Question

Let $z_1$ and $z_2$ be two complex numbers such that $\left|z_2\right|=1$ and $\frac{z_1-1}{z_1+1}=\left(\frac{z_2-1}{z_2+1}\right)^2$, then find the value of $\left| z _1-1\right|+\left| z _1+1\right|$.

Answer 1

$\because\left| z _2\right|=1 \Rightarrow z _2= e ^{ i \theta}=\cos \theta+ i \sin \theta$
$\therefore \frac{ z _2-1}{ z _2+1}=\frac{\cos \theta-1+ i \sin \theta}{\cos \theta+1+ i \sin \theta}=\frac{-2 \sin ^2 \frac{\theta}{2}+ i \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+ i \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\frac{ i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}= i \tan \frac{\theta}{2}$
$\therefore \frac{ z _1-1}{ z _1+1}=-\tan ^2 \frac{\theta}{2} \Rightarrow z _1-1=- z _1 \tan ^2 \frac{\theta}{2}-\tan ^2 \frac{\theta}{2}$
$\Rightarrow z _1=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}=\cos \theta$
which is purely real and lies in $[-1,1]$
$\therefore\left| z _1-1\right|+\left| z _1+1\right|=2$