Question

$z_1=a+ib$ and $z_2=c+id$ are complex numbers such that $\left|z_1\right|=\left|z_2\right|=1$ and $Re\left(z_1\overline{z_2}\right)=0$. If $w_1=a+ic$ and $w_2=b+id(a,b,c,d\in R),$ then

• A. $\left|w_1\right|=1$
• B. $\left|w_2\right|=1$
• C. $Re\left(w_1\overline{w_2}\right)=0$
• D. $Re\left(w_1\overline{w_2}\right)=1$

Answer 1

Answer: (A), (B), (C)

$Re\left(w_1\overline{w_2}\right)=0$
$\left|w_2\right|=1$
$\left|w_1\right|=1$

A,B,C

We have,

$|a+ib|=1\Rightarrow a+ib=\cos \theta +i\sin \theta$

$|c+id|=1\Rightarrow c+id=\cos \varphi +i\sin \varphi$

$a=\cos \theta ,b=\sin \theta ,c=\cos \varphi ,d=\sin \varphi$

Now, $z_1\overline{z_2}=(\cos \theta +i\sin \theta )(\cos \varphi -i\sin \varphi )$

$Re\left(z_1\overline{z_2}\right)=\cos \theta \cos \varphi +\sin \theta \sin \varphi$

Thus $\Rightarrow Re\left(z_1\overline{z_2}\right)=0\Rightarrow \cos (\theta -\varphi )=0$...............1

$\theta -\varphi =\frac{\pi }{2}or-\frac{\pi }{2}$

Now,

${\left|w_1\right|}^2=a^2+c^2={\cos }^2\theta +{\cos }^2\varphi ={\cos }^2(\varphi \pm \frac{\pi }{2})+{\cos }^2\varphi ={\sin }^2\varphi +{\cos }^2\varphi =1$

$\left|w_1\right|=1,\left|w_2\right|=1$

$w_1{w}_2=\cos \varphi \cos \theta +\sin \theta \sin \varphi =\frac{1}{2}(\sin 2\varphi +\sin 2\theta )=\sin (\theta +\varphi )\cos (\theta +\varphi )=0$