QJEEM22Maa7fe: JEE Main 2022 Mathematics - Class 11 JEEM22Maa7fe Solution: Sum of squares of modulus of all the complex numbers z…

Let $z=x+i y$

So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$

$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and

$$x^2-y^2+2xy=0\dots (ii)$$

From (i) and (ii) we get

$$x=0\text{ or }y=-\frac{1}{2}$$

When $x=0$ we get $y=0$

When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$

$\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$

So there will be total 3 possible values of $z$ , which are $0,\left(\frac{-1+\sqrt{2}}{2}\right)-\frac{1}{2} i$ and $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$

Sum of squares of modulus

$$\begin{array}{rl}& =0+{\left(\frac{\sqrt{2}-1}{2}\right)}^2+\frac{1}{4}+{\left(\frac{\sqrt{2}+1}{2}\right)}^2=+\frac{1}{4}\\ \\ & =2\end{array}$$