Question

If $z_1=a+ib$ and $z_2=c+id$ are two complex numbers such that $\left|z_1\right|=\left|z_2\right|=1$ and $Re(z_1.\overline{z_2})=0$ then for the pair of complex numbers ${\omega }_1=a+ic$ and ${\omega }_2=b+id$ :

• A. $Re\left({\omega }_1\overline{{\omega }_2}\right)=0$
• B. $Re\left({\omega }_1\overline{{\omega }_2}\right)=1$
• C. $\left|{\omega }_1\right|=1$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $Re\left({\omega }_1\overline{{\omega }_2}\right)=0$
C $\left|{\omega }_1\right|=1$

Let $z_1=a+ib,z_2=c+id.$

Then, $\left|z_1\right|=\left|z_2\right|=1\Rightarrow a^2+b^2=1$ and $c^2+d^2=1$ ...(i)

Also, $Re\left(z_1\overline{z_2}\right)=0\Rightarrow \left\{(a+ib)(c-id)\right\}=0$

or, $ac+bd=0$ ...(ii)

or, $ac=-bd$ or $\frac{a}{b}=\frac{d}{-c}=\lambda$ (say)

$\therefore a^2+b^2=1\Rightarrow b^2+b^2{\lambda }^2=1$

or, $b^2(1+{\lambda }^2)=1$ ...(iii)

and, $c^2+d^2=1\Rightarrow d^2(1+\frac{1}{{\lambda }^2})=1$

or, $d^2(1+{\lambda }^2)={\lambda }^2$ ...(iv)

$\therefore \frac{b^2}{d^2}=\frac{1}{{\lambda }^2}$ or $d^2=b^2{\lambda }^2$

Now, $\left|{\omega }_1\right|=a^2+c^2=b^2{\lambda }^2+\frac{d^2}{{\lambda }^2}=b^2{\lambda }^2+\frac{b^2{\lambda }^2}{{\lambda }^2}=b^2(1+{\lambda }^2)=1$

Also, $Re\left({\omega }_1\overline{{\omega }_2}\right)=Re\left\{(a+ic)(b-id)\right\}=(ab+cd)$

$=b^2\lambda +d(-\frac{d}{\lambda })=b^2\lambda -\frac{b^2{\lambda }^2}{\lambda }=0.$

Hence, $\left|{\omega }_1\right|=1$ and $Re\left({\omega }_1\overline{{\omega }_2}\right)=0$