complex integration along closed contour Let  $I_r=  \int dz/(z(z-1)(z-2))$ along $C_r$,  where $C_r = \{z\in\mathbb C : |z|=r\}$, $r>0$. Then

a.    $I_r= 2\pi i$  if $r\in (2,3)$
b.    $I_r= 1/2$  if $r\in (0,1)$
c.    $I_r= -2\pi i$  if $r\in (1,2)$
d.    $I_r= 0$ if $r>3$.

I am stuck on this problem . Can anyone help me please?
all options are looking wrong by using residue theorem......
 A: Define $f(z)=\frac{1}{z(z-1)(z-2)}$ then we have $$\forall r\in(0,1) :\; I_{r}=2\pi iRes[f;0]$$
$$\forall r\in(1,2) :\; I_{r}=2\pi i(Res[f;0]+Res[f,1])$$
$$\forall r\in (2,\infty) :\; I_{r}=2\pi i(Res[f;0]+Res[f,1]+Res[f;2])$$
Now  because $f(z)=\frac{1}{z(z-1)(z-2)}$ is at shape $f(z)=\frac{p(z)}{q(z)}$ when $p(z)=1$ and $q(z)=z(z-1)(z-2)$ that both are analytic in points $z_{0}=0,1,2$ and $p(z_{0})\neq 0$ , $q(z_{0})=0$ and $q'(z_{0})\neq 0$ for these three points. So $f$ has three simple pole there. And so $Res[f;z_{0}]=lim_{z\longrightarrow z_{0}}(z-z_{0})f(z)$ in those points.
$$Res[f;0]=lim_{z\longrightarrow 0}(z)\frac{1}{z(z-1)(z-2)}=\lim_{z\longrightarrow 0}\frac{1}{(z-1)(z-2)}=\frac{1}{2}$$
$$Res[f;1]=lim_{z\longrightarrow 1}(z-1)\frac{1}{z(z-1)(z-2)}=\lim_{z\longrightarrow 1}\frac{1}{z(z-2)}=-1$$
$$Res[f;2]=lim_{z\longrightarrow 2}(z-2)\frac{1}{z(z-1)(z-2)}=\lim_{z\longrightarrow 2}\frac{1}{z(z-1)}=\frac{1}{2}$$
And at the end;
$$\forall r\in(0,1) :\; I_{r}=\pi i$$
$$\forall r\in(1,2) :\; I_{r}=-\pi i$$
$$\forall r\in(2,\infty) :\; I_{r}=0$$
So the last choice "d" is correct choice.
