Is $\lim_{z \rightarrow z_0} zf(z) = 0$ sufficient to show $f(z)$ has a removable singularity or must we check $\lim f(z)$?

Question

The function $$f(z) = \frac{\exp(\frac{z}{z-5})-1}{z(z+1)}$$

Has the singularities $-1$, $0$ and $5$.

The singularity at $-1$ is a pole of degree one as

$$\lim_{z \rightarrow -1}(z+1)f(z) = -e^{\frac{1}{6}}+1 \ne 0$$

Now the singularity at $0$ is removable. To find the limit $\lim_{z\rightarrow 0}zf(z) = 0$ is easy, whereas finding the limit $\lim_{z\rightarrow 0}f(z)$ is much harder, but necessary for Riemann's Theorem on removable singularities in the strict form I have seen.

To show the singularity at 0 is removable, is the first limit $zf(z)$ sufficient? I think it should be (principal part of the Laurent series is 0), but I haven't seen it formally stated anywhere.

Answer 1

Yes, it is sufficient: $\lim_{z\to0}zf(z)=0$ if and only if $0$ is a removable singularity of $f$ (assuming that, as in this case, $f$ is analytic).

However, it is easy to compute $\lim_{z\to0}f(z)$. Take $g(z)=\exp\left(\frac z{z-5}\right)$. Then $g'(0)=-\frac15$ and therefore$$\lim_{z\to0}f(z)=\lim_{z\to0}\frac1{z+1}\times\frac{g(z)}z=g'(0)=-\frac15.$$

Answer 2

You don't need to check the limit of $f(z)$. The relevant theorem is:

Theorem (Riemann). If $U\subseteq \mathbb C$ is open, $z_0 \in U$, and $f: U \setminus \{z_0\} \to \mathbb C$ is homolorphic, then the following are equivalent:

1. $f$ admits a holomorphic extension to $U$;
2. $f$ is bounded in a neighborhood of $z_0$;
3. $\lim_{z \to z_0} (z-z_0)f(z) = 0$.