Question

If $z_1=a+ib$ and $z_2=c+id$ are complex numbers such that $\left|z_1\right|=\left|z_2\right|=1$ and $Re\left(z_1{\overline{z}}_2\right)=0$, then the pair of complex numbers ${\omega }_1=a+ic$ and ${\omega }_2=b+id$ satisfies

• A. $\left|{\omega }_1\right|=1$
• B. $\left|{\omega }_2\right|=1$
• C. $Re\left({\omega }_1\overline{{\omega }_2}\right)=0$
• D. ${\omega }_1{\overline{\omega }}^2=0$

Answer 1

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

The correct options are
A $\left|{\omega }_1\right|=1$
B $\left|{\omega }_2\right|=1$
C $Re\left({\omega }_1\overline{{\omega }_2}\right)=0$

$z_1=a+ib\&z_2=c+id$ such that $\left|z_1\right|=\left|z_2\right|=1$
Let $z_1=\cos A+i\sin A=cisA\&z_2=\cos B+i\sin B=cisB$
$\Rightarrow a=\cos A,b=\sin A,c=\cos B,d=\sin B$
$Re\left(z_1\overline{z_2}\right)=0\Rightarrow Re\left(cisAcis(-B)\right)=0\Rightarrow Re\left(cis(A-B)\right)=0\Rightarrow \cos (A-B)=0\Rightarrow A-B=\frac{\pi }{2}$
$w_1=a+ic=\cos A+i\cos B\Rightarrow w_1=\cos A+i\cos (A-\frac{\pi }{2})=\cos A+i\sin A=cisA\Rightarrow \left|w_1\right|=1$
$w_2=b+id=\sin A+i\sin B\Rightarrow w_2=\sin A+i\sin (A-\frac{\pi }{2})=\sin A-i\cos A=-icisA\Rightarrow \left|w_2\right|=1$
$w_1\overline{w_2}=-icisAcis(-A)=-i\Rightarrow Re\left(w_1\overline{w_2}\right)=0$
Ans: A,B,C