Question

Suppose z and $\omega$ are two complex numbers such that $|z|\le 1,|\omega |\le 1$, and $|z+i\omega |=|z-i\omega |=2$.

Which of the following is true for z and $\omega$?

• A. $Re\left(z\right)=Re\left(\omega \right)$
• B. $Im\left(z\right)=Im\left(\omega \right)$
• C. $Re\left(z\right)=Im\left(\omega \right)$
• D. $Im\left(z\right)=Re\left(\omega \right)$

Answer 1

The correct option is D $Im\left(z\right)=Re\left(\omega \right)$

Suppose $z=x+iy,\omega =\alpha +i\beta$

$|z+i\omega |=2$

$\Rightarrow |z{|}^2+|\omega {|}^2+i\omega \overline{z}-i\overline{\omega }z=4$

$\Rightarrow i\omega \overline{z}=i\overline{\omega }z=2$ ......... (1)

Also $|z-i\overline{\omega }|=2$

$\Rightarrow |z{|}^2+|\omega {|}^2+i\omega z-i\overline{\omega }\overline{z}=4$

$\Rightarrow i\omega z-i\overline{\omega }\overline{z}=2$

Adding (1) and (2), we get
$\Rightarrow i(\omega +\overline{\omega })(z-\overline{z})=4$

$\Rightarrow i\left(2i\beta \right)\left(2x\right)=4$

$\beta x=-1$....... (3)

Subtracting (1) and (2), we get

$i(\omega +\overline{\omega })(z-\overline{z})=0\Rightarrow \alpha y=0$........ (4)

$\Rightarrow$ either $\alpha =0$ or $y=0$

If $y=0$, then $x^2+y^2=1$

$\Rightarrow x=\pm 1$ and $z=1$ or $-1$

If $\alpha =0$ the $\alpha^2+\beta^2=1\Rightarrow \beta =\pm 1\Rightarrow \omega =\pm i$

So $Im\left(z\right)=Re\left(\omega \right)=0$