Question

In the Laurent expansion of $f\left(z\right)=\frac{1}{(z-1)(z-2)}$ valid in the region $1<|z|<2$, the coefficient of $\frac{1}{z^2}$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is D $-1$

$f\left(z\right)=\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}$
$=-\frac{1}{2}\left(\frac{1}{1-\frac{z}{2}}\right)-\frac{1}{z}\left(\frac{1}{1-\frac{z}{2}}\right)$
$(\because 1<|Z|<2)$
$=-\frac{1}{2}{[1-\frac{z}{2}]}^{-1}-\frac{1}{z}{[1-\frac{1}{z}]}^{-1}$
$=-\frac{1}{2}[1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+....]$
$-\frac{1}{z}[1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+....]$
$=(-\frac{1}{2}-\frac{z}{4}-\frac{z^2}{8}-\frac{z^3}{16}-....)-\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3}-....$
So coefficient of $\left(\frac{1}{z^2}\right)$ is $-1$