Let $R$ be a region and $z_0\in R$. Suppose $f$ is analytic on $R\setminus \{z_0\} $, is bounded, and has a isolated singularity at $z_0$. Show $z_0$ is a removable singularity
$f$ bounded implies $|f(z)|\leq M$ for some real number $M$. We know that $f$ has a removable singularity at $z_0$ if and only if $\lim_{z\to z_0}(z-z_0)f(z)=0$. $$\lim_{z\to z_0}|(z-z_0)f(z)|\leq \lim_{z\to z_0}|(z-z_0)|M = 0 $$ Thus, $\lim_{z\to z_0}(z-z_0)f(z)=0$ and we are done.
Is this correct?