Finding Residue of Complex Function at Infinity

[MartinKitty]

Hello everyone,
I have a problem with finding a residue of a function:
$f(z)={\frac{z^3*exp(1/z)}{(1+z)}}$ in infinity.
I tried to present it in Laurent series:
$\frac{z^3}{1+z} sum_{n=0}^\infty\frac{1}{n!z^n}$

I know that residue will be equal to coefficient $a_{-1}$, but i don't know how to find it.

 

[mathman]

Expand $\frac{z^3}{1+z}=z^3-z^4+z^5-z^6...$. The multiply the two series together to find the coefficient you want (as an infinite series).

 

[MartinKitty]

Expand $\frac{z^3}{1+z}=z^3-z^4+z^5-z^6...$. The multiply the two series together to find the coefficient you want (as an infinite series).

Then i get:
$\frac{z^3}{1+z}=(-1)^n*z^{n+3}$

and when i multiply I always get ${z^3}$ with some fraction

 

[Ssnow]

$\left(z^{3}-z^{4}+z^{5}-\cdots \right)\left(1+\frac{1}{z}+\frac{1}{2z^{2}}+\frac{1}{6z^{3}}+\frac{1}{24z^{4}}+\cdots \right)$, the only interested terms are of the forms $\frac{a_{-1}}{z}$, that are $\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\cdots$ so it is $e^{-1}-\frac{1}{2}+\frac{1}{3!}$ (if I did not make mistakes ...)