Singularity $z/\sin(z)$

Question

I'm trying to figure out the singularity of $$\frac z {\sin(z)}$$ I understand how it works with $$\frac {\sin(z)} z$$

But here I don't know what to do. I know, that the solution is that it has a pole of order $1$ and as well a removable singularity at zero.

Answer 1

That function has a removable singularity at $0$ (and no pole there), since$$\lim_{z\to0}\frac z{\sin z}=1.$$