MDoubt No 000010 - GATE

GATE [EC, 2010, 2 marks]
Question : The residues of a comlex function $\displaystyle X(z)=\dfrac{1-2z}{z(z-1)(z-2)}$ at its poles are

(a) $\displaystyle\dfrac{1}{2},-\dfrac{1}{2}$ and $1$
(b) $\displaystyle\dfrac{1}{2},\dfrac{1}{2}$ and $-1$
(c) $\displaystyle\dfrac{1}{2},1$ and $\displaystyle -\dfrac{3}{2}$
(d) $\displaystyle\dfrac{1}{2},-1$ and $\displaystyle\dfrac{3}{2}$

Solution : $X(z)=\dfrac{1-2z}{z(z-1)(z-2)}$
The poles are $z=0,1,2$
Residue at $z=0$
$\displaystyle =\lim_{z\to 0}(z-0)\frac{1-2z}{z(z-1)(z-2)}$
$=\dfrac{1-2\times 0}{(0-1)(0-2)}$
$=\dfrac{1}{2}$
Residue at $z=1$
$\displaystyle =\lim_{z\to 1}(z-1)\frac{1-2z}{z(z-1)(z-2)}$
$\dfrac{1-2\times 1}{1(1-2)}$
$=1$
Residue at $z=2$
$\displaystyle =\lim_{z\to 2}(z-2)\frac{1-2z}{z(z-1)(z-2)}$
$\dfrac{1-2\times 2}{2(2-1)}$
$=-\dfrac{3}{2}$
CORRECT ANSWER : C