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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?

  • 0
    By the residue theorem all option are looking wrong. Am I right????????/2012-12-18

2 Answers 2

1

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 $\pi i$.

0

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.