Show that the function $f$ has pole of order 1

Question

Problem: Show that the holomorphic function $f$, $$\frac{z-\pi}{(e^{iz}+1)^2}:D'(\pi;2\pi)\rightarrow \mathbb{C}$$ has a pole of order 1 at the point $\pi$.

This might be really simple but I cannot show that this has pole of order 1 but rather pole order 2, since the denominater is raised to the power of 2? Or am I wrong? When trying to show that it has a pole of order 1 I end with 0/0 but maybe this is sufficient and that 0/0 can be 1 if we compute the limit for $z\rightarrow\pi$?

I hope someone can clear this up for me. Notice that this is not homework. I'm just trying to get a better understanding of holomorphic functions. Thanks in advance.

Answer 1

The order of the numerator at $\pi$ is $1$, whereas the order of the denominator at $\pi$ is $2$. Therefore, the order of your function at $\pi$ is $-1(=1-2)$. In other words, your function has a simple pole at $\pi$.

Answer 2

Notice that

$$(z-\pi)\frac{z-\pi}{(e^{iz}+1)^2}=\frac{1}{\Big(\frac{e^{iz}+1}{z-\pi}\Big)^2}$$ Let $f(z)=e^{iz}$. Notice that $f(\pi)=-1$. Then $$\lim_{z\rightarrow\pi}\frac{e^{iz}+1}{z-\pi}=\lim_{z\rightarrow\pi}\frac{f(z)-f(\pi)}{z-\pi}=f'(\pi)=-i$$

Hence $$\lim_{z\rightarrow\pi}(z-\pi)\frac{z-\pi}{(e^{iz}+1)^2}=-1$$ From this one may conclude that $f$ has a pole of order one and the residue at $z=\pi$ is $-1$.