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

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?

• The function is not defined at $z=1$. – copper.hat May 1 at 16:35
• @copper.hat True, I guess it should be $\overline{D(0,1)}$ which makes more sense since as a result $\mathbb{C}\setminus \overline{D(0,1)}$will be an open set. – Julian Mejia May 1 at 18:05
• @JulianMejia: I was just being a penant :-) – copper.hat May 1 at 18:14
• Of course, a penant is just a poorly spelt pedant. – copper.hat May 1 at 18:24

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: $$\sum_{n=-\infty}^\infty\frac{a_n}{2n+1}z^{2n+1}$$.

• Interesting, I guess this technique works as long $a_{-1}=0$. – Julian Mejia May 1 at 18:08
• Indeed it does. – José Carlos Santos May 1 at 19:03