Question

1. Determine the poles and residue at each poles of

f(z)=2z+1/z²-z-2 over C=|z|=5/2 Hence Evaluate

∮c 2z+1/z²-z-2 dz over C=|z|=5/2

Answer 1

$z²-z-2=(z-2)(z+1)$

Poles:

$z_0=2,z_0=-1$

$res\ f(2)=\displaystyle{\lim_{z\to 2}}(f(z)(z-2))=\displaystyle{\lim_{z\to 2}}(\frac{2z+1}{z+1})=5/3$

$res\ f(-1)=\displaystyle{\lim_{z\to -1}}(f(z)(z+1))=\displaystyle{\lim_{z\to -1}}(\frac{2z+1}{z-2})=1/3$

$f(z_0)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z-z_0}dz$ , $z_0\isin D$

$\frac{1}{(z-2)(z+1)}=\frac{A}{z-2}+\frac{B}{z+1}$

$A(z+1)+B(z-2)=1$

$A+B=0$

$A-2B=1$

$B=-1/3,A=1/3$

$\oint_C\frac{f(z)}{z-z_0}dz=\frac{2\pi i}{3}\oint\frac{2z+1}{z-2}dz-\frac{2\pi i}{3}\oint\frac{2z+1}{z+1}dz=\frac{2\pi i}{3}(2z+1)|_{z=2}-\frac{2\pi i}{3}(2z+1)|_{z=-1}=$

$=\frac{10\pi i}{3}+\frac{2\pi i}{3}=4\pi i$