Suppose $f$ has a pole of order $m$, then $1/f$ has a removable singularity.
Basically my book say
$f(z) = \frac{g(z)}{(z-z_0)^m}$ and $1/f = \frac{(z-z_0)^m}{g(z)}$ with $g$ analytic at $z_0$ and $g(z_0) \neq 0$
But if $g(z_0) \neq 0$, what singularity is there in the first place?