Question

Determine the poles and residue at each pole of the function
f(z) = z2/(z-1)2 (z+2)

Answer 1

Answer:

sorry but I really don't know the answer

Answer 2

The function has poles at z =1 (of multiplicity 2), and z = -2 (of multiplicity 1). The residue at z = 1 is $\frac{5}{9}$ and residue at z = -2 is $\frac{-8}{81}$.

Given:

A function   $f(z)=\frac{z^2}{(z-1)^2(z+2)}$.

To find:

The poles and residue at each pole of the function.

Solution:

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

Finding residues at z = 1:

$Res_{z=1}f(z) =$$\frac{d}{dz}[(z-1)^2 f(z)]_{z=1}$

                  $= \frac{d}{dz}[\frac{z^2}{z+2} ]_{z=1}$

                  $= [\frac{(z+2)2z-z^2}{(z+2)^2}]_{z=1}$

                  $=\frac{(1+2)*2(1)-(1)^2}{(1+2)^2}$

                  $=\frac{6-1}{9}$

$Res_{z=1}f(z) =$ $\frac{5}{9}$

Residue at z = -2:

$Res_{z=-2}f(z) =$$\frac{d}{dz}[(z+2) f(z)]_{z=-2}$

                     $= \frac{d}{dz}[\frac{z^2}{(z-1)^2} ]_{z=-2}$

                     $= [\frac{(z-1)^2*(2z)-z^2*2(z-1)}{(z-1)^4}]_{z=-2}$

                     $=\frac{(-2-1)^2*(2(-2)-(-2)^2*2(-2-1)}{(-2-1)^4}$

                     $=\frac{9(-4)-(4)(-6)}{81}$

                     $=\frac{-36+24}{81}$

$Res_{z=-2}f(z) =$ $=\frac{-8}{81}$

Hence,the function has poles at z =1 (of multiplicity 2), and z = -2 (of multiplicity 1). The residue at z = 1 is  and residue at z = -2 is $\frac{-8}{81}$.

#SPJ2