Integrate complex function over unit circle

1. Mar 27, 2014

skrat

1. The problem statement, all variables and given/known data
Calculate $\int _Kz^2exp(\frac{2}{z})dz$ where $K$ is unit circle.

2. Relevant equations

3. The attempt at a solution

Hmmm, I am having some troubles here. Here is how I tried:

In general $\int _\gamma f(z)dz=2\pi i\sum_{k=1}^{n}I(\gamma,a_k)Res(f,a_k)$ where in my case $I(\gamma , a_k)=1$.

Now $Res(f,a_k)$:

$f(z)=z^2exp(\frac{2}{z})=z^2\sum_{n=0}^{\infty}\frac{1}{n!}(\frac{2}{z})^n=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{2^n}{z^{n-2}}$

To get $\frac{1}{z}$ clearly $n-2=1$ so $n=3$.

Which gives me $Res(f , 0)=\frac{4}{3}$.

Therefore $\int _Kz^2exp(\frac{2}{z})dz=2\pi i \frac{4}{3}$.

Or is that completely wrong? Thanks in advance!

2. Mar 27, 2014

electricspit

This is absolutely correct!

EDIT: Remember you can always verify your Laurent series (the hardest part in my opinion) using WolframAlpha :D

3. Mar 27, 2014

skrat

Thank you! =)

4. Mar 27, 2014

HallsofIvy

Staff Emeritus
You know, I presume that $e^x= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$.

So $e^{2/z}= 1+ 2/z+ 2/z^2+ 4/(3z^3)+ \cdot\cdot\cdot$ and then
$z^2e^{2/z}= z^2+ 2z+ 2+ 4/(3z)+ \cdot\cdot\cdot$

The integral around the closed circle of any power of z (positive or negative), except -1, is 0 and the integral of $z^{-1}$ is $2\pi i$ (which is where the idea of "residue" comes from) so the integral of $z^2e^{2/z}$ around the unit circle is $(4/3)(2\pi i)= (8/3)\pi i$.

5. Mar 27, 2014

skrat

hehe, nicely said Hallsoflvy. Everything is a bit more obvious now. Thank you!