Proving origin to be removable singularity(Proof verification)

Question

This particular question is from Ponnusamy Silvermann complex section 9.22 ( question no. 2) and I am unable to solve it.

Let f(z) be analytic in deleted neighborhood of origin and lim $z\to 0$ | zf(z) |=0 . Then show that origin is removable singularity of f(z).

I need to show that lim $z\to 0 f(z) $=M(finite) . But if it were infinite, then $z \to z_0$ |zf(z) | would be indeterminate.

So, it's removable.

Is my proof correct?

Answer 1

From $$\lim_{z\to 0}zf(z)=0,$$ we know than $z=0$ is removable singularity of $F(z):=zf(z)$. Define $F(0)=0$, then $F$ is analytic at $z=0$, and its Taylor series of $F$ at $z=0$ is $$F(z)=zf(z)=\sum_{n=1}^{\infty}a_nz^n,$$ ($a_0=0$ is due to $F(0)=0$). So the Taylor series of $f$ at $z=0$ is $$f(z)=\sum_{n=1}^{\infty}a_nz^{n-1},$$ hence, $$\lim_{z\to 0}f(z)=a_1.$$ So the origin is removable singularity of $f(z)$.

Answer 2

A counterexample for your proof would be $f(z)=1/\sqrt{z}$. I think you need to use the property of analyticity in some way to exclude such cases.