Question

Let $z$ and $\omega$ be two complex numbers such that $\left|z\right|\le 1,\left|\omega \right|\le 1$ and $|z-i\omega |=|z-i\overline{\omega }|=2$, and $z$ equals

• A. $1$ or $i$
• B. $i$ or $-i$
• C. $1$ or $-1$
• D. $i$ or $-1$

Answer 1

The correct option is B $i$ or $-i$

Given:

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

$⟹|z-i\omega {|}^2=4$

$\Rightarrow (z-i\omega )(\overline{z}+i\overline{\omega })=4$

$\Rightarrow z\overline{z}+iz\overline{\omega }-i\omega \overline{z}+\omega \overline{\omega }=4$ ....(1)

Similarly,

$⟹|z-i\overline{\omega }{|}^2=4$

$\Rightarrow (z-i\overline{\omega })(\overline{z}+i\omega )=4$

$\Rightarrow z\overline{z}+iz\omega -i\overline{\omega }\overline{z}+\omega \overline{\omega }=4$ ....(2)

From (2) - (1) , we get

$z(\omega -\overline{\omega })-\overline{z}(\overline{\omega }-\omega )=0$

$\Rightarrow (z+\overline{z})(\omega -\overline{\omega })=0$

Case I:

$\omega -\overline{\omega }=0$

$⟹\omega$ is an purely real number

Case II:

$z+\overline{z}=0$

$⟹z$ is an purely imaginary number

Now,

$\Rightarrow |z-i\omega |\le |z|+|i\omega |\le 2$

Because, it is given that $|z|,|\omega |\le 1$

But, given is $|z-i\omega |=2$

Which is only possible when

$|z|=|\omega |=1$

Since, $z$ is an purely imaginary number and its modulus is $1$.

Therefore, $z$ is either $i$ or $-i$

Hence, option B.