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. 
 A: Use Cauchy's Residue Theorem:$$\oint_\gamma f(z)dz=2\pi i\sum_{a_i\in A}\operatorname{Res}_{z=a_i}f(z)$$
When $\,A=\,$interior of the rectifiable curve $\,\gamma\,$ which meets no poles of $\,f\,$ .
Note that taking $\,r\in (2,3)\,$ or taking $\,r>3\,$ is the same regarding this integral (why?), and since all the function's poles are simple you can easily calculate its residue at pole $\,a_k\,$ by evaluating $$\lim_{z\to a_k}(z-a_k)f(z)$$  with $$f(z):=\frac{1}{z(z-1)(z-2)}$$
Added For any $\,r>0\,\,,\mathcal{C}_r\,$ is a circle centered at the origin and radius $\,r\,$, thus for instance:
$\,(2)\,$ For $\,r\in (1,2)\,\,,\,\mathcal{C}_r\,$ is a circle centered at the origin that intersects the $x-$axis at some point between $\,1\,$ and $\,2\,$, thus the inner part of this circle, $\,A\,$ (which is inclosed by the path $\,|z|=r\,$ , the circle's perimeter if you will) only contains the poles $\,0,1\,$of the function $\,f(z)\,$, and thus here $$I_r=2\pi i\sum_{a_i\in A}\operatorname{Res}_{z=a_i}f(z)=2\pi i\left(\frac{1}{2}+(-1)\right)=-\pi i$$
Why? Because for example, as stated above: $$\operatorname{Res}_{z=1}f(z)=\lim_{z\to 1}\left[(z-1)\frac{1}{z(z-1)(z-2)}\right]=\frac{1}{1\cdot (1-2)}=-1$$
Similarly, the residue at $\,z=0\,$ equals $\,1/2\,$, as you can readily check, and now you can try the other options...
Ps. The formula above to evaluate the residues works for simple poles ...!
A: 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 $r >3.$
