Question

$\text{If}{z}_1=a+ib\text{and}{z}_2=c+id$ are complex numbers such that $|z_1|=|z_2|=1\text{and}Re\left(z_1\overline{z_2}\right)=0$, then the pair of complex numbers $w_1=a+ic\text{and}{w}_2=b+id$ satisfies

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

Answer 1

The correct option is D All of these

$\text{Given}|z_1|=1,|z_2|=1$
$\text{Let}{z}_1=\cos {\theta }_1+i\sin {\theta }_1,z_2=\cos {\theta }_2+i\sin {\theta }_2$
$\Rightarrow a=\cos {\theta }_1,b=\sin {\theta }_1,c=\cos {\theta }_2,d=\sin {\theta }_2$
$\text{Given}Re\left(z_1\overline{z_2}\right)=0$
$\Rightarrow \cos ({\theta }_1-{\theta }_2)=0$
$\Rightarrow {\theta }_1-{\theta }_2=\frac{\pi }{2},\text{i.e.},{\theta }_1=\frac{\pi }{2}+{\theta }_2\cdots \left(i\right)$

$\text{Now},w_1=a+ic$
$=\cos {\theta }_1+i\cos {\theta }_2=\cos {\theta }_1+i\sin {\theta }_1\left[\text{Using}\right(i\left)\right]$
$\Rightarrow |w_1|=1$
$\text{Similarly},|w_2|=1$

$\text{Now},w_1\overline{w_2}=(\cos {\theta }_1+i\sin {\theta }_1)(\cos {\theta }_2-i\sin {\theta }_2)$
$=\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)$
$=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i$
$\Rightarrow Re\left(w_1\overline{w_2}\right)=0$