Prove that the complex function: $\frac{z}{ \sin(\pi/z)}$ has an anti-derivative on $\mathbb{C} \setminus D(0,1)$

Question

Prove that the function $\frac{z}{\sin(\pi/z)}$ of a complex variable has an anti-derivative on $\mathbb{C} \setminus D(0,1)$.

My attempt: I tried to develop the Laurent series at $z=0$ but without any success.

Any suggestions on how to prove that using Laurent series?

Answer 1

Your function is an even function. Therefore, its Laurent series is of the form $\sum_{n=-\infty}^\infty a_nz^{2n}$, which has an antiderivative: $\sum_{n=-\infty}^\infty\frac{a_n}{2n+1}z^{2n+1}$.