If $f$ has a pole at $z_0$, then $1/f$ has a removable singularity 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?

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.
