Value of the integral : $ I_r$ =$\int_{C_r}$ $\frac{dz}{z(z-1)(z-2)}$ It is given that $$ I_r =\int_{C_r}\frac{dz}{z(z-1)(z-2)}$$ 
where $ C_r = \{z\in \Bbb{C}: |z|=r\}$ , $ r >0 $, $r\neq 1,2$ . Then which of the following holds: 


*

*$ I_r = 2 \pi\ i $ if $r\in(2,3)$

*$ I_r = -2 \pi\ i $ if $r\in(1,2)$

*$ I_r = 0 $ if $r >3$


Please suggest which option is correct. 

Suppose $f(z)={1\over{z(z-1)(z-2)}}.$
$\implies {1\over z^2}f({1\over z})={1\over z^2}{z\over ({1\over z}-1)({1\over z}-2)}={z\over (1-z)(1-2z)}={{(1-z)-(1-2z)}\over(1-z)(1-2z)}={1\over {1-2z}}-{1\over {1-z}}$
$\implies(1-2z)^{-1}-(1-z)^{-1}$
$\implies$ coefficient of ${1\over z}$ in the expansion is $0\implies I_r = 0 $ if $r >3.$
