Question

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

• A. number of solutions for $z$ is $1$
• B. number of solutions for $z$ is $2$
• C. number of solutions for $w$ is $1$
• D. number of solutions for $w$ is $2$

Answer 1

Answer: (B), (D)

The correct options are
B number of solutions for $z$ is $2$
D number of solutions for $w$ is $2$

$\because |z+iw|\le |z|+|iw|$
$\Rightarrow 2\le |z|+|i||w|$
$\Rightarrow |z|+|w|\ge 2$
But from given condition, $|z|+|w|\le 2$
$\therefore |z|=|w|=1$

Let $z=\cos {\theta }_1+i\sin {\theta }_1$ and $w=\cos {\theta }_2+i\sin {\theta }_2(\because |z|=|w|=1)$
$|z+iw|=2$
$\Rightarrow |(\cos {\theta }_1-\sin {\theta }_2)+i(\sin {\theta }_1+\cos {\theta }_2)|=2$
$\Rightarrow {\cos }^2{\theta }_1+{\sin }^2{\theta }_2-2\sin {\theta }_2\cos {\theta }_1+{\sin }^2{\theta }_1+{\cos }^2{\theta }_2+2\sin {\theta }_1\cos {\theta }_2=4$
$\Rightarrow 2(\sin {\theta }_1\cos {\theta }_2-\sin {\theta }_2\cos {\theta }_1)=2$
$\Rightarrow \sin ({\theta }_1-{\theta }_2)=1$
$\Rightarrow {\theta }_1-{\theta }_2=\frac{\pi }{2},\frac{-3\pi }{2}\dots \left(1\right)$
$\because {\theta }_1,{\theta }_2\in (-\pi ,\pi ]$
$\Rightarrow {\theta }_1\pm {\theta }_2\in (-2\pi ,2\pi ]$

$|z-i\overline{w}|=2$
$\Rightarrow |(\cos {\theta }_1-\sin {\theta }_2)+i(\sin {\theta }_1-\cos {\theta }_2)|=2$
$\Rightarrow {\cos }^2{\theta }_1+{\sin }^2{\theta }_2-2\sin {\theta }_2\cos {\theta }_1+{\sin }^2{\theta }_1+{\cos }^2{\theta }_2-2\sin {\theta }_1\cos {\theta }_2=4$
$\Rightarrow -2(\sin {\theta }_1\cos {\theta }_2+\sin {\theta }_2\cos {\theta }_1)=2$
$\Rightarrow \sin ({\theta }_1+{\theta }_2)=-1$
$\Rightarrow {\theta }_1+{\theta }_2=-\frac{\pi }{2},\frac{3\pi }{2}\dots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$
$({\theta }_1,{\theta }_2)=(\pi ,\frac{\pi }{2}),(0,-\frac{\pi }{2}),(0,\frac{3\pi }{2}),(-\pi ,\frac{\pi }{2})$
$\because {\theta }_1,{\theta }_2\in (-\pi ,\pi ]$
$\therefore ({\theta }_1,{\theta }_2)=(\pi ,\frac{\pi }{2}),(0,-\frac{\pi }{2})$

$\text{Case-1:}$
${\theta }_1=\pi ,{\theta }_2=\frac{\pi }{2}$
$z=-1,w=i$

$\text{Case-2:}$
${\theta }_1=0,{\theta }_2=-\frac{\pi }{2}$
$z=1,w=-i$

Hence, there are $2$ possibilities for $z$ and $w$.