I am having troubles finding the residue of f(z) around $z=0$. I am expanding my Laurent series around $z = 0$ but the coefficient of $z^{-1}$ is not $0$ as it should be.
$$ f(z) = \frac{z^3e^\frac{1}{z^2}}{z^5+4} $$
The answer is $$ Res(f(z),0)=0 $$
$$e^{\frac 1{z^2}}=\sum _{n\ge0}\dfrac {z^{-2n}}{n!}$$
Also $\dfrac 1{z^5+4}=\dfrac 14\dfrac 1{1-(-\frac {z^5}4)}=\dfrac 14\sum _{n\ge0}(-\dfrac {z^5}4)^n$
Putting things together, one sees that the Laurent series indeed has infinitely many $\dfrac 1z$ terms. Because we have $$\dfrac14(\sum _{n\ge0}\dfrac {z^{-2n+3}}{n!})(\sum _{n\ge0}(\dfrac {-z^5}4)^n)$$.
So every solution to the linear diophantine equation $-2k+3+5l=-1$ gives us a $\dfrac 1z$ term. The solutions are $(k,l)=(2+5t,2t)$.
So the residue is $$\dfrac 14\sum_{t\ge0}\dfrac 1{(2+5t)!}\dfrac 1{16^t}\gt\dfrac 18$$.