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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.

  • 1
    @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.2012-06-08

2 Answers 2

3

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 ...!

  • 0
    @Preeti Indeed, it is the third one. Good for you.2012-06-10
2

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 $r >3.$