Question

find the residue of z3/(z2-1) at z= [infinity]

Answer 1

Answer:

your answar;-

f(z) = z + 1 z3(z2 + 1) has isolated singularities at z = 0,±i and a zero at z = −1. We will ... pole of order n, a function behaves like 1/(z − z0)n. ... We define the residue of f at infinity by.

Answer 2

Answer:

The residue of the given function at $z=\infty$ is $Res[f(z),z=\infty]=-1$

Step-by-step explanation:

Let the given function be $f(z)=\frac{z^{3} }{z^{2} -1}$

we are required to find the residue of the given function at z=∞.

The given function can be written in the following way by factorizing the denominator-$f(z)=\frac{z^{3} }{(z -1)(z+1)}$

Hence we first find the residues at z=1 and z=-1.They are calculated as shown below-

$Res[f(z),z=1]= \lim_{z\to 1} (z-1)f(z)=\lim_{z\to 1 }\frac{z^{3} }{z+1} =\frac{1}{2}$  Similarly we have

$Res[f(z),z=-1]= \lim_{z\to -1} (z+1)f(z)=\lim_{z\to- 1 }\frac{z^{3} }{z-1} =\frac{1}{2}$

We know that for a function,sum of residues at all the points must be equal to zero.Therefore,Mathematically

$Res[f(z),z=-1]+Res[f(z),z=1]+Res[f(z),z=\infty]=0$

Substituting the values of the residues in above relation,we get

$\frac{1}{2} +\frac{1}{2}+Res[f(z),z=\infty]=0= > Res[f(z),z=\infty]=-1$

Therefore,the residue of the given function is$Res[f(z),z=\infty]=-1$

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