JEE Main 2022 — Mathematics Algebra

Given,

$$z+\bar{z}={\mathrm{izz}}^{2}+{z}^{2}$$z+zˉ=izz2+z2

Consider $z=x+iy$z=x+iy

$$2x=(i+1)({x}^{2}-{y}^{2}+2xyi)$$2x=(i+1)(x2−y2+2xyi)

On comparing real and imaginary part we get,

$\Rightarrow 2x={x}^{2}-{y}^{2}-2xy$⇒2x=x2−y2−2xy and ${x}^{2}-{y}^{2}+2xy=0$x2−y2+2xy=0

$$\Rightarrow 2x=-4xy$$⇒2x=−4xy

$\Rightarrow x=0$⇒x=0 or $y=\frac{-1}{2}$y=2−1​

Case 1: $x=0\Rightarrow 2\times 0=(i+1)({0}^{2}-{y}^{2}+2\times 0\times yi)y=0$x=0⇒2×0=(i+1)(02−y2+2×0×yi)y=0

Here $z=0$z=0

Case 2: $y=\frac{-1}{2}$y=2−1​

$$\Rightarrow 4{x}^{2}-4x-1=0$$⇒4x2−4x−1=0

$${(2x-1)}^{2}=2$$(2x−1)2=2

$$2x-1=\pm \sqrt{2}$$2x−1=±2​

$$x=\frac{1\pm \sqrt{2}}{2}$$x=21±2​​

Here $z=\frac{1+\sqrt{2}}{2}-\frac{i}{2}$z=21+2​​−2i​ or $z=\frac{1-\sqrt{2}}{2}-\frac{i}{2}$z=21−2​​−2i​

Sum of squares of modulus of $z$z

$$=0+\frac{{(1+\sqrt{2})}^{2}+1}{4}+\frac{{(1-\sqrt{2})}^{2}+1}{4}=\frac{8}{4}=2$$=0+4(1+2​)2+1​+4(1−2​)2+1​=48​=2