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

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