Question

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

Given, $|z+i\omega |=|z-i\overline{\omega }|=2\Rightarrow |z-(-i\omega )|=|z-\left(i\overline{\omega }\right)|=2\Rightarrow |z-(-i\omega )|=|z-\left(i\overline{\omega }\right)|$
$\therefore$ z lies on the perpendicular bisector of the line joining $-i\omega andi\overline{\omega }.$ Since $i\overline{\omega }$ is the mirror image of $-i\omega$ in the x-axis, the locus of z is the x-axis.
$Letz=x+iyandy=0.Now,|z|\le 1\Rightarrow x^2+0^2\le 1\Rightarrow -1\le x\le 1.$
$\therefore$ z may take values given in option (c).