Question

If z and $\omega$ are two unimodular complex number such that $z^2+{\omega }^2=\frac{3}{2},$ then ${|z+\overline{z}|}^2+{|\omega +\overline{\omega }|}^2$ is

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

Let $z=\cos \theta +i\sin \theta$
$w=\cos \varphi +i\sin \varphi$
$z^2+w^2=\cos 2\theta +\cos 2\varphi +i(\sin 2\theta +\sin 2\varphi )$
$z^2+w^2=\frac{3}{2}\Rightarrow \cos 2\theta +\cos 2\varphi =\frac{3}{2},\sin 2\theta +\sin 2\varphi =0$
Now, ${|z+\overline{z}|}^2+{|w+\overline{w}|}^2=4{\cos }^2\theta +4{\cos }^2\varphi$
$2(1+\cos 2\theta )+2(1+\cos 2\varphi )$
$=4+2\frac{3}{2}=7$