A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Ch 9.2

Cor 9.6 of Prop 9.5(*)

Suppose $f$ is holomorphic in $\{0<|z-z_0| < R\}$. Then $f$ has a pole at $z_0 \iff \exists m \in \mathbb N$ and holomorphic $g: \{0<|z-z_0| < R\} \to \mathbb C, g(z_0) \ne 0$ and $$f(z) = \frac{g(z)}{(z-z_0)^m} \ \forall 0<|z-z_0| < R$$


  1. Is this proof for $\Leftarrow$ right?

$$f(z)=\frac{g(z)}{(z-z_0)^m} \iff (z-z_0)^{m+1}f(z)=\frac{(z-z_0)g(z)}{1}$$

$$\implies \lim (z-z_0)^{m+1}f(z)=\lim \frac{(z-z_0)g(z)}{1}=0$$

$\therefore, z_0$ is a pole if not removable. To show $z_0$ is not removable:

$\lim \frac{(z-z_0)f(z)}{1}=\lim \frac{g(z)}{(z-z_0)^{m-1}}$, w/c dne because $\lim g(z) = g(z_0) \ne 0$ while $\lim (z-z_0)^{m-1} = 0$ if $m=1$ and dne if $m>1$.

$\therefore, z_0$ is not removable and thus a pole with order $n=m$. QED for $\Leftarrow$

  1. How do I do $\Rightarrow$ ? Follow pf of Prop 9.5? Use Laurent series?

(*) Prop 9.5

enter image description here

up vote 0 down vote accepted
  1. Right

  2. Yes to Laurent. Also yes to follow pf (**)

Holomorphic implies Analytic for Laurent as well as Taylor. Since multiplying by a power of $n+1$ makes the limit zero, $f$'s Laurent series expansion must start with a negative exponent of $n$ (after you multiply, the Laurent series begins with a positive power of $1$).

$$\therefore, f = \frac{1}{(z-z_0)^{n}}(c_{-n}+c_{-(n-1)}(z-z_0) + \dots) =: \frac{g(z)}{(z-z_0)^{n}}$$

We take out $n$ and not $n+1$ because we must have $g(z_0) \ne 0$.

(**) Actually just before Cor 9.6 and right after Prop 9.5, book says 'We underline one feature of the last part of our proof:'

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.