Question

determine the poles and the residues at each pole of the function f(z)=z²-2z/(z+1)²(z²+4)

Answer 1

$f(z)=\frac{z^2-2z}{(z+1)^2(z^2+4)}=\frac{z^2-2z}{(z+1)^2(z+2i)(z-2i)}$

This function is rational, thus, its poles are roots od the denominator. Therefore, this function has poles at $z=-1$ (of multiplicity 2), $z=2i$ and $z=-2i$ (of multiplicity 1).

${\rm res}_{z=1}f(z)=\frac{d}{dz}[(z+1)^2f(z)]_{z=1}=\frac{d}{dz}[\frac{z^2-2z}{z^2+4}]_{z=1}$

$\frac{d}{dz}[\frac{z^2-2z}{z^2+4}]=\frac{(2z-2)(z^2+4)-(z^2-2z)\cdot 2z}{(z^2+4)^2}=\frac{2z^2+8z-8}{(z^2+4)^2}$, therefore,

${\rm res}_{z=1}f(z)=[\frac{2z^2+8z-8}{(z^2+4)^2}]_{z=1}=\frac{2}{25}$

${\rm res}_{z=2i}f(z)=\lim\limits_{z\to2i}[(z-2i)f(z)]=\lim\limits_{z\to2i}[\frac{z^2-2z}{(z+1)^2(z+2i)}]$

$=\frac{(2i)^2-2\cdot 2i}{(2i+1)^2(2i+2i)}=\frac{i-1}{(2i+1)^2}=\frac{(i-1)(2i-1)^2}{(2i+1)^2(2i-1)^2}=\frac{7+i}{25}$

${\rm res}_{z=-2i}f(z)=\lim\limits_{z\to-2i}[(z+2i)f(z)]=\lim\limits_{z\to-2i}[\frac{z^2-2z}{(z+1)^2(z-2i)}]$

$=\frac{(-2i)^2-2\cdot(- 2i)}{(1-2i)^2(-2i-2i)}=\frac{-i-1}{(1-2i)^2}=\frac{(-i-1)(2i+1)^2}{(1-2i)^2(2i+1)^2}=\frac{7-i}{25}$