Question

Let z and w be two complex numbers such that

$|z|\le 1$, $|w|\le 1$ and |z+iw|= |z- i$\overline{w}$|= 2. Then z is equal to

• 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

Let z=a+ib, $|z|\le b$ $\Rightarrow$ $a^2+b^2\le 1$

And w=c+id, $|w|\le b$ $\Rightarrow$ $c^2+d^2\le 1$

|z+iw|=|a+ib+i(c+id)|=2

$\Rightarrow$ $(a-b{)}^2$ + $(b+c{)}^2$ = 4 $\cdots$$\cdots$(i)

|z+i$\overline{w}$|=|a+ib-i(c-id)|

$\Rightarrow$ $(a-b{)}^2$ + $(b-c{)}^2$ = 4 $\cdots$$\cdots$(i)

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