Question

Let z and w betwo complex numbers such that $|z|\le 1,|w|\le 1and|z+iw|=|z-\overline{iw}|=2$ Then z is equal to

• A. 1 or i
• B. i or -1
• C. 1 or -1
• D. i or -1

Answer 1

The correct option is C

1 or -1

$Letz=a+ib,|z|\le 1\Rightarrow a^2+b^2\le 1andw=c+id,|w|\le 1\Rightarrow c^2+d^2\le 1$
$|z+iw|=a+ib+i(c+id)|=2\Rightarrow (a-d{)}^2+(b+c{)}^2=4.........\left(i\right)$
$|z-i\overline{w}|=|a+ib-i(bc-id)|\Rightarrow (a-d{)}^2+(b-c{)}^2=4.........\left(ii\right)$
From (i) and (ii), we get bc = 0
$\Rightarrow$ Either b =0 or c = 0
If b = 0, then $a^2\le 1$. Then, only possibility is a =1 or -1