Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Christopher Xu 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
(https://image.ibb.co/j6v2qV/20181121170205.png)

$\cos\theta=\frac{e^{i\theta}+e^{i\theta}}{2}$
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}}} $

Ye: $\pm$ is \pm.

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

Yeah I got the same answer too