Question

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

Given, $|ziw|=|z- \overline {iw|}| = 2$ $\Rightarrow \, \, \, \, |z-(-iw)|z-(\overline {iw|})=2$ $\Rightarrow \, \, \, \, |z-(-iw)|=|z-(-\overline {iw)|}$ $\therefore $ z lies on the perpendicular bisector of the line joining - iw and $-\overline {iw|}$. Since,$-\overline {iw|}$ is the mirror image of - iw in the X-axis, the locus of z is the X-axis. Let z = x + i y and y = 0. Now, $\, \, \, \, \, \, \, \, |z|\le 1 \, \Rightarrow \, \, x^2+0^2 \le 1\, \Rightarrow \, \, -1 \le \, x \, \le \, 1$ $\therefore $ 2 may take values given in option (c). Alternate Solution $\, \, \, \, \, \, \, \, \, \, \, \, |z+iw|\le |z|+|iw|$ $\, \, \, \, \, \, \, \, \, \, \, \, =|z|+|w|$ $\, \, \, \, \, \, \, \, \, \, \, \, \le 1+1=2$ $\therefore \, \, \, \, \, \, \, \, \, \, \, |z+iw|\le 2$ $\Rightarrow \, \, \, \, \, \, \, \, \, \, |z+iw|= 2$ holds when $ \, \, \, \, \, \, \, \, \, $ arg z - arg iw = 0 $\Rightarrow \, \, \, \, arg\frac{z}{iw}=0$ $\Rightarrow \, \, \frac{z}{i w}$ is purely real. $\Rightarrow \, \, \frac{z}{w}$ is purely imaginary. Similarly, when $| z - i \overline{w}|=2 , \, then \frac{z}{\overline{w}}is$ purely imaginary Now, given relation $|z+iw|=|z-i\overline{w}|=2$ Put w = t, we get $ \, \, \, \, \, \, \, \, \, \, \, \, \, |z+i^2|=|z+i^2|=2$ $\Rightarrow \, \, \, \, \, \, \, |z-1|=1$ $\Rightarrow \, \, \, \, z=-1 \, \, \, \, \, [\because |z| \le 1]$ Put w = - i , we get $\, \, \, \, \, \, \, \, \, \, \, |z-i^2|=|z-i^2|=1$ $\Rightarrow \, \, \, \, \, \, \, |z+1|=1$ $\Rightarrow \, \, \, \, z=1 \, \, \, \, \, [\because |z| \le 1]$ $\therefore $ 2 = 1 or - 1 is the correct option.