Question

Consider the following complex function:
$f\left(z\right)=\frac{9}{(z-1)(z+1{)}^2}$
Which of the following is one of hte residues of the above function?

• A. $-1$
• B. $\frac{9}{16}$
• C. $2$
• D. $-\frac{9}{4}$

Answer 1

The correct option is D $-\frac{9}{4}$

$f\left(z\right)=\frac{9}{(z-1)(z+1{)}^2}$
Poles are $z=1,-1$
Res {$f\left(z\right);z=1$}$=\frac{9}{(1+1{)}^2}=\frac{9}{4}$
Res {$f\left(z\right);z=-1$} $=\frac{g^{\prime }(-1)}{1!}$
[Where $g\left(z\right)=\frac{9}{z-1}\Rightarrow g^{\prime }\left(z\right)=\frac{-9}{(z-1{)}^2}$]
$=-\frac{9}{(-1-1{)}^2}$
$=-\frac{9}{4}$