If a function $f$ is not analytic at a point $a$, can we conclude that $f^2$ will not be analytic at $a$?

Question

I am trying to show that an isolated singular point of a function $f$ is also an isolated singular point of $f^2$. Can we show that $f^2$ is also not analytic or do I have to go about it using Laurent Series?

Answer 1

An analytic function with an non-removable isolated singularity is unbounded near the singularity. Hence, so is $f^2$.

Of course, if the singularity is removable, it is quite different: Consider $f(z)=1$ everywhere except $f(0)=-1$, for example.