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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 $$f(z) = \frac{g(z)}{(z-z_0)^m} \ \forall 0<|z-z_0| < R$$

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  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

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  • 13,459

1 Answers1

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  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:'

BCLC
  • 13,459