Question

Sum of squares of modulus of all the complex numbers z satisfying
$\overline{z}=iz^2+z^2–z$
is equal to ________.

Answer 1

The correct answer is 2
Let z = x + iy
So 2x = (1 + i)(x2 – y2 + 2xyi)
⇒ 2x = x2 – y2 – 2xy …(i) and x2 – y2 + 2xy = 0 …(ii)
From (i) and (ii) we get
x = 0 or y$=−\frac{1}{2}$
When x = 0 we get y = 0
When y$=−\frac{1}{2}$
we get $x^2−x−\frac{1}{4}=0$
$⇒x=\frac{−1±\sqrt2}{2}$
So there will be total 3 possible values of z, which are
$0,(\frac{−1+\sqrt2}{2})−\frac{1}{2}i$ and $(\frac{−1−\sqrt2}{2})−\frac{1}{2}i$
Sum of squares of modulus
$=0+(\frac{\sqrt2−1}{2})^2+\frac{1}{4}+(\frac{\sqrt2+1}{2})^2=+\frac{1}{4}$
= 2