Answer to Question #232175 in Complex Analysis for Fozia Sayda

Question

Determine the poles and the residue at each pole of the function f(z)=1/(z-1)²(z+2)

Answer 1

 If $f(z)$ has a pole of order $k$ at $z=z_0$ then

$rez(f,z_0)=\lim\limits_{z\to z_0}\big(\dfrac{1}{(k-1)!}\dfrac{d^{k-1}}{dz^{k-1}}((z-z_0)^kf(z))\big)$

$z=-2,$ simple pole

$rez(f,-2)=\lim\limits_{z\to-2}((z+2)f(z))$

$=\lim\limits_{z\to-2}((z+2)(\dfrac{1}{(z+2)(z-1)^2}))$

$=\lim\limits_{z\to-2}(\dfrac{1}{(z-1)^2})=\dfrac{1}{(-2-1)^2}=\dfrac{1}{9}$

$z=1,$ order $2$ pole

$rez(f,1)=\lim\limits_{z\to 1}\big(\dfrac{1}{(2-1)!}\dfrac{d^{2-1}}{dz^{2-1}}((z-1)^2f(z))\big)$

$=\lim\limits_{z\to 1}\big(\dfrac{d}{dz}((z-1)^2(\dfrac{1}{(z+2)(z-1)^2}))\big)$

$=\lim\limits_{z\to 1}\big(\dfrac{d}{dz}(\dfrac{1}{z+2})\big)$

$=\lim\limits_{z\to 1}\big(-\dfrac{1}{(z+2)^2}\big)$

$=-\dfrac{1}{(1+2)^2}=-\dfrac{1}{9}$

$z=-2,$ simple pole

$rez(f,-2)=\dfrac{1}{9}$

$z=1,$ order $2$ pole

$rez(f,1)=-\dfrac{1}{9}$