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Find and classify all isolated singularities of $\displaystyle f(z)=\frac{1}{1+\sqrt{z}}$.

So if $1+\sqrt{z}=0$ then $\sqrt{z}=-1$. Therefore $z=1$. Hence, $1$ is an isolated singularity and it is a pole. I am not quite sure whether my argument is correct here. Is it correct? If not, how may I do it?

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  • $\begingroup$ yes it is correct if you've defined $\sqrt{z}$ such that it is holomorphic in the neighborhood of $z=1$. how would you do that ? $\endgroup$
    – reuns
    Apr 28, 2016 at 0:39

1 Answer 1

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That depends on the branch of $\sqrt z$. Usually, $\sqrt z$ is taken to be the branch defined on $\mathbb C\setminus(-\infty,0]$ and $\sqrt 1=1$. In that case, $f$ does not have isolated singularities.

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  • $\begingroup$ So can we say that $1$ is a removable singularity? $\endgroup$
    – Extremal
    Apr 29, 2016 at 13:57
  • $\begingroup$ It is not a singularity. $\endgroup$ Apr 29, 2016 at 15:32

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