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 f is entire which satisfies |f(z)| \leq A|z|^{k} + B for every z\in {\mathbb C} and every A,B > 0.  Prove that f is a polynomial of degree at most k.


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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 C be the unit circle oriented counterclockwise.  Find the integral

\displaystyle\int_{C}\frac{\exp(1 + z^{2})}{z}\, dz.


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:


= \displaystyle\int_{C} \frac{1 + (1+z^{2}) + (1 + z^{2})^{2} + \cdots}{z}\, dz


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


= \displaystyle\int_{C} \frac{1 + (1+z^{2}) + \frac{\sum_{n=0}^{2}\binom{2}{n}z^{n}}{2!} + \frac{\sum_{n=0}^{3}\binom{3}{n}z^{n}}{3!} + \cdots}{z}\, dz


And at this point we note that everything is going to die off when we take the integral except the coefficient of the \frac{1}{z} term.  Our residue (the coefficient) will be:

1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots

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

\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots

which we should recognize as the Taylor expansion of e^{z} at the point z = 1.  Nice!  Now we note that plugging in e^{i\theta} to take the contour integral (ignoring all those terms which don’t matter) will force us to integrate

\displaystyle e \int_{0}^{2\pi}\frac{1}{e^{i\theta}}ie^{i\theta}\, d\theta = ie\int_{0}^{2\pi}\, d\theta = 2\pi i e.

Cutely, if we think of the Greek letter \pi 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 z - z_{0}.  This screams Cauchy Integral Formula.  Indeed, according to the CIF, we should get the solution as

2\pi i \exp(1 + 0^2) = 2\pi i e

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.

Limits play an important role in standard calculus: it is the limit that allows us to get “infinitely small” and to be able to take derivatives and integrals. It would be nice, then, to see when our functions behaved nicely when we took limits of them.

In fact, we have a relatively nice theorem regarding sequences of measurable functions and their limits. Why do I pick this theorem to prove for you out in particular, as opposed to the many others above which I simply state? I feel that the proof of this theorem is neither a “follow-your-nose” proof, nor an “ah-ha!” proof that you will “get” once you read it; no, this proof is more of a “how was I supposed to think of that!” proof. What’s more is that the proof is not complicated; in fact, after explaining exactly what is going on (which I will do, following the proof) it will seem nearly obvious to you.

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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. 

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I’m a bit sad that I don’t have a catchy title for this post, but at least we have a neat topic: measurable functions.  In pure, unadulterated generality, these things (to me, at least) look exactly like continuous functions in topology: the idea is that if you "pull back" a measurable set, you ought to get a measurable set.  Specifically, our "pulling back" in this case is doing to be taking the pre-image of the function (not to be confused with the inverse, which does not always exist; the pre-image, on the other hand, exists and is a set). Does this look familiar?  If I said, maybe, "open" instead of "measurable"?  Yeah.  Yeah. 

In this post, we will work with {\mathbb R} with the standard sigma-algebra generated by the open sets of the standard topology on {\mathbb R}.  But, that’s still kind of a lot of sets to consider.  Do we have to look at them all, or do we have a basis for this sigma-algebra?

For all you lazy mathematicians, you’re in luck.  In fact, the sets (-\infty, a) for each a\in {\mathbb R} generate the sigma-algebra we’re considering (try to prove this; it’s not that hard!)  Thus, it suffices just to check out the pre-images f^{-1}((-\infty,a)) for a\in {\mathbb R} in order to see if f is measurable.

Last, note that we could use the other open or closed semi-infinite intervals, but I like (-\infty, a) the best.


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When you’re on facebook, you’ll see a lot of people posting a lot of silly things.  It would be interesting to see what kinds of things are posted when, but this is a subject for some kind of cybersociologists.  Nonetheless, if you are awake at around 11pm and on facebook (and, I’ll admit it, I’m one of these people), you might see a string of comments that essentially say: "omg, it’s 11:11, make a wish."

If you’re superstitious or obsessive compulsive, this might be a lucky reminder to make a wish at this time, but besides its symmetry and visually appealing form, what’s so special about 1111? 

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