# 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......

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For rє(2,3), if I take r=2.5 etc……………. All the options are wrong??? –  prakash Dec 17 '12 at 9:18
For rє(2,3), if I take r=2.5 etc............ and i used residue theorem but none of the options is correct in my calculaton.......am i right???? –  prakash Dec 17 '12 at 9:53
For rє(2,3),can I take r=2.5 etc. or not????????? –  prakash Dec 17 '12 at 10:48
For rє(2,3), if I take r=2.5 etc……… using residue theorem Ir = 2Πi(sum of residue)=2Πi(2-1+2)=6Πi so option a is wrong am i right?????????? –  prakash Dec 17 '12 at 11:22

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.