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

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.

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.

• You can also prove it with out computing Laurent series. Apr 3, 2018 at 23:30

$\lim_{z\to z_0}|f(z)|=\infty$ 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.
• also is not so clear to me that $1/f$ is analytic in the punctured disc Apr 4, 2018 at 0:26