Question

If $z+1/z=2cos\theta$.

Then prove that $\left|\right(z^{2n}-1\left)/\right(z^{2n}+1\left)\right|=\left|tann\theta \right|$.

Answer 1

$z+\frac{1}{z}=2cos\theta$
or $z^2-2cos\theta z+1=0$
or $z=\frac{2cos\theta \pm \sqrt{4co{s}^2\theta -4}}{2}$
$=cos\theta \pm isin\theta$
Taking positive sing, we get
$z=cos\theta +isin\theta$
$\therefore \frac{1}{z}=(cos\theta -isin\theta )$
$\therefore \frac{z^{2n}-1}{z^{2n}+1}=\frac{z^n-\frac{1}{z^n}}{z^n+\frac{1}{z^n}}$
$=\frac{(cos\theta +isin\theta {)}^n-(cos\theta -isin\theta {)}^n}{(cos\theta +isin\theta {)}^n+(cos\theta -isin\theta {)}^n}$
$=\frac{2isinn\theta }{2cosn\theta }$
$=i\tan n\theta$
Taking negative sign, we get
$\frac{z^{2n}-1}{z^{2n}+1}=\frac{-2isinn\theta }{2cosn\theta }=-\tan n\theta$
$⟹\left|\frac{z^{2n}-1}{z^{2n+1}}\right|=|\pm i\tan \theta |=\tan n\theta$
Ans : 1