MAT334-2018F > MAT334--Lectures & Home Assignments

2.3 Q7


Christopher Xu:
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?

Ye Jin:
Let z=$e^{i\theta}$$\Rightarrow$$\cos\theta=\frac{z^2+1}{2z}$ $\Rightarrow$ $a+b\cos\theta=a+\frac{b}{2}(z+\frac{1}{z})$,$d\theta=\frac{dz}{iz}$




$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

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

      =$2\pi i f(z_0)$
      =$2\pi i \frac{2}{\frac{2i\sqrt{{a^2}-{b^2}}}{b}}$
      =$\frac{2\pi b}{\sqrt{{a^2}-{b^2}}} $

Jingxuan Zhang:
Ye: $\pm$ is \pm.

Ye Jin:
Fixed it. Thx. And my answer is different from the textbook answer but I cannot see mistakes in my steps.

Christopher Xu:
Yeah I got the same answer too


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