Isolated Singularities of $f(z)$ and $1/z^2 f(1/z)$

Question

Let f be holomorphic $f:\mathbb{C} \rightarrow \mathbb{C}$ except for finitely many isolated singularities. Prove that $h(z)=\frac{1}{z^2} f(\frac{1}{z})$ is holomorphic in a punctured neighborhood of $0$.

—- In order for $h(z)$ to be holomorphic in a punctured neighborhood of 0 the singularity must be isolated So $f(z)$ does not have any non-isolated singularity, however that same thing can not be said about $h(z)=\frac{1}{z^2} f(\frac{1}{z})$ (can we take $f(z)=\sin(z)$ and $h(z)=\sin(1/z)$ as example?). But how could I prove that is in analytic in a punctured neighborhood of zero?

Answer 1

Hint:If it has a finite number of singularities, then outside a ball of large enough radius $f(z)$ is holomorphic. So for if for $|z|> R$ , $f(z)$ is holomorphic , then for $0<|z|<\frac{1}{R}$ , $f(\frac{1}{z})$ is holomorphic and also $\frac{1}{z^{2}}$ is holomorphic there. So their product is holomorphic.

Answer 2

Notice it suffices to show that the function $g(z)=f(\frac{1}{z})$ is holomorphic in a punctured neighborhood of 0.

Now to do this, notice that the singularities of $g$ correspond exactly to the singularities of $f$, more precisely, if $f$ has a singularity at $z_0$, then $g$ will have the corresponding singularity at $1/z_0$ (counting the singularity at infinity). From the fact that $f$ only has finitely many singularities, there must exist some $R>0$ such that it doesn't have any more singularities outside of the disk of radius $R$( except the singularity at infinity).

But from this we can deduce the following about the singularities of $g$ , by recalling our earlier remark: $g$ has no singularities inside the disk of radius $1/R$ (except from the singularity of $f$ at infinity, which corresponds to the singularity of $g$ at zero).