If $f(z) = \frac{z^3e^{1/z}}{1+z^3}$ and $C:=\{z \in \mathbb{c} :|z|=3\}$ be in a positive sense (anti-clockwise), then $$\int_{C}f(z)dz = $$ (a) $2\pi i$
(b) 0
(c) $-2\pi i$
(d) $-3\pi i$
Attempt:
The singularities are at the 4 points given by $z^3 =-1$ (we obtain 3 singularities from here) and $z=0$. So, the solution can be found out by Cauchy's integral formula given by $\sum_{i}2\pi i f(a_i)$ where $a_i$ denotes the singularities.
But the above mentioned method is very time consuming and won't help me in a competitive exam. Is there any other way out to solve this possibly quicker than the above mentioned way?
Please help. Thanks in advance!
The change of variable $w=\frac 1 z$ reduces the integral to $-\int_{C'} \frac {e^{w}} {w^{2}(1+w^{3})} dw$ where $C'=\{w: |w|=\frac 13\}$. Now there is only one pole. By expanding the exponential and using the expansion of $\frac 1 {1+w^{3}}$ as $1-w^{3}-w^{6}-\cdots$ you can quickly see that the coefficint of $\frac 1 w$ is $-1$ so the answer is $-2\pi i$.