I have this complex function:
$$f(z)=\frac{z}{\sin\left(\frac{\pi}{z+1}\right)}$$
I'd like to compute residues in all isolate singularities.
If I'm not mistaken $f$ has poles in $z=\frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $\xi$, where $\xi$ is $z-\frac{1}{k}+1$.
The best I can obtain is this:
$$\frac{z}{\sin\left(\frac{k\xi+1-k}{k\xi+1}\right)}$$
Can you help me?
Best Answer
Let $f(z)=\sin\left(\frac\pi{z+1}\right)$. Then$$f'(z)=-\frac{\pi\cos\left(\frac\pi{z+1}\right)}{(z+1)^2}$$and therefore$$f'\left(\frac1k-1\right)=(-1)^{k-1}k^2\pi.$$So,$$\operatorname{res}_{z=\frac1k-1}\left(\frac z{\sin\left(\frac\pi{z+1}\right)}\right)=\frac{\frac1k-1}{(-1)^{k-1}k^2\pi}=(-1)^{k-1}\frac{1-k}{k^3\pi}.$$