Question

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

• A. $|w_1|=1$
• B. $|w_2|=1$
• C. $Re\left(w_1\overline{w_2}\right)=0$
• D. none

Answer 1

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

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

$z_1=e^{i{\theta }_1}=a+ib{z}_2=e^{i{\theta }_2}=c+idRe\left(z_1\overline{z_2}\right)=0Re\left(e^{i({\theta }_1-{\theta }_2)}\right)=0$

$Re\left(\cos \right({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2\left)\right)=0\cos ({\theta }_1-{\theta }_2)=0{\theta }_1-{\theta }_2=\frac{\pi }{2}\Rightarrow {\theta }_2={\theta }_1-\frac{\pi }{2}$

$z_1=a+ib=\cos {\theta }_1+i\sin {\theta }_1\Rightarrow a=\cos {\theta }_1;b=\sin {\theta }_1{z}_2=c+id=\cos {\theta }_2+i\sin {\theta }_2{z}_2=cos({\theta }_1-\frac{\pi }{2})+i\sin ({\theta }_1-\frac{\pi }{2})z_2=\sin {\theta }_1-\cos {\theta }_1\Rightarrow a=\sin {\theta }_1;b=-\cos {\theta }_1$

$w_1=a+ic=\cos {\theta }_1+i\sin {\theta }_1\Rightarrow |w_1|=1w_2=b+id=\sin {\theta }_1-i\cos {\theta }_1\Rightarrow |w_2|=1$

$Re\left(w_1\overline{w_2}\right)=Re\left(\right(\cos {\theta }_1+i\sin {\theta }_1\left)\right(\sin {\theta }_1+i\cos {\theta }_1\left)\right)=Re\left(i\right)=0$

So options (A) , (B) and (C) are correct.