Question

If $z^{2}-2 z \cos \theta+1=0$, 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)

We have,
$z^{2}-2 z \cos \theta+1=0$
$\Rightarrow z=\frac{2 \cos \theta \pm \sqrt{4 \cos ^{2} \theta-4}}{2}$
$=\cos \theta \pm \sqrt{\cos ^{2} \theta-1}$
$=\cos \theta \pm \sqrt{-\sin ^{2} \theta}$
$=\cos \theta \pm \sqrt{i^{2} \sin ^{2} \theta}$
$=\cos \theta \pm i \sin \theta$
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 $