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If a function is holomorphic on $\Bbb D$\ ${0}$, does that mean 0 is a removable singularity of the function if the residue at 0 is 0?

I think this is false because I thought I did some practice problems and sometimes the residue at poles can be 0 too. But can someone please offer me a proof? Thanks.

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    $\begingroup$ Consider $f(z) = 1/z^2$. $\endgroup$ – Ethan Alwaise Dec 14 '16 at 9:06
  • $\begingroup$ That's a counterexample. Could you tell me how to prove it? I could think of some counterexamples, but don't know how to prove it.. Thanks! $\endgroup$ – J.doe Dec 14 '16 at 9:07
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    $\begingroup$ Do you mean prove that it is a counterexample? $\endgroup$ – Ethan Alwaise Dec 14 '16 at 9:08
  • $\begingroup$ @EthanAlwaise No I meant is there a way to prove it directly, instead of providing counterexamples. $\endgroup$ – J.doe Dec 14 '16 at 9:09
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    $\begingroup$ The statement is false, so providing a counterexample is a proof. $\endgroup$ – Ethan Alwaise Dec 14 '16 at 9:14
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All kinds of singularities are possible:

$f(z)=\frac{\sin z}{z}$ has a removable singularity at $0$

$f(z)=\frac{1}{z^{123456}}$ has a pole at $0$

$f(z)=e^{1/z}$ has an essential singularity at $0$

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  • $\begingroup$ you meant $ e^{1/z^2}$ or $e^{1/z} - 1/z$ such that $Res(f(z),0) = 0$ $\endgroup$ – reuns Dec 14 '16 at 15:04
  • $\begingroup$ yes, this is a better example $\endgroup$ – Fred Dec 14 '16 at 15:18

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