# If $z_0$ is a pole of $f$ then $\lim_{z \rightarrow z_0} |f(z)| = \infty$

Say $f$ is analytic at on $B(z_0,R)-\{z_0\}$. Then $f$ has a Laurent series expansion $$f(z)=\sum_{k=-\infty}^{\infty}\alpha_k(z-z_0)^k$$ We say $z_0$ is a pole of $f$ if $~\exists~~ m \geq 1$ such that $\alpha_m \neq 0$ and $\alpha_l=0~~\forall l <m$.

Let $z_0 \in \mathbb{C}$ and $R>0$. Suppose that $f$ is analytic on $B(z_0,R)-\{z_0\}$ and has a pole of order $m \geq 1$ at $z_0$. Then I need to show $|f(z_0)|=\infty$.

I have done the following: Since $f$ has pole of order $m$ $$f(z)=\sum_{k=-m}^{\infty}\alpha_k(z-z_0)^k$$ Hence $f(z)=(z-z_0)^{-m}g(z)$, where $g(z)$ is analytic at $z_0$ and $g(z_0)=\alpha_{-m}$. Now can we take the limit $z \rightarrow z_0$ on $|f(z)|$ and say the limit goes to $\infty$? We know the limit is bounded for $g(z)$ and $\lim_{z \rightarrow z_0} \frac{1}{|z-z_0|}=\infty$.

• You forgot to mention that $g(z)\to g(0) \neq0$. – Kabo Murphy Jul 28 '18 at 23:09
• yes, that is true. But I wanted to ask since $|g(z)|$ is bounded and the $\frac{1}{|z-z_0|}$ goes to $\infty$, can we say $|f(z)|$ is infinity? – Arindam Jul 28 '18 at 23:12
• No. The product of two numbers, one going to infinity and the other going to $0$ need not even have a limit; if it has a limit the limit can be any number. – Kabo Murphy Jul 28 '18 at 23:14
• But $\alpha_{-m} \neq 0$ as $z_0$ is a pole of order $m$. so one limit is non-zero finite and another goes to $\infty$. In that case their product goes to infinity. Thanks I understand the point. – Arindam Jul 28 '18 at 23:18
• By definition (see Ahlfors, p. 127) an isolated singularity $z_0$ with $\lim_{z\to z_0}|f(z)|=\infty$ is a pole of $f$. – Christian Blatter Jul 29 '18 at 14:00

Perhaps we can show that if $z_0$ is a pole, then the Laurent series has a finite number of negative powers. Then it is a direct consequence that the limit is unbounded.
Let $z_0$ be a zero of order $l$ of a function $f(z)$. Let $g(z)=\frac1{f(z)}$, so that $z_0$ is a pole of order $l$ of g(z).
We have that $$g(z)=(z-z_0)^l h(z)$$ and $$h(z_0)\neq 0$$
Then, $$f(z)=\frac{1}{(z-z_0)^l} \frac{1}{h(z)}$$ but $h(z)$ is analytic at $z=z_0$ so we have $$f(z)=(z-z_0)^{-l} \sum_{n=0}^\infty a_n(z-z_0)^n = \sum_{n=-1}^\infty a_{n+l}(z-z_0)^n$$
so that $f(z)$ has finitely many negative powers.
Now, if we take the limit as $z\rightarrow z_0$ we see that $|f(z)|\rightarrow \infty$ since it has at least one term $\frac1{z-z_0}$.