Jordan’s Lemma.

[This post is for those of you who are already comfy with doing some basic contour integrals in complex analysis.]

 

So you’re sitting around, evaluating contour integrals, and everything is fine.  Then something weird comes up.  You’re asked to evaluate an integral that looks like

for is continuous.  Eek.  Don’t panic though, because Camille Jordan’s gonna help you out.

 

Applying the Maximum Modulus Principle and Cauchy: A Common Question.

I’ve been a bit busy moving and packing my things, but I’ve tried to keep my problem-doing up to practice for my analysis qual.  One problem that seems to come up quite a bit in the complex analysis part of the exam is something like the following:

 

Question.  Suppose that is entire which satisfies for every and every .  Prove that is a polynomial of degree at most .

 

Why thinking ahead is important: a complex integral.

I’ve been up for a while doing practice qualifying exam questions, and sometimes I hit a point where I just do whatever it is that comes to my mind first, no matter how tedious or silly it seems.  This is a bad habit.  I’ll show why with an example.

 

Here’s the question.  Let be the unit circle oriented counterclockwise.  Find the integral

.

 

The sophisticated reader will immediately see the solution, but humor me for a moment.  I attempted to do this by Taylor expansion.  The following calculations were done:

 

 

To which the binomial theorem was applied to the numerator terms to obtain:

 

 

And at this point we note that everything is going to die off when we take the integral except the coefficient of the term.  Our residue (the coefficient) will be:

which can also be written (slightly more suggestively) as:

which we should recognize as the Taylor expansion of at the point .  Nice!  Now we note that plugging in to take the contour integral (ignoring all those terms which don’t matter) will force us to integrate

.

Cutely, if we think of the Greek letter as being a "p", this solution spells out "2pie".

 

But now, readers, let’s slow down.  This is, indeed, the correct answer.  But if I had just looked at the form of the integrand, I would have seen an everywhere analytic function divided by a form of .  This screams Cauchy Integral Formula.  Indeed, according to the CIF, we should get the solution as

which is exactly what we got before, but only took about 4 seconds to do.  It’s nice to be able to check yourself by doing something two different ways, but when time isn’t on your side (like in a qualifying exam situation, for example!) then remember:

Think before you Taylor Expand.

Harmonic Conjugates: Order Matters!

I was skimming over Brown’s Complex Analysis book, and I came across a neat little exercise I thought I would share.  The solution is not difficult — indeed, it is just a manipulation of equations — but the idea is interesting and especially telling about the strange kinds of not-so-symmetric things that go on in complex analysis. 

The Argument Principle, The Winding Number, and Rouché’s Theorem (Part 2.)

(I’ve decided against giving a proof of Rouché’s theorem until such a time as I find one that doesn’t use algebraic topology or isn’t tedious as hell.)

 

Let’s simply state Rouché’s theorem, and then we’ll talk about how to actually apply Rouché’s theorem.

 

The Argument Principle, The Winding Number, and Rouché’s Theorem (Part 1.)

(In this part: The Argument Principle and the Winding Number.)

Each of these three theorems (the argument principle, the winding number theorem, and Rouché’s theorem) are all interesting in their own right, but something really special happens when you put them into a cocktail mixer and shake them up together.  Really; I’m not a fan of analysis, but what we’re doing in this post I think of as almost magical. 

Complex Analysis Study Guide: Brown and Churchill.

Brown and Churchill (8th ed) was the book I used for the second complex analysis class I’ve had to take so far (the first was Lang).  My class went over the first six chapters and half of the seventh: so, up to the middle of the section on applications of residues. 

To prep for the final, I compiled a quick, slightly-shorter list of things that I feel the complex student should know if they’ve used this book and have gotten to around the same point.  I’ve excluded the chapter on applications of residues, since it’s a relatively short chapter with better pictures in the text than ones I could draw at 5am.  Because sharing is caring, below is a link to the pdf.  Enjoy!

http://www.scribd.com/doc/45170305

You say Holomorphic, I say Analytic.

In another post, I noted something a bit strange at first reading: some authors use the word holomorphic to describe a power series expansion and reserve analytic for complex differentiable while other authors swap those terms.  I then noted that “this doesn’t matter.”  Well, why not?  I mean, definitions are pretty important in mathematics!  My reasoning is: these are really the same thing.  If is holomorphic then it is also analytic, and vice versa.  I’ve been putting off doing this proof for way too long, so let’s just get it over with.  It’s not hard, it’s just an analysis proof, which means that it’s extremely easy to describe (“take a little ball and do something in it”) but extremely tedious to work out (“take an epsilon such that this epsilon is less than the sum of the minimum of the supremums of…”) but I’m going to try to have a complete proof and motivate every step.

After this, I’ll give a short proof that if a function is complex differentiable (holomorpic, to me.) once, then it is complex differentiable infinitely many times.  It’s not a direct corollary, but it’s a nice fact to know.

Morera’s Theorem.

Morera’s theorem, named after the mathematician Giacinto Morera whose name is pretty sweet but is second only to his ultra-fly mustache

is an extremely important result in complex analysis: it states that if is a continuous function defined on an open set in the complex plane, and, in addition, we have that the integral around every closed curve is zero for every closed curve in , then, in fact, must be complex differentiable everywhere in .  This is a common tool to use in the proofs of other theorems, as well as in its own right showing that a continuous function is actually much nicer than “just” continuous.

Entire Bump Functions? Where?

Here’s the cute proof of the week.  Liouville’s Theorem (in complex analysis) is coming up (because this theorem is so important to me, I’m trying to scrounge up a lot of applications for it) and so I wanted to just give a cute little corollary that comes directly from Liouville’s Theorem.  First, let me just state Liouville really quickly:

 

Theorem (Liouville).  A function from which is bounded (in the sense that for some real number ) and entire (complex differentiable) everywhere is constant.

 

Now, let’s talk about Bump Functions for a second.  A bump function is a function which is bounded, smooth, and has compact support (it is zero everywhere but on compact set), but we can generalize this to complex functions by simply changing the domain and range to the complex numbers .  So why don’t we ever hear about complex bump functions? 

A short story before tossing the corollary at you: bump functions are nice, because sometimes we need to partition up spaces and nicely distribute functions about.  For example, there are things call partitions of unity which allow us to talk about functions on manifolds nicely.  While working through one of my books for diffi-manifolds today, I noticed that while most of the time the book worked in an arbitrary field, one particular theorem only was only stated for the reals.  I wasn’t sure if it was a typo or not, so I attempted to adapt the proof for complex numbers (at least!) but I got to a point where I needed to use the complex equivalent of a bump function.  No matter where I looked, I couldn’t find any mention of them.  And then it hit me.

 

Corollary.  Suppose that is a bump function as we’ve defined above.  Then is the zero function.

 

Proof.  By Liouville’s theorem, since is bounded, entire, and smooth (which, in this case, means holomorphic as complex functions differentiable once are infinitely differentiable) it is constant.  Because it is zero at least on some non-compact set, it follows that it must be zero everywhere.

 

So there’s not much bump in bump functions on the complex plane.  That’s kind of sad!  Nonetheless, it’s a good (or at least cute) application of Liouville’s theorem.