Author Topic: 2.3 Q7  (Read 764 times)

Christopher Xu

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2.3 Q7
« on: November 21, 2018, 05:11:26 PM »
I found the condition that a>b>0 was not enough to determine the singularities, unless we consider seperated cases. However, the answer suggests otherwise. How do we solve it?
Cheers

Ye Jin

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Re: 2.3 Q7
« Reply #1 on: November 21, 2018, 06:23:05 PM »
 $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$
Let z=$e^{i\theta}$$\Rightarrow$$\cos\theta=\frac{z^2+1}{2z}$ $\Rightarrow$ $a+b\cos\theta=a+\frac{b}{2}(z+\frac{1}{z})$,$d\theta=\frac{dz}{iz}$

$\int_{0}^{2\pi}\frac{d\theta}{a+b\cos\theta}$

=$\int_{|z|=1}\frac{dz}{iz(a+\frac{b}{2}(z+\frac{1}{z}))}$

=$\int_{|z|=1}\frac{2dz}{i(2az+bz^2+b)}$

$2az+bz^2+b=0$$\Rightarrow$ $z=\frac{-a\pm\sqrt{a^{2}-b^{2}}}{b}$

a>b>0$\Rightarrow$$\frac{-a}{b}<-1$ $\Rightarrow \frac{-a+ \sqrt{a^{2}-b^{2}}}{b} >-1 $which is inside |z|=1 and $\frac{-a- \sqrt{a^{2}-b^{2}}}{b}$<-1 which is outside |z|=1

$\therefore$ $\int_{|z|=1}\frac{2dz}{i(2az+bz^2+b)}$
     
      =$\int_{|z|=1}\frac{2/i(z+\frac{a+\sqrt{a^{2}-b^{2}}}{b})}{(z-\frac{-a+\sqrt{{a^2}-{b^2}}}{b})}dz$

      =$2\pi i f(z_0)$
     
      =$2\pi i \frac{2}{\frac{2i\sqrt{{a^2}-{b^2}}}{b}}$
     
      =$\frac{2\pi b}{\sqrt{{a^2}-{b^2}}} $
« Last Edit: November 21, 2018, 10:03:44 PM by Ye Jin »

Jingxuan Zhang

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Re: 2.3 Q7
« Reply #2 on: November 21, 2018, 07:51:46 PM »
Ye: $\pm$ is \pm.

Ye Jin

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Re: 2.3 Q7
« Reply #3 on: November 21, 2018, 10:10:15 PM »
Fixed it. Thx. And my answer is different from the textbook answer but I cannot see mistakes in my steps.

Christopher Xu

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Re: 2.3 Q7
« Reply #4 on: November 21, 2018, 10:15:08 PM »
Yeah I got the same answer too