Question

If $z^2-2z\cos \theta +1=0$ , then $z^2+z^{-2}$ is equal to

• A. $2\cos 2\theta$
• B. $2\sin 2\theta$
• C. $2\cos \theta$
• D. $2\sin \theta$

Answer 1

Answer: (A)

$2\cos 2\theta$

We have, $z^2-2z\cos \theta +1=0$

$\Rightarrow z=\frac{2\cos \theta \pm \sqrt{4{\cos }^2\theta -4}}{2}=\cos \theta \pm \sqrt{{\cos }^2\theta -1}$

$=\cos \theta \pm \sqrt{-{\sin }^2\theta }=\cos \theta \pm \sqrt{i^2{\sin }^2\theta }=\cos \theta \pm i\sin \theta$

when $z=\cos \theta +i\sin \theta$

$z^2+z^{-2}=\cos 2\theta +i\sin 2\theta +(\cos 2\theta -i\sin 2\theta )=2\cos 2\theta$

and when $z=\cos \theta -i\sin \theta$

$z^2+z^{-2}=\cos 2\theta -i\sin 2\theta +(\cos 2\theta +i\sin 2\theta )=2\cos 2\theta$