Originally Posted by

**vsywod** Well, actually I'm not sure it's a simple pole. I guess it is, but how can I show it?

I'ld have to give a holomorphic function $\displaystyle g$ thats doesn't vanish in 0 and which satisfies

$\displaystyle f(z) = z^{-1} \cdot g(z)$.

This would mean $\displaystyle g(z) = \frac{z^n}{\sin^n(z)}$ which is clearly holomorphic outside of 0. But why is $\displaystyle \lim_{z\to 0} g(z) = 1$?

I think this is a stupid question, but my head is really spinning at the moment...

Thank you!