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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?

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    $\begingroup$ The function is not defined at $z=1$. $\endgroup$ – copper.hat May 1 at 16:35
  • $\begingroup$ @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. $\endgroup$ – Julian Mejia May 1 at 18:05
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    $\begingroup$ @JulianMejia: I was just being a penant :-) $\endgroup$ – copper.hat May 1 at 18:14
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    $\begingroup$ Of course, a penant is just a poorly spelt pedant. $\endgroup$ – copper.hat May 1 at 18:24
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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: $\sum_{n=-\infty}^\infty\frac{a_n}{2n+1}z^{2n+1}$.

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

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