# Problem of the Week #31 - October 29th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $f(z)$ be a complex valued function that has a pole of order $m\geq 1$ at $z=z_0$. Prove that the residue of $f$ at this point is given by

$\text{res}(f,z_0)=\lim_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{\,d^{m-1}}{\,dz^{m-1}} \left[ (z-z_0)^m f(z) \right].$

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Hint:
Consider the Laurent series characterization of poles for $f(z)$, i.e. let
$f(z)=\frac{a_{-m}}{(z-z_0)^m}+\cdots+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\ldots$
Find a way to extract the residue $a_{-1}$ from this.

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his solution below.

$f(z)=\frac{a_{-m}}{(z-z_0)^m}+\cdots+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\ldots$

$\Rightarrow (z-z_0)^m f(z)=a_{-m}+\cdots+a_{-1}(z-z_0)^{m-1}+a_0 (z-z_0)^m+a_1(z-z_0)^{m+1}+a_2(z-z_0)^{m+2}+\ldots$

$\Rightarrow \frac{\,d^{m-1}}{\,dz^{m-1}} \left[(z-z_0)^m f(z) \right]=a_{-1}(m-1)! +a_0 (m-1)! (z-z_0)+ a_1(m-1)! (z-z_0)^{2}+ a_2(m-1)! (z-z_0)^{3}+\ldots$

$\Rightarrow \frac{1}{(m-1)!}\frac{\,d^{m-1}}{\,dz^{m-1}} \left[(z-z_0)^m f(z) \right]=a_{-1} +a_0 (z-z_0)+ a_1 (z-z_0)^{2}+ a_2 (z-z_0)^{3}+\ldots$

$\Rightarrow \lim_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{\,d^{m-1}}{\,dz^{m-1}} \left[(z-z_0)^m f(z) \right]=a_{-1}$

$\therefore \text{res}(f,z_0)=a_{-1}=\lim_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{\,d^{m-1}}{\,dz^{m-1}} \left[(z-z_0)^m f(z) \right]$

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