Show that $\lim_{z\to z_0}|f(z)|=\infty$ if and only if $z_0$ is a pole

Question

Im stuck with this exercise

Let $f$ holomorphic on $U\setminus\{z_0\}$ where $U$ is open. Show that $\lim_{z\to z_0}|f(z)|=\infty$ if and only if $z_0$ is a pole.

One direction is easy to show, Im stuck trying to show that

$$\lim_{z\to z_0}|f(z)|=\infty\implies z_0\text{ is a pole}$$

In particular Im having a hard time to figure how to show that if $\lim_{z\to z_0}|f(z)|=\infty$ then $z_0$ cannot be an essential singularity.

My work so far: if $\lim_{z\to z_0}|f(z)|=\infty$ then $f$ have a non-removable singularity at $z_0$. If $z_0$ would be an essential singularity then I know that (this is my definition of essential singularity) the principal part of the Laurent expansion of $f$ around $z_0$ have infinitely many non-zero coefficients, that is

$$\lim_{z\to z_0}|f(z)|=\lim_{z\to z_0}\left|\sum_{k=1}^\infty c_{-k}(z-z_0)^{-k}\right|\tag1$$

where there are infinitely many $c_{-k}\neq 0$. Then to show that the limit of $(1)$ doesnt exists is enough to show that there is some $\theta\in[0,2\pi)$ such that

$$\lim_{r\to\infty}\left|\sum_{k=1}^\infty c_{-k}e^{ik\theta}r^k\right|<\infty\tag2$$

That is: if there is some linear path $z\to z_0$ such that $\lim_{z\to z_0}|f(z)|<\infty$ we are done. However Im not sure if this approach is useful or not. In any case I get stuck here, I dont have a clue about how to show that $z_0$ cannot be an essential singularity if $\lim_{z\to z_0}|f(z)|=\infty$.

Some help will be appreciated, thank you.

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EDIT: I think I see a path for a solution. We can write $\omega_k(\alpha_k)^k$ for $\omega_k\in\mathrm S^1$ and $\alpha_k\in(0,\infty)$ such that $\lim \alpha_k=\infty$ instead of $c_{-k}e^{ik\theta}r^k$ (because $\lim c_{-k}= 0$) what give us great freedom to choose suitable sequences $(\omega_k)$ and $(\alpha_k)$ to try to prove $(2)$.

Indeed, if Im not wrong, choosing $\omega_k=(-1)^k$ we are done.

Answer 1

$\lim_{z\to z_0}|f(z)|=\infty$ only if $z_0$ is a pole.

Proof: Suppose $z_0$ is an isolated singularity of $\ f$ such that $\lim_{z\to z_0}|f(z)|=\infty$. Then you can find a $\delta>0$ such that $|f(z)|>1$ for $0<|z-z_0|<\delta$. Thus the function $\frac{1}{f(z)}$ is analytic, bounded and nonzero on that punctured disc and hence $z_0$ is a removable singularity of $\frac{1}{f(z)}$ (by Riemann's theorem). Now consider the function, $$g(z)=\begin{cases}\frac{1}{f(z)}& \text{if}\ 0<|z-z_0|<\delta \\0& z=z_0 \end{cases}$$ Then $\ g$ is analytic on that disc and has a zero at $z_0$ (say, of order $m \ge 1$) . Then $g(z)=(z-z_0)^m \phi(z)$, where $\phi$ is analytic and $\phi(z_0) \ne0$. Thus $f(z)=\frac{1}{(z-z_0)^m \phi(z)}$, for $0<|z-z_0|<\delta$.

This shows $z_0$ is a pole of $\ f$ of order m.