Question

Let $z$ & $\omega$ be two complex numbers such that $|z|=1,|\omega |=1$ 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 C 1 or -1

Let $z=cos\theta +i.sin\theta ,\omega =cos\alpha +i.sin\alpha$
So, $|z+i\omega |=\left|\right(cos\theta -sin\alpha )+i(sin\theta +cos\alpha \left)\right|$
Also, $|z-i\overline{\omega }|=\left|\right(cos\theta -sin\alpha )+i(sin\theta -cos\alpha \left)\right|$
Since these are equal, $cos\alpha =0=>sin\alpha =1$
Since the values of the modulus is 2, we have:
${(cos\theta -sin\alpha )}^2+{\left(sin\theta \right)}^2=4=>{cos}^2\theta +{sin}^2\alpha +{sin}^2\theta -2cos\theta sin\alpha =4$
Thus, $cos\theta \ast sin\alpha =1=>cos\theta =sin\alpha =\pm 1$
This implies $sin\theta =cos\alpha =0$
Hence (c) is correct.