Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$Since,|z_1|=|z_2|=1,wehave{z}_1=cos{\theta }_1+isin{\theta }_1,z_2=cos{\theta }_2+isin{\theta }_{2}where{\theta }_1=arg\left(z_1\right)and{\theta }_2=arg\left(z_2\right)Also,z_1=a+iband{z}_2=c+id.Thereforea=cos{\theta }_1,b=sin{\theta }_1,c=cos{\theta }_{2}andd=sin{\theta }_2$ $Also,R\left(z_1\overline{z_2}\right)=0\Rightarrow R\left[\right(cos{\theta }_1+isin{\theta }_1\left)\right(cos{\theta }_2-isin{\theta }_2\left)\right]=0\Rightarrow R\left[\right(cos({\theta }_1-{\theta }_2)+isin({\theta }_1-{\theta }_2)\left)\right]=0\Rightarrow cos({\theta }_1-{\theta }_2)=0\Rightarrow {\theta }_1-{\theta }_2=\frac{\pi }{2}\Rightarrow {\theta }_1={\theta }_2+\frac{\pi }{2}$ $Now,w_1=a+ic=cos{\theta }_1+icos{\theta }_2=co{s}_1+isin{\theta }_1\Rightarrow |w_1|=1Similarly,|w_2|=1Next{w}_1\overline{w_2}=(cos{\theta }_1+isin{\theta }_1)(cos{\theta }_2-isin{\theta }_2)=cos({\theta }_1-{\theta }_2)+isin({\theta }_1-{\theta }_2)\Rightarrow |w_1\overline{w_2}|=1Finally,R\left(\overline{w_1}{w}_2\right)=R\left(w_1\overline{w_2}\right)=R\left[\right(cos{\theta }_2+isin{\theta }_2\left)\right(cos{\theta }_1-isin{\theta }_1\left)\right]=R\left[cos\right({\theta }_2-{\theta }_1)+isin({\theta }_2-isin{\theta }_1\left)\right]=cos({\theta }_2-{\theta }_1)=cos\left(\frac{-\pi }{2}\right)=0$