Show that $\text{Im}(\tfrac{e^{i \theta}}{1-xe^{i \theta}})=\frac{\sin(\theta)}{x^2-2x \cos(\theta)+1}$ Let $0<\theta<\frac{\pi}{2}$ and $0\le x \le1$, show that :
$$\text{Im}\left(\frac{e^{i \theta}}{1-xe^{i \theta}}\right)=\frac{\sin(\theta)}{x^2-2x \cos(\theta)+1}$$
I tried to look for the exponential form of $1-xe^{i \theta}$, so i've found $1-xe^{i\theta}=re^{i(-\alpha)}$ where $r = \sqrt{1+x^2-2x\cos(\theta)}$ and $\alpha= \arctan(\frac{x\sin(\theta)}{1-x\cos{\theta}})$. Thus $\frac{e^{i \theta}}{1-xe^{i \theta}}=\frac{1}{r}e^{i(\theta-\alpha)}$, and i'm stuck here. Is there a way to easily compute $\alpha$, or another way to solve this ?
 A: Note that\begin{align*}\frac{e^{i\theta}}{1-xe^{i\theta}}&=\frac{\cos(\theta)+i\sin(\theta)}{1-x\cos(\theta)-xi\sin(\theta)}\\&=\frac{(\cos(\theta)+i\sin(\theta))(1-x\cos(\theta)+xi\sin(\theta))}{(1-x\cos(\theta))^2+x^2\sin^2(\theta)}\\&=\frac{\cos(\theta)-x+i\sin(\theta)}{x^2-2x\cos(\theta)+1}.\end{align*}
A: Obtain a fraction with a real denominator by multiplying the top and bottom of the fraction 
by the complex conjugate of the denominator:
$$\text{Im}\left(\frac{e^{i \theta}}{1-xe^{i \theta}}\right)=\text{Im}\left(\dfrac{e^{i\theta}(1-xe^{-i\theta})}{(1-xe^{i\theta})(1-xe^{-i\theta})}\right)=\text{Im}\left(\dfrac{e^{i\theta}-x}{1-x(e^{i\theta}+e^{-i\theta})+x^2}\right).$$
Can you take it from here?
A: Well if $z=a+bi$, we note that $2bi = (z-\bar z)$. So using the properties of conjugation
$$\frac{e^{i\theta}}{1-xe^{i\theta}}-\frac{e^{i\cdot(-\theta)}}{1-xe^{i\cdot(-\theta)}}$$
Now for the sake of algebraic simplicity, I'll let $e^{i\theta}=a$, so
$$\frac{a}{1-xa}-\frac{1/a}{1-x/a}=\frac{a}{1-xa}-\frac{1}{a-x}$$
$$=\frac{a(a-x)+xa-1}{(a-x)(1-xa)}=\frac{a^2-1}{a-xa^2-x+ax^2}=\frac{a-1/a}{x^2-x(a+1/a)+1}$$
Now $2\cos(\theta)=a+1/a$ (left as an exercise by converting $a=\cos(\theta)+i\sin(\theta)$) and $2i\sin(\theta)=a-1/a$ gives the desired conclusion.
