JEE Main 2023 — Complex Number Question with Solution

Given,

$\overline{z}=i\left(z^2+Re\left(\overline{z}\right)\right)$

Now, let $z=x+iy$, so $\overline{z}=x-iy$

Now putting the value in, $\overline{z}=i\left(z^2+Re\left(\overline{z}\right)\right)$ we get,

$\overline{z}=i\left(z^2+Re\left(\overline{z}\right)\right)$

$\Rightarrow x-iy=i\left(x^2-y^2+2ixy+x\right)$

$\Rightarrow x-iy=i\left(x^2-y^2+x\right)-2xy$

Now comparing real part we get,

$\Rightarrow x=-2xy\Rightarrow x(2y+1)=0$

$\Rightarrow x=0,y=\frac{-1}{2} .....\left(1\right)$

And imaginary part we get,

$-y=x^2-y^2+x .......\left(2\right)$

Now taking Case (I) when $x=0$ in equation $\left(2\right)$ we get,

$\Rightarrow -y=-y^2$

$\Rightarrow y^2-y=0\Rightarrow y=0,1$

So, $z=0, i$

Now taking Case (II) when $y=\frac{-1}{2}$ in equation $\left(2\right)$ we get,

$\Rightarrow \frac{1}{2}=x^2-\frac{1}{4}+x$

$\Rightarrow x^2+x-\frac{3}{4}=0$

$\Rightarrow 4x+4x-3=0$

$\Rightarrow (2x-1)(2x+3)=0$

$\Rightarrow x=\frac{1}{2},\frac{-3}{2}$

So, $z=\frac{1}{2}-\frac{1}{2}i,\frac{-3}{2}-\frac{1}{2}i$

Now finding, $\sum |z{|}^2=0+1+\frac{1}{2}+\frac{5}{2}=4$