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

Correct Answer: 2

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