Question

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

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

Answer 1

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

The correct options are
A $|w_1|=1$
B $|w_2|=1$
C $Re\left(w_1\overline{w_2}\right)=0$

$z_1=a+ib\left|z_1\right|=1z_2=c+id\left|z_2\right|=1Re\left(z_1\overline{z_2}\right)=0$
$z_1=a+ib=cis\left(A\right)=\cos A+i\sin A$ ..{$\because \left|z_1\right|=1$}
$z_2=c+id=cis\left(B\right)=\cos B+i\sin B$ ...{$\because \left|z_2\right|=1$}
$⟹a=\cos A\&b=\sin A$
$Re\left(cis\left(A\right)cis(-B)\right)=0⟹\cos (A-B)=0⟹A-B=\frac{\pi }{2}$
$⟹z_2=\cos (A-\frac{\pi }{2})+i\sin (A-\frac{\pi }{2})=\sin A-i\cos A$
$⟹c=\sin A\&d=-\cos A$
$w_1=a+ic=\cos A+i\sin A=cisA{w}_2=b+id=\sin A-i\cos A=-i\left(i\sin A+\cos A\right)=-icis\left(A\right)$
$⟹\left|w_1\right|=1\&\left|w_2\right|=1w_1\overline{w_2}=icisAcis(-A)=i⟹Re\left(w_1\overline{w_2}\right)=0$
Hence, options A,B and C are correct.