I must prove that the singularities of $\displaystyle f$ are poles. I also must determine their order and the corresponding residue.

$\displaystyle f(z)=\frac{z}{\cos z}$.

My attempt :

I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).

So I wrote $\displaystyle \cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$.

So if I can find the order of the zeros of $\displaystyle \frac{1}{f(z)}$ I'm done.

I've found out that $\displaystyle \frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$

The first term of the series really bothers me. If it wasn't there then $\displaystyle \frac{1}{f(z)}$ would be equal to $\displaystyle z\cdot h(z)$ with $\displaystyle h(0) \neq$ $\displaystyle 0$ and so $\displaystyle z=0$ would be a $\displaystyle 0$ of order $\displaystyle 1$ of $\displaystyle \frac{1}{f}$, thus a pole of order $\displaystyle 1$ of $\displaystyle f$. But of course I should take in count the $\displaystyle \frac{1}{z}$ term and look for $\displaystyle z=k \frac{\pi}{2}$, $\displaystyle k \in \mathbb{Z}$, but I'm stuck.