Gammaf(z) dz=i Omega[fx+ify]/2 dxdy.
Here Gamma is the whole boundary curve of a region Omega, oriented so that the region is the left of the curve Gamma as we more along Gamma. This is a Green's Theorem changed so that we can use it in complex analysis. The fx and fy are the partial derivatives with respect to x and y. We can check it for the case we just computed. The double integral's integrand, [fx+ify]/2, is 1/2, and the area of the square is 1, and the i out in front makes everything come out correctly.

A sort of proof
If we wanted to check more generally (whisper the word, proof) then ... well, since both sides are linear we could start with looking at i Omega[fx]/2 dxdy. We can use the Fundamental Theorem of Calculus to convert the partial with respect to x into f(1,y)-f(0,y) because the integral over the unit square has both x and y going from 0 to 1. Then the double integral becomes iy=0y=1[f(1,y)-f(0,y)]/2 dy. This then breaks up (and I'll change y to t): t=0t=1f(1,t) i dt-t=0t=1f(0,t) i dt. The first of these integrals is the integral of f(z) over the right-hand side of the boundary of the unit square, going up, and the second is the integral of f(z) over the left-hand side of the unit square, going down. The other two sides are gotten from the fy term in a similar manner.

Now look at the important integral ...
We previous computed the most important line integral. Let's try it again. Here f(z)=1/z=1/(x+iy), and the region, Omega, is the circle of radius R centered at the origin. The corresponding double integral has as integrand [fx+ify]/2. I'll use the Chain Rule and get fx(z)=-[1/z2]1 and fy(z)=-[1/z2]i. Therefore the integrand [fx+ify]/2= -[1/z2]1+i-[1/z2]i=0 if we recognize that i2 is -1. So use of Green's Theorem in this most important example yields 0. This is distressing since we previously computed it and got 2(Pi)i. What is wrong? The use of Green's Theorem here is not valid.

Better state Green's Theorem more precisely
The functions in Green's Theorem need to be differentiable ("smooth") at every point on and inside the curve. Even one "lousy" point (a singularity) makes it possible for the theorem to be not correct. This is subtle, and you should reinforce it by thinking, again and again, that 2(Pi)i is not the same as 0.

Now look at what "inside" means using a weird computation
Let's try to compute B(1/z) dz. Here B is the orinted boundary of the square whose sides are parallel to the coordinate axes, centered at 0, and whose side length is 4. A direct computation by parameterization of this integration is (barely) possible. I've done it and it is very tedious and very unpleasant. We can be, instead, a bit devious, and compute the integral quite easily.

As indicated in the picture, the curve A will be the positively orineted unit circle with center 0. The region R is the subset of C between these curves. Notice that the correctly oriented boundary of R is B-A. The minus sign is very necessary. We can apply Green's Theorem to this R with 1/z, because the only singularity, 0, is outside of R and A and B. But we saw that the integrand of the double integral for this function is 0 before (when we computed [fx+ify] for f(z)=1/z. Therefore the double integral term in Green's Theorem is 0. What's the other side? It is B-A1/z dz. Thus B-A1/z dz=0, or B1/z dz-A1/z dz=0 or B1/z dz=A1/z dz. But the A integral is 2(Pi)i, and therefore the B integral is 2(Pi)i. This is quite tricky, and I will essentially use ideas like this frequently in the remainder of the course.

What's inside?
The logic of the preceding computation can be generalized, so that if C is a closed curve (oriented positively) and p is a point not on C, the C[1/(z-p)] dz is either 0 or 2(Pi)i. In the first case, p is outside of C and in the second case, p is inside C. Maybe this is how to figure out what inside and outside are!

### Tuesday, February 8 (Lecture #7)

Here is an interesting and startling experiment which I tried in class. Imagine this story: a Mad Scientist, Count Ogus von Brandenburg of Bavaria, learned some complex analysis in 1830. He then devoted the remainder of his life and the resources of his principality to constructing mechanical devices which imitated complex mappings. In particular, he fell in love with the exponential mapping. He constructed a linkage which would allow a point in the range to be attached to several points in the domain, and the domain points would all be mapped to the given range point by the exponential mapping. A sort of drawing of part of the count's mechanism is shown above. I tried to simulate the mechanism of the crazy count in class.

Four of the wonderful students in the class volunteered (always better than the truth, which is closer to were dragooned) to draw some pictures with me. I only show the results of three of the students. I set up the exponential function mapping from a domain copy of C to a range copy of C. Then I stood at the range copy, with my chalk at w=1, which I also called HOME. I asked each of the students, each with their individually colored chalk, to choose a value of log(1). Since this is a complex analysis course, there were certainly enough distinct values to choose from. Each student could have their own value of log(1). In my head (sometimes small, sometimes large, frequently empty) I saw a sort of mechanical linkage reaching backwards from my 1 to their choices of log(1). The linkage would "automatically" respond to any changes of position in my location by moving to nearby places whose value of exp would be where my chalk moved to. Sigh. There's no pathology in this course: things will be continuous, things will move nicely, etc. Much more subtle problems will be met.

 The first trip Then I carefully and slowly (well, almost) moved my chalk from HOME in a simple closed curve. The curve was made up of circular arcs centered at the origin and pieces of rays which started at the origin. I used these curves because last time we had previously computed how vertical and horizontal lines in the domain were mapped. The inverse images of my "trip" from HOME to HOME were carefully drawn. They were congruent, because exp is 2Pii periodic. Each started from each student's choice of log(1) and returned to that choice.

 The second trip Well, now I left HOME and traveled in a circle around 0, and returned to HOME. The students attempted to imitate the Mad Count's device, and there was some controversy. If we tried to make everything continuous, then, even though my path in the range was a simple closed curve, each student draw a path going from one choice of log(1) to another choice of log(1). This is very, very mysterious. I drew a few more paths in the range and mentioned what would happen in the domain, and said we would need to explain all this is great detail later. This is one of my goals in the course.

Right now, no sine, no cosine!
Section 1.5 of the text discusses sine and cosine. I will save these functions until later. I want to get into the "core" of the course (and the subject) as soon as possible.

This stuff, which must be understood by 403 students, should have been covered in any calc 3 course, and is discussed in section 1.6 of the text.

PhraseProper definitionPossible picture
Curve A curve is a function whose domain is an interval, [a,b], in R, and whose range is in C. If C(t) is the curve, we could write C(t)=x(t)+iy(t). In this course, these functions will generally be at least continuous.
Smooth (differentiable) curve C(t)=x(t)+iy(t) should be differentiable, so that x(t) and y(t) are differentiable. Such a curve has a nice tangent vector at every point.
Piecewise smooth curve The curve C(t) is certainly continuous. But additionally, we can divide the interval [a,b] into a finite number of pieces, and in each of these subintervals C(t) is smooth. This will allow us to take care of curves with finite numbers of corners, for example.
Simple curve A simple curve is one which does not intersect itself except maybe ... at the ends (see the next plus one definition). So if the domain of C(t) is [a,b], and if a<s<t<b, then C(s) is not equal to C(t), and C(a) is not C(s) and C(b) is not C(s).
Closed curve "Head=Tail" -- that is, if the domain of C(t) is {a,b], then C(a)=C(b).
Simple closed curve A closed curve which is also simple. Here C(s)=C(t) with s<b implies either s=t or a=s and b=t.
Adding curves Suppose C1(t) is a curve with domain [a,b] and C2(t) is a curve with domain [c,d]. If C1(b)=C2(c) (that is, C1's "Head"=C2's "Tail", perhaps) then the curve C(t) whose domain is [a,b+d-c] and defined by C(t)=C1(t) if t<b and C(t)=C2(t-b+c) (I hope!) is called the sum of C1 and C2. We will write C=C1+C2.
Reversing curves If C(t) is a curve with domain [a,b], then the "reversal of C(t)", also called -C, will be the curve, let me temporarily call it D(t), whose domain will also be [a,b], defined by D(t)=C(b+a-t). -C is just C backwards. The picture is slightly weird (how should I have drawn it?) but I hope you get the idea.

I hope the darn formulas for C1+C2 and -C(t) are correct. I don't think I will ever be so formal again in this course. I will usually be rather casual. The curves in this course will all be piecewise differentiable, and, actually, they will turn out to be sums of line segments and circular arcs: HAH! to general theory!

Actually I should acknowledge that what we are defining are called parameterized curves. For example, the curves C(t)=t and C(t)=t2 (use [0,1] as domains in both cases) are really different, even though they "look" the same. It turns out that we will mostly be interested in line integrals and a remark below about the values of the integrals explains why we will not care about different parameterizations.

A few examples
 This curve is a straight line segment starting at -4i and going to -2i, followed by a quarter circle whose center seems to be 0, starting at -2i and going to 2. If I wanted to be totally strict, I guess I could define a C(t) on the interval from [0,2+(Pi/2)] Let's see: for t in [0,2), let C(t) be (t-4)i, and for i in [2,2+(Pi/2)], define C(t) to be 2ei(t-2-(Pi/2)). Some remarks Almost always in this course I will parameterize circles and circular arcs using the complex exponential. My first guess will always be CENTER+RADIUSei(stuff). Another remark: what I've just done is excessively precise and maybe slightly silly. I want to write curves in a way that is convenient. So this curve should be written as a sum of two curves, say C1+C2. Here C1(t)=ti where t is in the interval [-4Pi,-2Pi], and C2(t)0+2eit where t is in the interval [-Pi/2,0]. It is just easier to do this than to try to figure out the total length, how to combine and adjust parameterizations, etc. Here I will write the curve again as a sum, C1+C2. C1 seems to be a semicircle of radius 2 and center -1. It goes backwards. So C1(t)=-1+2ei([Pi]-t) where t is in the interval [0,Pi]. C2 is a straight line segment. This I know how to parameterize from thinking about convexity. C2(t)=t(1)+(1-t)(-3i) where t is in the interval [0,1]. A remark Parameterizing a circular arc in a clockwise fashion feels close to obscene to me. Almost always in this course our circles will be parameterized positively, which will mean counterclockwise.

Line integrals on curves
If C(t) is a differentiable curve and u(z) is continuous in some open set which includes the image of the parameterized curve C(t), then Cu(z) dz is t=at=bu(z(t))C´(t)dt (here C´(t)=x´(t)+iy´(t) and u(z(t)) is itself a complex number for each t).

Parameterized curves and line integrals
Two different parameterizations give the same values.
Example Consider u(z)=z4, and the two curves C1(t)=t and C2(t)=t2, with both domains equal [0,1]. Then:

• C1u(z) dz=01[t]41dt=1/5.
• C2u(z) dz=01[t2]42t1dt=inv012t9dt=2(1/[10])=1/5.
So the results are the same. The examples do not verify the general claim! But it turns out that this is always and is a consequence of the one-variable chain rule for differentiation of composite functions. The proof is actually written out in detail in {all|many|most} calc 3 textbooks.

The most important line integral
Suppose C(t) is the circle of radius R>0 and center 0. Suppose u(z)=1/z (continuous away from 0, of course. We compute C[1/z] dz. The circle is {oriented/parameterized} positively: C(t)=Reit and t is in [0,2Pi]. Then z=Reit so dz=Reiti dt. The limits on the integral are 0 and 2Pi. The integral itself becomes 02Pi[1/{Reit}]Reiti dt which is 02Pii dt and has value 2Pi i.
The value of this integral and the fact that it is not zero are related to the exercise with the Mad Count done at the beginning of this lecture.

### The complex exponential function

It's beautiful, it's wonderful ...

What should the exponential function be?

1. Certainly exp(0)=1.
2. I think exp(z1+z2)=exp(z1)exp(z2) for all possible complex z1 and z2.
3. When we've got a good theory, certainly exp should be differentiable, and exp´(z) should be exp(z).
These attributes are not logically independent. A good calc 1 textbook will substantiate the following: any differentiable function satisfying A and B together must satisfy C, while any function which satisfies A and C must satisfy B. A, by the way, is almost decoration -- it just gives an initial value for the function. Thus B and C are nearly equivalent. I would be happy to discuss these implications with you if you

The textbook asserts that exp(z)=exp(x+iy)=excos(y)+iexsin(y).

Standard notation which I will use is that the domain variable be called z=x+iy and the range be called w=u+iv. Here u(x,y) is excos(y) and v(x,y)=exsin(y).
Why does the textbook assert that the complex exponential should be this (initially, at least) weird combination of functions? There are several ways to motivate the definition. Let me go over a few of them.

• The ODE point of view
If exp(z)=exp(x)exp(iy), then we just need to understand exp(iy). If f(y)=exp(iy), then f(0)=exp(0)=1. Also, if we believe in the chain rule, f´(y)=exp´(iy)i, and then f´´(y)=exp´(iy)i2=-exp´(iy). So f(y) is a function which is its own second derivative, and we have the initial conditions f(0)=1 and f´(0)=i. Well, we know from our differential equations course that any function which is minus its second derivative is a linear combination of sine and cosine, so exp(iy) must be Acos(y)+Bsin(y) for some A and B. The initial conditions allow us to conclude that A=1 and B=i.
• The power series point of view
If exp(z)=exp(x)exp(iy), then since exp(w) should be SUMn=0infinitywn/n! (well, by absolute convergence ==> convergence and using the Ratio Test, the series converges for all w's in C) I conclude that exp(iy) should be SUMn=0infinity(iy)n/n!. Now a slight amount of thought shows that the positive integer powers of i cycle from i to -1 to -i to 1. When we rearrange the series by taking first the imaginary terms (those with +/- i's, and then factoring out the i's) and then the real terms, the two sums turn out to be cos(y)+isin(y).
• The complex differentiability viewpoint
A remarkably powerful theorem we will verify later in the course has a special case which applies here: if f is a complex differentiable function defined in all of C, and if for z's on the real axis (z's in R) I know that f(z)=exp(z) (the "real" exponential function) then f(z) must be given by the formula f(x+iy)=excos(y)+iexsin(y). I will show you this in a few weeks.
So for these and for lots of other reasons, the complex exponential function must be excos(y)+iexsin(y).

Notice please that this function is made of sums and products of well-known functions which are certainly continuous, and therefore it is nice and continuous, also.

 The geometry of the exponential function This is what I will concentrate on during this lecture. The complex exponential function has many pleasant and some irritating aspects. It is a bit difficult to draw a true graph of this function. The domain points are in C which is R2, as are the range points. Therefore the graph, the pairs of points (z,ez) would be elements of R4, and I don't know any neat ways of showing four-dimensional graphs. (Many attempts of been made. For example, we could sketch (x,y,u(x,y)) and change this standard graph of a surface in R3 to the graph (x,y,v(x,y)) by some geometric process. Or one could "draw" the graph of u(x,y) and put colors on the graph according to the values of v(x,y). You can easily find these methods and others on the web. I'll stick to more ordinary methods here. I'll have a complex plane representing the domain to the left, and then a range complex plane to the right. I'll try to sketch objects on the left in the domain and then sketch the corresponding objects on the right, objects which "correspond" under the exponential mapping.

 One horizontal line Consider the horizontal axis. We could image a bug moving on the axis in the domain at uniform velocity, from left to right. Here y=0, and x goes from -infinity to +infinity. The image in the w-plane is excos(0)+iexxsin(0) which is just ex. Notice that this is a ray, a halfline, whose image covers the positive x-axis. The image of uniform motion is no longer uniform motion. Near "-infinity" the image bug moves very slowly, and as the image goes from left to right along the positive x-axis, the image bug moves ever faster, exponentially faster.

 One vertical line Now consider the vertical imaginary axis, where x=0, and a bug crawling at uniform velocity from -infinity (in y) to +infinity (in y). Now with x=0, the image is e0cos(y)+ie0sin(y). So the image is the unit circle, described in the positive (counterclockwise) fashion, with the circle drawn at uniform velocity (1 radian per timestep).

 Other horizontal lines If we look for the image of a horizontal line which is just slightly above the x-axis, what will we get? Here we have excos(y)+iexsin(y) with y a small positive number. In fact, in terms of vectors, this is all positive (ex) multiples of the unit vector [cos(y),sin(y)]. So the image is again a halfline, a ray, emanating (originating?) from 0 at angle y to the positive x-axis. You could imagine as the horizontal lines go up or go down, the images will rotate around the origin.

 Other vertical lines If we look for the image of a vertical line which is to the right of the y-axis, what will we get? Here we have excos(y)+iexsin(y) with x a positive number. In fact, this describes a circle whose center is the origin, and whose radius is ex, always a positive real number. You could imagine as the vertical lines go left or right, the images will all be circles centered at the origin of varying radii.

The algebra
We can try to "solve" ez=w and we will get some algebra which resembles the real case, and some startling changes. If excos(y)=u and exsin(y)=v, we will get the same u+iv if we increase y by 2Pi.
The complex exponential is 2Pii periodic.
Notice that we will not be able to get w=0. If u=o then either ex=0, not possible from the real exponential, or cos(y)=0. But also sin(y) would have to be 0, and since the sum of their squares is always 1, they can't be 0 together!
The complex exponential is never 0.
Also, if w is any non-zero complex number, we can solve ez=w. The modulus of excos(y)+exsin(y)i is ex, and this must be |w|, so x=ln(|w|). Also v/u is sin(y)/cos(y) since the ex's cancel, so y is arctan(v/u). A better way to think about y is that it must be any value of arg(w). Remember that the argument of 0 is not defined.
If w is not zero, then z=x+iy, when x=ln(|w|) and y=arg(w) is a solution of w=ez, and we will write z=log(w). When that equation has one solution, it has an infinite number of solutions, separated by integer multiplies of 2Pii.

I remared that the image of a "tilted" line (not vertical or horizontal) turns out to be an equiangular spiral.

The exponential function takes any horizontal strip of "height" 2Pi and compresses the left infinity "edge" to a point, and spreads out the right edge so that the result covers all of C except for 0.

I then computed what happens to the image a straight line under the exponential mapping. I got another parameterized curve. I establishes that the mapping rotated tangent vectors by a constant amount, the result of multiplying by exp(z). Thus the exponential mapping is "conformal". If this multiplication represents the complex derivative (which it does, as we will see later) then we have established that exp'=exp.

 Here is the result of the Maple command plot3d(exp(x)*cos(y),x=-2..3,y=0..2*Pi,axes=normal); which will draw the real part of the complex exponential function. Possibly this drawing, a different kind from what I did above, may help you understand the exponential function.

Comment The problem is "easy" but irritating. Maple could help you with part of it. For example, the command evalf(I^I); gets the response .2078795764 which might be an approximation to part of one answer above. Note, though, that I'd like exact answers in terms of traditional mathematical constants and values of calc 1 functions, not approximations.

1.6: 1, 2, 4, 5

### Tuesday, February 1 (Lecture #5)

Starting on Thursday, February 3, the class will meet in Hill 423.

What is a continuous function?
Suppose f:D-->C. We'll call f continuous if for every sequence {zn} which converges to w (and all the zn's and w are in D) then the sequence {f(zn} converges to f(w).

Examples of continuous functions
Polynomials are continuous. Rational functions, in their implied domains (where the functions "make sense" are continuous.

An example of a function which is not continuous
Here I looked at Q(z)=z/conjugate(z). This function is bounded. The modulus of z and the conjugate of z are the same, so |Q(z)|=1 always. What happens as z-->0? In class I tried to look at the real and imaginary parts of Q(z). I did this in the following way:

``` x+iy     (x+iy)(x+iy)    (x2-y2)       2xy
------ = ------------- = -------- + i ------
x-iy     (x-iy)(x+iy)    (x2+y2)     (x2+y2)```
The real and imaginary parts are commonly used examples in calc 3. On the real and imaginary axes away from the origin these functions have different values. For example, the real part is {x2-y2]/[(x2+y2)]. When y=0 this is 1 and when x=0 this is -1. So there is no unique limiting value as (x,y)-->{0,0).
In complex notation the modulus of Q(z)=z/conjugate(z) is 1, and the argument is twice the argument of z. In fact, Q(z)=cos(2arg(z))+i sin(2arg(z)).

Several intermediate value theorems, some of which are not correct
So I will try to duplicate the wonderful dynamic (?) atmosphere of the classroom (clashroom?) as the instructor challenged the students to find a valid intermediate value theorem (IVT).

1. Suppose D is an open subset of C, and f:D-->R is a real-valued continuous function, and there are points p and q in D so that f(p)<0 and f(q)>0. Then f must have a root in D. (A root is a number w so that f(w)=0).
False Take D to be the union of two disjoint open discs, say D is A, a disc of radius 1 centered at 3i, union with B, a disc of radious 1 centered at 7. Then define f to be -7 on A and 38 on B. f is continuous, because convergent sequences in D must ultimately end up either in A or in B. But f is never 0. This version of the IVT is false because the domain is "defective".
2. Suppose D is a connected open subset of C, and f:D-->C is a continuous function, and there are points p and q in D so that f(p) is real and negative and f(q) is real and positive. Then f must have a root in D.
False Here the catch is that I made f complex-valued and the complex numbers are sort of two-dimensional and the image of f could go "around" 0. Here's an example which is a bit nicer than the one I offered in class. Suppose D is the open right halfplane, those z's for which Re(z)>0. Take f(z) to be defined by f(z)=z3. Certainly the only complex number for which z3=0 is z=0. But let's see: f(1)=1, so f has a positive value on this D. And f(1+sqrt(3)i)=(1+sqrt(3)i)3. I chose this z (1+sqrt(3)i) exactly because its argument is Pi/3. Now when I cube the number, the argument become 3·Pi/3=Pi so that the result is negative. The modulus of this z is 2 and its cube is 8 so the exact result is -8. So here the function has both positive and negative values, and is never 0.
3. Suppose D is a connected open subset of C, and f:D-->R is a real-valued continuous function, and there are points p and q in D so that f(p) is real and negative and f(q) is real and positive. Then f must have a root in D.
True We will get much finer root-finding results in this course, but this one does work. How could we prove this? Well, create a polygonal path from p to q inside D. For example, we could start by thinking of a line segment inside D connecting p and q. On this line segment, f is a real-valued continuous function defined on the intervale parameterizing the line segment, and its values are positive at one end and negative at the other end. By the standad Calc 1 IVT, f must have a root somewhere in the segment. What if the polygonal line segment has two pieces? Well, let's look at the middle vertex. So f could be 0 there, in which case certainly a root exists. If f is either positive or negative, we will have a line segment where the endpoints have different signs. So we can use the previous argument. In general, I guess we could use an induction argument based on the number of line segments in the connecting polygonal path.

What follows is not in the textbook in exactly this form. It does appear later in the textbook.

What is a real differentiable function?
This is a function f so that the limit as h-->0 of [f(x+h)-f(x)]/h exists. Untwisting the definition a bit, what this means is that f(x+h)=f(x)+f´(x)h+Err(?)h, where Err(?)-->0 as h-->0 and Err(?) is a possibly complicated function depending on f and x and h. The multiplication of Err by h guarantees that the Err(?)h is a "higher-order" term as h-->0. The local picture depends mostly on f(x) (the old value) and f´(x)h, a linear approximation. For example, if f´(x)=3, then locally f just stretches things by a factor of 3 near x.

What is a complex-differentiable function?
Well, I change x to z. And more or less everything else stays the same. All I have used in the previous analysis was the ability to add/subtract/multiply/divide. And to learn what big/small were (done with | |, which is absolute value for R and modulus for C). But now I can try to think about the following situation:
What if f is complex-differentiable, and I know that f(7+2i)=3-i and f´(7+2i)=1+i? Then locally, near 7+2i, f multiplies little perturbations h by the complex number 1+i. This means that the little perturbation h, a small complex number (or, equivalently, a small two-dimensional vector h) is changed to (1+i)·h. This takes the vector and stretches it by a factor of sqrt(2) (just |1+i|) and rotates it by Pi/2, a value of arg(1+i). Is this o.k.? This is a rather strict geometric transformation. In fact, if we took a pair of h's, say h1 and h2, they would both be magnified by sqrt(2) in length by f, and both be rotated by Pi/4. Hey: the angle between the h's in the domain and the angle between their images in the range must be the same! Angles are preserved. Such a mapping is called conformal. Indeed, I will establish this formally later in the course, but if f is complex-differentiable, and if the derivative is not 0, then f is conformal: angles are preserved.

A specific example of a function: C(z)=(z-i)/(z+i)
This is a remarkable function and I will try to not explain where it comes from. The domain I want to fix for C is H, the upper halfplane, those z's with Im(z)>0. One remarkable fact about this remarkable function is that if D is the unit disc (center 0, radius 1) then C maps H 1-1 and onto D. This is not obvious!

Well, if |(z-i)/(z+i)|<1, this is the same as
|z-i|<|z+i| (the bottom is not 0 for z in H).
We can square: |z-i|2<|z+i|2.
If z=x=iy, then z-i=x+i(y-1) and z+i=x+i(y+1), so the inequality with the squares is: x2+(y-1)2<x2+(y+1).
We can "expand" and get: x2+y2-2y+1<x2+y2+2y+1.
Simplify by cancelling to get:-2y<2y.
This is equivalent to 0<y.

All of these algebraic steps are reversible. The only one which might slow you down is the squaring step, but if we know A and B are non-negative, then A<B is logically equivalent to A2<B2. The algebra indeed says that C establishes a 1-1 correspondance between H and D. If we now compute (and assume all of the differentiation algorithms are still valid, which indeed is correct) then C´(z)=2i/[(z+i)2]. So C´(z) is never 0. Indeed, the mapping C is conformal. It takes the upper halfplane to the unit disc is such a way that all angles between curves are preserved. This is not supposed to be obvious. In conformal geometry, where measurement is with angles, H and D are the same. Certainly, if we were concerned with distances between pairs of points (metric geometry) D and H are not the same. So this is somewhat remarkable.

Thursday, January 27 (Lecture #4)
I returned the Entrance Exam, and went over it hastily. I tried to reinforce the idea that the skills "tested" in this exam would be needed in the course. I urged students whose work on various problems was not satisfactory to review material.

I reviewed convergence of an infinite sequence introduced last time and made a very poor pedagogical decision. I decided to state the Cauchy Criterion for convergence of a sequence. I tried to motivate this in several ways. First, the definition of a convergent series used the limit of the sequence as part of the defintion. I wanted to show students that there could be an "internal" condition, only referring to the terms of a sequence itself, which would be equivalent to convergence. But I ignored my own exprience, much of it obtained by teaching Math 311. This experience showed that developing and understanding the Cauchy criterion for convergence takes most people weeks of struggle (and some students seem hardly able to understand it even then!). So I talked "around" the Cauchy Criterion for a while, and then finally stated it:
A sequence {zn} satisfies the Cauchy Criterion if and only if for every epsilon>0, there is a positive integer N *usually depending on epsilon) so that for n and m >=N, |zn-zm|<epsilon.

Any sequence which converges satisfies the Cauchy Criterion, because the terms, when they are "far out" all get close to the limit of the sequence, and thus, by the Triaqngle Inequality, they get close to each other. That the converse is also true is not obvious, as I mentioned above, and several weeks of discussion in Math 311 are devoted to this converse.

The reason for the digression which was the mention of the Cauchy Criterion has to do with convergence of series. I defined the partial sum of a series, and as usual defined the convergence of an infinite series to be the limit of the sequence of partial sums, if such a limit existed. Then I remarked that a (non-zero!) geometric series would converge if the (constant) ration between successive terms had modulus less than 1. I could even then write a formula for the sum of such a series.

Most series do not have constant ratios between successive terms, and are not geometric series. The simplest way to test them for convergence is to use absolute convergence. A series converges absolutely if the series consisting of the moduli of the original series converges. This fact uses the Cauchy Criterion, and I should have followed the textbook, which cleverly sidesteps the Cuachy Criterion by mentioning that a complex series converges if and only if the real and imaginary parts converge, and then uses the real form of the comparison test to develop the result just quoted. This would have been much better!

A function f is continuous if f takes convergent sequences to convergent sequences.

We're almost at complex analysis. I'll get a version of the Intermediate Value Theorem next time, and then start differentiating.

### Tuesday, January 25 (Lecture #3)

The lecturer was ferociously attacked by alien germs from Andromeda, and therefore had hardly any energy. This should be regretted.

### The Zoo of Sets!

• Open disc Suppose z0 is a fixed point in C, and r is a positive real number. Then the "open disc with center z0 and radius r" is the collection of complex numbers, z, so that |z-z0|<r. This is, of course, the points inside the appropriate circle. The text has no specific notation for this set. I will call it D(z0,r). My best example of such an object is what I will call the "unit disc", whose center is 0 and whose radius is 1.

• Interior point If S is a subset of C and z0 is in S, then "z0 is an interior point of S" if there is some positive r so that D(z0,r) is a subset of S. That is, if every point in C closer to z0 than r is in S. Maybe some examples will help.
• Notice that R is a subset of C. No point of R is an interior point! That's because any disc we take with positive radius centered around any point of R will have points whose imaginary part is positive and whose imaginary part is negative, so the whole disc can't be in R.
• Consider the unit disc. I claim that every point of the unit disc is an interior point. Well, if z0 is in D(0,1), then take r=1-|z0|. I will convince you that D(z0,r) is in D(0,1). If w is in D(z0,r), then |w-z0|<1. But |w|=|(w-z0)+z0|<r+|z0|=1-|z0|+|z0|=1. There's one "<" in that chain which shows that the desired conclusion is correct. In fact, every point in every open disc is an interior point.
• Look at the closed unit disc. This is the collection of numbers, z, so that |z|<=1. Which points of this set are interior points? Certainly any point of D(0,1) is (repeating the previous argument). But no point with |z|=1 is an interior point. Let me argue about this (sorry, more politely, "verify this") for the example z=1. If we take r>0, then any disc D(1,r) will have the point 1+(r/2). But that point is not in the closed unit disc. So 1 cannot be an interior point.

• Open set A subset S of C is open if every point of S is an interior point of S. I tried to make the argument, which I mostly believe, that experimentally, all one would every "observe" of a set S would be its interior points, because things in real life are imprecise and wiggly. I don't know if anyone believed me. Certainly any open disc is an open set. The union of open sets is open (put more points in and things become even more interior [?]). Thus starting with discs one can construct rather complicated looking open sets (take the figure 8 in the plane and "thicken it up") with lots of holes.

• Closed set A subset S of C is closed if the collection of points not in S is open. For example, R is a closed subset of C. The collection of points not in S is sometimes called the complement of S and written C\S.

 Warning Although doors are either open or closed and are not both, sets may be neither. That is, there are sets which are not closed and also are not open. This is only a linguistic tragedy and I hope will confuse you only at the start. Here's an EXAMPLE Look at the closed unit disc and take out the origin. This set is not open because (as before) 1 is not an interior point. This set is not closed because its complement consists of these points: 0 and all z's with |z|>1. 0 is not an interior point of the complement.

• Boundary point Suppose S is a subset of C. z0 is a boundary point of S if every open disc with center z0 must have at least one point of S and one point of the complement of S.

 Comments on all this I'm going through this string of definitions at jet speed. Luckily we don't need to be intimate with all of them. But students who are seeing this for the first time may feel i) Submerged or overwhelmed (who needs all this vocabulary?) ii) Resentful (why the heck are there all these special words?). A beginning of possible answers I will be more precise shortly about the terms we'll use most often. The vocabulary began to arrive about 1880 and was in place, mostly, by the 1920's. Now everyone in the whole world who studies this subject and many others uses these words. In fact, just studying the implications of what we are introducing here is the introduction to a whole subject, topology. Also, it turns out that I am being quite precise (quantifiers, etc.) in what I am writing because the precision is needed: statements are wrong, examples are irritating, etc., without some real precision. This will become clearer as we go forward in the course. The precision will pay off with some amazing consequences.

• Line segment Suppose p and q are points in C. Then the set of all complex numbers which can be written tp+(1-t)q for t in the (real) unit interval, [0,1], is called the (straight) line segment from p to q. The t sort of says "travel from q, when t=0, to p, when t=1." If a name is necessary, I will call the line segment L(p,q).

• Convex set A set S in C is convex if for every pair of points p,q in S, L(p,q) is in S. I think this definition is known to almost everyone. Can we get an example of
• A convex open set: yes, the unit disc (or any disc!).
• An open set which is not convex: yes. Take the set to be the union of, say, the unit disc and the disc of radius 1 centered at 10. The line segment connecting 0 and 10 has real points which are farther away than 1 unit from both 0 and 10, and so cannot be in the union of the two discs.
• A convex closed set: yes, the closed unit disc. Closed "halfplanes" such as Im(z)>=0 (the closed upper halfplane) are also convex.
• A closed set which is not convex. Uhhhh ... two closed discs which don't overlap?
• A convex set which is neither closed nor open: I think the following contribution was made a student: z's so that |z|<1 and Re(z)>=0. Part of the boundary (the part on the imaginary axis) is in the set and part is not. The set is also convex.

• Connected open set or domain This is a very important concept. SUch a set is called a domain. We will like it because it is open, and therefore we can make little errors and still be in the set, and because of connectedness, results which resemble the Intermediate Value Theorem will still be true. So how should we define domain? A subset S of C is a domain if S is open and if, for every p and q in S, there is a finite sequence of line segments in S which connect p and q. Precisely, there is a set of points z0, z1, ..., zn so that L(zj,zj+1) is contained in S for j=0, 1, ..., up to n-1, and z0=a and zn=b. This is much harder to write than to draw. I will try to draw some pictures later.

• Convergent sequence of complex numbers Suppose {zn} is a sequence of complex numbers. We say the sequence converges to a complex number w if given epsilon>0, there is a positive integer N so that for n>N, |zn-w|<r. That is, given any open disc centered at w, eventually all the elements of the sequence are in the open disc.
Of course my first example was the sequence zn=in, the sequence of positive powers of i. This isn't the simplest example, and does show not great judgement. In fact, I should certainly have mentioned that all real sequences are examples of complex sequences. The sequence {in} does not converge (I didn't handle this rigorously, but asked what it might converge to and tried to discuss the consqeuences), and, in fact, the sequence {qn} (for fixed q with |q|=1) converges only when q=1: if qn-->w then surely qn+1-->qw and qn+1-->w (because you can multiply convergent sequences and because you can sort of move sequences along). Therefore qw=1, so (q-1)w=0 and either q=1 or w=0. Now |qn|=|q|n=1, so we would expect that the modulus of w to also be 1. But then w is not 0, so q=1. Whew! I built is some cute observations in that sequence of sentences, so you may want to look at them and discuss with me things you don't understand.

### Thursday, January 20 (Lecture #2)

Please read carefully sections 1.1, 1.1.1, 1.2, and 1.3 of the text.

Let's review the vocabulary: complex number, real part, imaginary part, complex conjugate, addition of complex numbers, multiplication of complex numbers, modulus, argument, ...

What does multiplication by i do to the (complex) plane? It rotates by Pi/2 (or 90o) in the positive sense (clockwise, orientation preserving). Multiplication by z=a+bi? If z is the 0 complex number, the effect is rather simple: all points go to the origin. Otherwise, vectors get clockwise rotated by arg(z) and get stretched by |z|. If theta is arg(z) and r=|z|, this takes a vector (x,y) which we could write as a column vector and then multiplies it by two matrices.

```      Rotation              Dilation

( cos(theta) -sin(theta) )   ( r  0 )
(                        )   (      )
( sin(theta)  cos(theta) )   ( 0  r ) ```
Mr. Siegal was correct. I did have the minus sign in the wrong place for the standard matrix representation of a linear transformation.

Both of these preserve orientation. How could we reverse orientation? Just consider complex conjugation.

One of the wonders and irritations of the subject is that there can be lots of different information concealed in one "chunk" of complex stuff. Even the equality |z| |w|=|zw| has several guises. First, it asserts that the distance from the origin of the product of two complex numbers is the product of their distances from the origin. Second, if we square the equation, and if z=a+bi and w=c+di, we see that (a2+b2)(c2+d2)=(ac-bd)2+(ad+bc)2. This equation is very useful in deciding, for example, which integers can be written as the sum of the squares of two integers. 5 is the sum of two squares (12+22) and so is 13 (32+22) therefore the product, 13·5=65 also must be (and it is: 12+82).

 THIS  IS  NOT  COMPLEX  ANALYSIS  SO  YOU  CAN  DISREGARD  IT! Real division algebras This is a slight diversion, which I hope will give students some perspective. A division algebra is a real vector space which has a distributive multiplication with a multiplicative identity and multiplicative inverses. People searched for examples of these starting soon after the complex numbers became well-known. It turns out (and good proofs of this were only completed in the mid-20th century!) that such objects exist in Rn only for n=1 and 2 and 4 and 8. These examples are wonderful: the real numbers, the complex numbers (the subject of this course) the quaternions (a 4-dimensional division algebra), and the octonions or Cayley numbers (in R8). A general discussion of the latter two kinds of "numbers" is available in the quite recent book, On Quaternions and Octonions by Conway and Smith. Quaternions definitely have applications in physics and geometry. A book covering the whole subject of number systems, and whose early chapters are quite understandable, is Numbers by Ebbinghaus etc. Why don't we have courses about the quaternions and the octonions? It turns out that as the dimensions increase, some nice property is lost. For example, going from the complex numbers to the quaternions: commutativity is no longer necessarily true. And from the quaternions to the octonions, in general, multiplication is no longer necessarily associative. One natural question for students in a complex analysis course might be, "What's lost going from R to C?" Math 311 discusses the theory of the real numbers, and establishes that R is a complete ordered field. Here order is used in a technical sense: we can tell when one number is larger than another. The ordering has the following desirable properties: Any non-zero number is either positive (larger than 0) or its additive inverse is positive and these alternatives are exclusive. The sums and products of positive numbers are positive. All of the familiar rules for manipulating inequalities can be derived from these properties. In particular, Math 311 verifies that if x is not zero, then x2 is always positive: that is, squares are always positive. Therefore, 1=12 is positive. Hypothetically, suppose we could "order" C. Then look at i. It isn't 0, so its square must be positive. But that square is -1, which we already know is not positive since 1 is positive. Therefore C is not an ordered field, and there is no way, no ordering obeying the rules above. Pedagogical consequence If you write any statement which directly connects "<" and complex numbers, I can stop reading or paying attention. Your reasoning is incorrect.

So what one gives up going from R to C is ordering. But this is a course which will do lots and lots of estimation. We will need to discuss continuity and convergence, for example. Even if we aren't totally formal and rigorous, we will need to estimate sizes. The words BIG and small will be used constantly. How do we know what they mean?

I remark that rigor is defined as

1. a sudden feeling of cold with shivering accompanied by a rise in temperature, preceding a fever etc.
2. rigidity of the body caused by shock or poisoning etc.
while I would like it to mean, here,
the quality of being logically valid
Pathology versus mathematics!

A measure of the size of a complex number z=a+bi will be its modulus. We already know what happens to modulus when we multiply:|z| |w|=|zw|. But what about addition? How can we relate |z| and |w| and |z+w|? Well, since |z| is the absolute value of z if z is real, we know that the sum |z|+|w| is rarely equal to |z+w| (hey: z=3 and w=-3). But the most powerful "tool" dealing with modulus is the Triangle Inquality.
|z+w|2=... (p.12 of the text) ...<=(|z|+|w|)2.
(I got from one side to the other by "straightforward" (!) manipulation which I frequently get confused about.)
So If I tell you that |z|=30 and |w|=5, then |z+w| will be at most 35.
Trick question How big can |z-w| be?
In fact, |z-w|<=|z|+|-w|, and all we can actually conclude is that |z-w| is bounded by 35 also! I did not include any information about the "direction" of the complex numbers! (Notice that |-w|=|-1| |w|=1|w|=|w|.)

• Exercise 1: If |z|=1, find useful over- and under-estimates of |z5+z+7|.
How to do it Certainly I haven't defined "useful" in this context, and we won't see the real uses of the estimations for a few weeks. But let me show the techniques that are usually tried first. They almost always involve the Triangle Inequality and the Reverse Triangle Inequality. So:
|z5+z+7|<=|z5|+|z|+|7|=|z|5+|z|+7=1+1+7=9. So we have an over-estimate.
How about the other side? Here 7 seems to be a good candidate for BIG. Therefore
|7|-|z5+z|<=|z5+z+7|. Now the largest that |z5+z| can be is 2 (again using the Triangle Inequality). Therefore the smallest that |7|-|z5+z| can be is 7-2=5.
The simple estimates are: if |z|=1, then 5<=|z5+z+7|<=9.

• Exercise 2: If |z|=2, find useful over- and under-estimates of |z5+z+7|.
How to do it I changed the modulus of z. The over-estimate part works much the same:
|z5+z+7|<=|z5|+|z|+|7|=|z|5+|z|+7=32+2+7=41.
But for the under-estimate, the BIG term now seems to be z5. Therefore
|z5|-|z+7|<=|z5+z+7|. Here |z5|=|z|5=35, and the largest that |z+7| can be when |z|=2 is 9. Therefore 35-9<=|z5+z+7|.
The simple estimates are: if |z|=2, then 26<=|z5+z+7|<=41.

• Exercise 3: If |z|=3, find useful over- and under-estimates of |(7z-1)/(2z2-7)|.
How to do it I found over- and under-estimates for the Top and the Bottom separately.
Top: |7z-1|<=7|z|+|-1|<=7(3)+1=22. Also, |7z|-|-1|<=|7z-1|, so that 20<=|7z-1|.
Bottom: |2z2-7|<=2|z2|+|-7|=2|z|2+7=18+7=25. And |2z2|-|-7|<=|2z2-7| so that 18-7<=|2z2-7|.
Therefore we know the following:
20<=Top<=22
11<=Bottom<=25
What can we conclude about Quotient=Top/Bottom? To over-estimate a quotient, we must over-estimate the top and under-estimate the bottom. The result is that Quotient<=22/11. For an under-estimate of a quotient, we need to under-estimate the top and over-estimate the bottom. So 20/25<=Quotient.

Mr. Wolf in class and Mr. Bittner after class expressed some doubts about how well these methods might work with some other examples (Shouldn't we worry about division by 0? What happens if we encounter a situation where the BIG part is not obvious, or maybe there isn't a BIG part?). Their misgivings are indeed correct. My random (random?) examples were actually chosen fairly carefully so that such problems didn't occur.

We need to discuss the notion of interior point, interior of a set, an open set, boundary point, and closed set. Also convex set, and connected open set. This is all in 1.3.

### Tuesday, January 18 (Lecture # 1)

I mentioned the extravagant simplicity of "calculus with complex numbers". This must certainly be justified. I will try to do this. I should also have mentioned the utility of complex analysis. In many areas of application, just thinking through the implications of replacing a real parameter by a complex parameter ("complexifying") turns out to lead directly to useful conclusions.

The Euclidean plane can be thought of as the playground for two diemnsional vectors (forces, velocities, etc.). It can also be regarded as a two-dimensional real vector space, R<2. Or we can think of it as C, the complex numbers. The vector i (1,0) could be 1 and the vector j (0,1) could be i. I ask only that i2=-1.

If all of the routine rules of computation (distributivity, commutativity, associativity, etc.) are satisfied, then for example:
(3+2i)x(5+7i)=2(5)+2(7)i2+[2(5)+7(3)]i=-4+31i.

In fact, it is true that C, numbers written a+bi with a and b real numbers, are a field when addition and multiplication are defined appropriately. Verification of this is tedious and maybe not really interesting.

I asked how the multiplicative inverse of a number like 3+2i would be found. So we need a+bi so that (3+2i)x(a+bi)=1=1+0i. This means that
3a-2b=1
2a+3b=0
which has a unique solution (Cramer's Rule) if and only if (exactly when!) the determinant of the coeffients is not zero. Here that determinant is 32+22. Generally, if we looked at a+bi, the determinant would be a2+b2.
Or we could look at 1/(3+2i) and multiply ("top and bottom") by 3-2i. Things work out nicely.

Vocabulary If z=a+bi, then a is the real part of z, Re(z), and b is the imaginary part of z, Im(z). The number sqrt(a2+b2) is called the modulus of z, and written |z|. The number a-bi is called the complex conjugate of z, and written overbar(z) (z with a bar over it).

I then discussed polar representation of complex numbers. We found that z=|z|(cos(theta)+i sin(theta)). Here theta was an angle defined only up to a multiple of 2Pi ("mod 2Pi"). Thus 1+i is sqrt(2)[cos(Pi/4)+i sin(Pi/4)], for example.

I then computed the product of two complex numbers written in their polar representation. I found that:

• The modulus of the product zw is the product of |z|, the modulus of z, and |w|, the modulus of w.
• The argument of zw, that is, the angle used in the polar representation, is the sum of the argument of z and the argument of w.
But the second equation should be understood "mod 2Pi". The sine and cosine functions only pay attention to the values of functions up to a sum or difference of integer multiples of 2Pi. To illustrate this, I looked at 1+i tried to understand (1+i)7 and (1+i)13.

I wrote De Moivre's Theorem, which states that if z=r[cos(theta)+i sin(theta)] then zn must be rn[cos(n theta)+i sin(n theta)].

As a consequence I attempted to "solve" z7=1 (the solutions are a regular heptagon [seven-sided] inscribed in a unit circle centered at 0 with one vertex at 1). We discussed how many solutions such an equation could have (at most 7). Then I tried to "solve" (describe or sketch the solutions) of z3=1+i (a equilateral triangle of numbers).