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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: $$ a_{-1}=\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^mf(z)\right]_{z=z_0} $$ ?

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up vote 4 down vote accepted

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.

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You mean $g(z)=(z-z_0)^m f(z)$ – M. Strochyk Nov 7 '12 at 14:28
@M.Strochyk: Thank you for correcting this typo. – 23rd Nov 7 '12 at 14:32
@richard and putting $n-m=x$ we can find any $a_x$ using $$a_x=\frac{1}{(m+x)!}g^{(m+x)(z_0)}$$. Is that correct? – Ivan Lerner Nov 28 '12 at 14:11
@IvanLerner: Yes, you are right! – 23rd Nov 28 '12 at 16:13

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