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

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

  2. $I_r= 1/2$ if $r\in (0,1)$

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

  4. $I_r= 0$ if $r>3$.

I am stuck on this problem . Can anyone help me please...... I can't solve it with residue theorem......

I don't know where to begin?

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Where are the poles? Around each pole that within the circle of radius $r$, assuming you are integrating CCW, you get the residue of the pole, times $2\pi i$. Remember that the residue is the value of the rest of the function at the pole, if you ignore the factor that is going to infinity. What do you get when you try that? –  Mario Carneiro Dec 18 '12 at 9:11
    
By the residue theorem all option are looking wrong. Am I right????????/ –  pankaj Dec 18 '12 at 11:45

2 Answers 2

You can directly apply the residue theorem. You have a function with three simple poles at $z=0,1,2$. Since they are all simple poles, and the function is of the form $1/f$, you can get the residues just by evaluating $(z-z_{pole})/f$ at each pole.

For example, if $r \in (0, 1)$ then it winds around only the pole at $0$. Plugging $0$ in, you get residue $\frac{1}{(-1)(-2)}=\frac{1}{2}$. By the residue theorem the integral around that contour is $2\pi i \frac{1}{2}$ or $\pi i$.

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As @jshin47 explained @pawan you can directly apply the residue theorem. For $r>3$ you get residue $ \frac{1}{2} + (-1) + \frac{1}{2} = 0 $. By the residue theorem the integral around that contour is $2πi \times 0 $ or $0.$

So, option 4 is right answer.

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