Question

Let $\left|z\right|=2$ and $w=\frac{z+1}{z-1}$, $z,w\in C$ (where C is the set of complex numbers). Then product of least and greatest value of modulus of w is

Answer 1

Given that $\left|z\right|=2$

let $z=2(\cos \theta +i\sin \theta )$

now, ${\left|w\right|}^2=w\overline{w}=\left(\frac{z+1}{z-1}\right)\left(\frac{\overline{z}+1}{\overline{z}-1}\right)=\frac{{\left|z\right|}^2+z+\overline{z}+1}{{\left|z\right|}^2-z-\overline{z}+1}=\frac{5+4\cos \theta }{5-4\cos \theta }=\frac{10}{5-4\cos \theta }-1$

Since, $-1\le \cos \theta \le 1$

$\therefore max\left({\left|w\right|}^2\right)min\left({\left|w\right|}^2\right)=(\frac{10}{5-4}-1)(\frac{10}{5+4}-1)=1$
Ans: 1