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I tried a few examples and I think that the following in complex analysis holds: If a function $f$ has a pole at $z_0$, then $1/f$ has a removable singularity at this point. Is this correct?

user1551
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1 Answers1

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If $f$ has a pole at $z_0$ then there is an open $U$ about $z_0$ small enough that$$f(z)=\frac{g(z)}{(z-z_0)^k}$$where $k$ is the order of the pole and $g(z)$ is analytic and non-vanishing on $U$

So on $U\setminus\{z_0\}$,$$\frac1f=\frac{(z-z_0)^k}{g(z)}$$ which is analytic on $U$ since we chose $U$ small enough that $g$ doesn't vanish there. Since $\lim_{z\rightarrow z_0}(z-z_0)\frac1{f(z)}=0$, the singularity is removable.

Karthik C
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  • Can't we conclude from $\lim_{z\to z_0}\frac{1}{g(z)}}(z-z_0)^k=0$ that singularity is removable? –  May 22 '19 at 01:35