Question

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

• A. $1ori$
• B. $ior-i$
• C. $i-1$
• D. $-1,1$

Answer 1

The correct option is D $-1,1$

Given $2=|z+iw|\le \left|z\right|+\left|w\right|$ $\therefore \left|z\right|+\left|w\right|\ge 2$ ......(i)

Also $\left|z\right|\le 1,\left|w\right|\le 1\Rightarrow \left|z\right|+\left|w\right|\le 2$ .....(ii)

$\therefore$ From (i) &(ii) $\left|z\right|=\left|w\right|=1$ and $|z+iw|=|z-i\overline{w}|$

$\Rightarrow {|z+iw|}^2={|z-i\overline{w}|}^2$

$\Rightarrow (z+iw)\left(\overline{z+iw}\right)=(z-iw)\left(\overline{z-i\overline{w}}\right)$ [$\because |z{|}^2=z.\overline{z}$]

$\Rightarrow (z+iw)(\overline{z}-i\overline{w})=(z-i\overline{w})(\overline{z}+iw)$ [$\because \overline{z_1+z_2}={\overline{z}}_1+{\overline{z}}_2\overline{z_1.z_2}={\overline{z}}_1.{\overline{z}}_2\overline{i.w}=-i\overline{w}$]

$\Rightarrow w\overline{z}-z\overline{w}+\overline{z}\overline{w}-wz=0$ $\Rightarrow (\overline{z}-z)(w+\overline{w})=0\Rightarrow \overline{z}=z,w+\overline{w}=0\Rightarrow Imz=0,Re\left(w\right)=0$

Also $\left|z\right|=1,\left|w\right|=1\Rightarrow z=1,-1$ & $w=i,-i$