JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Complex Numbers

Let $z=x+iy$. Then $\overline{z}=x-iy$ and $z^2=x^2-y^2+2ixy$.

The given equation becomes:
$x-iy=(i+1)(x^2-y^2+2ixy)-(x+iy)$.

Expanding the right side and equating real and imaginary parts:
Real: $2x=x^2-y^2-2xy$
Imaginary: $0=x^2-y^2+2xy$.

From the imaginary part: $x^2-y^2+2xy=0$.
Substituting into the real part: $2x=-4xy$, so $x(1+2y)=0$.

Case 1: $x=0$ ⇒ $y=0$ ⇒ $z=0$, $|z{|}^2=0$.

Case 2: $y=-\frac{1}{2}$. From $x^2-y^2+2xy=0$:
$x^2-\frac{1}{4}-x=0$ ⇒ $4x^2-4x-1=0$ ⇒ $x=\frac{1\pm \sqrt{2}}{2}$.

For both values: $|z{|}^2=x^2+\frac{1}{4}=\frac{(1\pm \sqrt{2}{)}^2+1}{4}=\frac{3\pm 2\sqrt{2}+1}{4}=\frac{4\pm 2\sqrt{2}}{4}=1\pm \frac{\sqrt{2}}{2}$.

Sum = $0+\left(1+\frac{\sqrt{2}}{2}\right)+\left(1-\frac{\sqrt{2}}{2}\right)=2$.

Answer: 2.