Maximum modulus principle

Here’s a fact you probably never noticed: Holomoprhic functions have no local maxima. Okay, constant functions do, but those are lame.

Theorem 1 (Maximum modulus principle) Let . Then if has for every , then is constant.

Proof: If has a local maximum at , then in a small ball around , consider the image of . It looks something like . We don’t really care about the lower bound. The important point is that the upper bound is attained. This tells us that, the image of the small ball under is completely contained in , and that it touches the boundary. Thus, the image of this ball has a boundary and thus cannot be an open set. But by the open mapping theorem, this is only possible if were in fact constant.

Open mapping theorem

Today we’ll prove the open mapping theorem:

Theorem 1 (Open mapping Theorem) Let , for some open set in . Then , the set of all possible images of is either constant, or is open in .

Let , and let be a preimage. That is, . Since is open, there must be a small closed ball around completely contained in . Let be it’s radius. Then . Let . What do we know about ?

We know that the roots of are isolated points. After all, if not, we would have a convergent sequence of roots, contradiction one of the theorems we proved here. So we can choose to be even smaller so that the only root of in is . Since the boundary of is a circle, and hence compact, and is continuous it attains its minimum when restricted to the boundary. Let be the minimum value of for on the boundary of .

Now, by Rouché’s theorem, will have the same number of roots in in as for any in . (Application of Rouché’s theorem is left as an exercise.)

In particular, this means that for every complex number in , the function has at least one root in , and so is contained in , meaning our arbitrarily chosen point is in the interior of . Thus is open.

Rouche’s Theorem

Today we prove Rouché’s theorem. The gist is that it helps us count the number of roots of a holomorphic function, given some bounds on its values.

Theorem 1 Suppose and are holomorphic functions inside and on the boundary of some closed contour . If

on , then and have the same number of zeros on the interior of .

Before we begin proving this, it should be emphasized that we count with multiplicity. We would count the number 1 as a root of twice. Most root counting we every do will be done this way. I feel confident in saying that it is the correct way to count roots, even if at first it is unintuitive.

Proof: By hypothesis, has no roots on the boundary . Define . The roots of are the roots of . The poles of are the roots of . So it suffices to use the argument principal to show that . That is, we need to show that

is zero.

But from our hypotheses, we can conclude that

That is, never takes on values more than away from . Imagine a dog tethered by a leash of length less than to the point . That dog can’s reach the origin, but more importantly, he can’t walk a loop around the origin. So computing the winding number about the origin must give us zero. This proves the theorem.

Argument Principle

Let be a zero of a meromorphic  with multiplicity . Then we can write where . Taking derivatives yields

Hence, for , we get

The residue of this sum is simply the sum of the residues of the two parts. The first term has residue at . The second part has no pole at , and hence zero residue. Thus, the residue of at is .

What if we pick a pole of of ? Then by a similar construction, if is an order pole, we can write and compute

yielding a total residue of .

If a point is neither a pole nor a zero, then is holomorphic at , and has residue zero at .

If we take a big contour around all of these points, then the integral will be the sum of the residues inside the contour, which we have just shown is the number of roots minus the number of poles (counted with multiplicity and order, respectively).

That is, if denotes the number of zeros counted with multiplicity inside a contour , and denotes the number of poles counted with order, then

This result is known as the argument principle.

A joke

Today, a definition, then a joke.

I defined poles of meromorphic functions, but we can be a bit more descriptive. Suppose we have a meromorphic function which is is undefined at some point . We can expand it as a Laurent series, and get something like:

It may be that we can make arbitrarily negative and still have be nonzero. This is basically the worst situation possible. We don’t in fact call it a pole. We say that it is an essential singularity at

In a better situation, it might be that is nonzero, but with any smaller (more negative) index, is zero. Then, we would say that has a pole  at  of order . Of course there’s nothing special about seventeen.

If a pole has order 1, we say that it is a simple pole.

Now for a joke.

An airplane is on its way out of Warsaw, and the pilot suffers a heart attack and dies. A passenger is asked to navigate the plane to safety. He looks worried, so the stewardess asks “what’s wrong?” He responds “I’m just a simple Pole in a complex plane!”

Laugh, damn it!

Residue calculus

There’s a neat trick we can use to integrate real integrals using complex analysis. These can be made arbitrarily complicated, but I’ll give you a simple example. Compute the integral:

This is another way to write down

Let’s change all the s to s, and pretend we’re integrating along the path in the complex plane. Our notion of integrating in is defined in such a way that this makes sense. Okay, but we normally integrate loops, not paths, so let’s complete a full loop is the semi-circle in the upper half-plane with base and radius . Then we can break the path into two pieces:

The first piece is the integral we want to compute, and the second is the curved part of the semi-circle. Together they make a closed loop whose integral we can actually calculate using the residue theorem. Since our function is meromorphic on all of , we need to simply figure out which poles are inside the semicircle , and find their residues. We can tell that the only poles are at and , of which only is in the semicircle (for every ). It’s residue we compute as .

Thus, the left hand side, via the residue theorem, must be .

Now we just need to show that the semi-circular arc’s contiribution is neglibible for large . Then, taking yields

Indeed,

which decreases to zero as .

I’m awful at integration in general, so I’m not sure if this integral can be done using integration by parts, or a tricky -substitution. There are definitely some integrals though, for which those standard methods just don’t cut it.

Meromorphic functions and residues

Last time, we discussed Laurent series, which are essentially two-way power series. They are almost as nice as holomorphic functions, but not quite. Maybe we can recoup some of the lost beauty of holomorphicity by imposing a reasonable condition.

We want to allow ourselves to have points where the function isn’t defined, but let’s limit these points. Let’s require them to be isolated points. We say that a function is meromorphic on an open set if is holomorphic on , except at some number of isolated points. We write . These isolated points are called poles. We obviously shouldn’t expect to have a power series centered at one of these poles, but it does have a Laurent series.

Let and let be a pole of . Then as we saw last time, has a Laurent series centered at :

Moreover, the series has inner radius of convergence 0, so this representation is valid for all close enough to .

Now if we take a loop in , and integrate, if has no poles on the interior of , then the integral is zero. This is the Cauchy integral theorem. What if it does have a pole? We can use the generalized Cauchy integral formula we saw last time:

Theorem 1 (Residue theorem) Let and let be a pole of in . Expand as a Laurent series as above. Let be a small counter-clockwise circle about such that the only pole in its interior is . Then

Proof:The generalized Cauchy integral formula we saw last time said that

Let .

If we take a loop that goes around several poles, the integral must be the sum of the integrals, as we can build a homotopy as shown in the images below (click to blow up).

The value (for an expansion about ) is called the residue of at the pole . In other words, the theorem states that the value of an integral about a contour is the sum of the residues of the poles inside.