Question

The residue of $f\left(z\right)=\frac{z^3}{(z-1{)}^4(z-2\left)\right(z-3)}$ at $z=3$ is

• A. $-8$
• B. $\frac{101}{16}$
• C. $0$
• D. $\frac{27}{16}$

Answer 1

The correct option is D $\frac{27}{16}$

Poles of $f\left(z\right)=\frac{z^3}{(z-1{)}^4(z-2\left)\right(z-3)}$ are $z=2,3,$ & $1$
$2$ and $3$ are smiples poles and $1$ is pole of order 4
$\underset{(z=3)}{Resf\left(z\right)}=\underset{z\to 3}{lim}(z-3)f\left(z\right)$$={lim}_{z\to 3}\frac{z^3}{(z-1{)}^4(z-2)}=\frac{27}{16}$