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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:

  1. $ I_r = 2 \pi\ i $ if $r\in(2,3)$
  2. $ I_r = -2 \pi\ i $ if $r\in(1,2)$
  3. $ I_r = 0 $ if $r >3$

Please suggest which option is correct.

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You have a typo in your definition of $C_r$? What have you tried? –  Jonas Meyer Jun 8 '12 at 17:21
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I think you mean $C_r=\{z\in \mathbb{C}:\ |z|=r\}$, right? –  Potato Jun 8 '12 at 17:23
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Hint: Compute the residues of $\frac{1}{z(z-1)(z-2)}$ at each of its three poles. The value of the integral is ($2\pi i$ times) a sum of some of these residues (which ones?). –  mrf Jun 8 '12 at 18:33
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@preeti: It is not, in fact, differentiable everywhere, since it has poles at $0,1,2$. The region is the open disk of radius $r$ about the origin. Do you know how to find the residues at each of the three singularities? If so, the hint given by mrf is really all you need. If you don't, there are other ways to go about it, using integration by parts, continuous deformations of $C_r$, and winding numbers –  Cameron Buie Jun 8 '12 at 20:23
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@preeti Draw a picture. A circle centered at the origin of radius between 2 and 3 will enclose all of the poles. (So will a circle of radius larger than 3.) In other words, the integrand is not holomorphic on the interior of the circle, so Cauchy's theorem cannot be used directly. –  mrf Jun 8 '12 at 20:24
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2 Answers

up vote 3 down vote accepted

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 ...!

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Thanks for your reply. Please see my doubts above. –  preeti Jun 9 '12 at 17:24
    
@Preeti See my added coments in my above answer –  DonAntonio Jun 9 '12 at 20:46
    
@DanAntonio: Many thanks for your detailed reply. I have understood my mistake. If r > 3, then Res f(z) at z = 0 is (1/2), Res f(z) at z = 1 is (- 1) , Res f(z) at z = 2 is (1/2). Therefore, the sum of the residues is 0 and hence the value of the integral is 0. Hence, the third option is correct. –  preeti Jun 10 '12 at 18:20
    
@Preeti Indeed, it is the third one. Good for you. –  DonAntonio Jun 10 '12 at 18:25
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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 $r >3.$

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