Question

If z1=a+ib and z1=c+id are complex numbers such that |$z_1$|=|$z_2$|=1 and R($z_1\overline{z_2}$) =0, then the pair of complex numbers w1=a+ic and w2=b+id satisfies

• A.
• B.
• C.
• D. All the above

Answer 1

The correct option is D

All the above

Since|$z_1$|=|$z_2$|=1, we have

$z_1$=cos${\theta }_1$+isin${\theta }_1$,$z_2$=cos${\theta }_2$+isin${\theta }_2$

where ${\theta }_1$=arg($z_1$) and ${\theta }_2$=arg($z_2$)

Also, $z_1$=a+ib and $z_2$=c+id

Therefore a=cos${\theta }_1$, b=sin${\theta }_1$, c=cos${\theta }_2$ and d=sin${\theta }_2$,

Also, R($z_1\overline{z_2}$) =0

$\Rightarrow$ R[(cos${\theta }_1$+isin${\theta }_1$)(cos${\theta }_2$-isin${\theta }_2$)]=0

$\Rightarrow$ R[cos(${\theta }_1$-${\theta }_2$)+isin(${\theta }_1$-${\theta }_2$)]=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${\theta }_2$=cos${\theta }_1$+isin${\theta }_1$

$\Rightarrow$ |$w_1$|=1

similarly, |$w_2$|=1

Next $w_1\overline{w_2}$ = (cos${\theta }_1$+isin${\theta }_1$) (cos${\theta }_2$-isin${\theta }_2$)

=cos(${\theta }_1$-${\theta }_2$)+isin(${\theta }_1$-${\theta }_2$) $\Rightarrow$ |$w_1\overline{w_2}$|=1

Finally, R($\overline{w_1}{w}_2$) = R($w_2\overline{w_1}$)

= R[cos${\theta }_2$+isin${\theta }_2$) (cos${\theta }_1$-isin${\theta }_1$]

=R[cos(${\theta }_2$-${\theta }_1$)+isin(${\theta }_2$-${\theta }_1$)]

=cos(${\theta }_2$-${\theta }_1$)=cos$\left(\frac{\pi }{2}\right)$=0