Answer to Question #146560 in Complex Analysis for Usman

Question

Q: Evaluate the following integral using residue theorem
∫ Z^2 e^z dz ; C : |z| = 1 ,Answer(πi/3)

Answer 1

$\int_C z^2 e^z dz$ , $C: |z| = 1$

The residue theorem:

$\int_C f(z)dz = 2\pi i \sum_{k=1}^m res_{z= a_k} f(z)$, where $a_k$ - singularity point.

So we have found in which poin function has a singularity:

$f(z) = z^2 e^z,$ has a singularity only in $z = \infin$

In our case, for $z = \infin$ formula from residue theorem will take the next form:

$\int_C f(z)dz = -2\pi i \sum_{k=1}^m res_{z= a_k} f(z)$

So we have to find $res_{z=\infin} f(z)$

Also we know $res_{z=\infin} f(z)= - C_{-1}$, where $C_{-1}$ - coefficient near $z^{-1}$

Find the Laurent series of $f(z)$. Series in negative power of z:

$f(z)=z^2e^z = z^2(\sum_{n=0}^\infin \frac{1}{n!z^n}) = z^2(1+\frac{1}{z} + \frac{1}{2!z^2} + ... +\frac{1}{n!z^n}+...)=\\= z^2 + z + \frac{1}{2} + \frac{1}{3!z}+ ...+ \frac{1}{n!z^{n-2}}+...$

We can see that $C_{-1} = \frac{1}{3!} = \frac{1}{6}$, so $res_{z=\infin} f(z) = -\frac{1}{6}$ and

$\int_C z^2 e^z dz = -2\pi i (-\frac{1}{6}) = \frac{\pi i}{3}$