1
$\begingroup$

How can I prove that the coefficient of $(z-z_0)^{-1}$ when $f(z)$ has a pole of order $m$ at $z=z_0$ is given by: $ a_{-1}=\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^mf(z)\right]_{z=z_0} $ ?

1 Answers 1

4

Let $f(z)=\sum\limits_{n=-m}^{\infty}a_n(z-z_0)^n$ be the Laurent expansion of $f$ at $z_0$. Then $g(z)=(z-z_0)^mf(z)$ is holomorphic around $z_0$ and its Taylor expansion at $z_0$ is $g(z)=\sum\limits_{n=0}^{\infty}a_{n-m}(z-z_0)^n$. Therefore, $a_{-1}=\frac{1}{(m-1)!}g^{(m-1)}(z_0)$, and the conclusion follows.

  • 0
    @IvanLerner: Yes, you are right!2012-11-28