Taylor series and singularities of complex function

Question

I have been struggling with the following problem. I have the function $$f(z)=\frac{e^{z}}{(z-1)cos(\pi z)}$$  I want to find all the singularities. It is easy to see that this function has an editable singularity at $z=1$ but I don't know what happens when $z=n$ odd integer number. I would like to find the Taylor series of $f(z)$ around $z=0$. Thank you for your time.

Answer 1

$$f(z)=\frac{e^{z}}{(z-1)\cos(\pi z)}$$ Around $z=0$, we have $$f(z)=\frac{1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+O\left( z^{5}\right) }{(z-1)\left(1-\frac{\pi ^2 z^2}{2}+\frac{\pi ^4 z^4}{24}+O\left(z^{6}\right) \right)}$$ Now, long division to get $$f(z)=1+2 z+\frac{1}{2} \left(5+\pi ^2\right) z^2+\left(\frac{8}{3}+\pi ^2\right) z^3+\frac{5}{24} \left(13+6 \pi ^2+\pi ^4\right) z^4+O\left(z^5\right)$$