Why thinking ahead is important: a complex integral.

July 24, 2011

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.