Question

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

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

Answer 1

The correct option is A $1$ or $i$

Let $z=a+ib$ and $w=c+id$.

Then, $z+iw=(a-d)+i(b+c)$ and $z-i\overline{w}=(a-d)+i(b-c)$

So, $Re(z+iw)=Re(z-i\overline{w})$

And, $|z+iw|=|z-i\overline{w}|$

ie, Case 1: $z+iw=z-i\overline{w}$

Then, There are infinitely many solutions.

Case 2: $\overline{z+iw}=z-i\overline{w}$

Then, $z=a$

As $z$ only has real part.

Correct option is C.