JEE Main 2023 — Mathematics Algebra

Given,

$$\bar{z}=i({z}^{2}+Re(\bar{z}))$$zˉ=i(z2+Re(zˉ))

Now, let $z=x+iy$z=x+iy, so $\bar{z}=x-iy$zˉ=x−iy

Now putting the value in, $\bar{z}=i({z}^{2}+Re(\bar{z}))$zˉ=i(z2+Re(zˉ)) we get,

$$\bar{z}=i({z}^{2}+Re(\bar{z}))$$zˉ=i(z2+Re(zˉ))

$$\Rightarrow x-iy=i({x}^{2}-{y}^{2}+2ixy+x)$$⇒x−iy=i(x2−y2+2ixy+x)

$$\Rightarrow x-iy=i({x}^{2}-{y}^{2}+x)-2xy$$⇒x−iy=i(x2−y2+x)−2xy

Now comparing real part we get,

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

$$\Rightarrow x=0,y=\frac{-1}{2}.....(1)$$⇒x=0,y=2−1​.....(1)

And imaginary part we get,

$$-y={x}^{2}-{y}^{2}+x.......(2)$$−y=x2−y2+x.......(2)

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

$$\Rightarrow -y=-{y}^{2}$$⇒−y=−y2

$$\Rightarrow {y}^{2}-y=0\Rightarrow y=0,1$$⇒y2−y=0⇒y=0,1

So, $z=0,i$z=0,i

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

$$\Rightarrow \frac{1}{2}={x}^{2}-\frac{1}{4}+x$$⇒21​=x2−41​+x

$$\Rightarrow {x}^{2}+x-\frac{3}{4}=0$$⇒x2+x−43​=0

$$\Rightarrow 4x+4x-3=0$$⇒4x+4x−3=0

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

$$\Rightarrow x=\frac{1}{2},\frac{-3}{2}$$⇒x=21​,2−3​

So, $z=\frac{1}{2}-\frac{1}{2}i,\frac{-3}{2}-\frac{1}{2}i$z=21​−21​i,2−3​−21​i

Now finding, $\displaystyle \sum |z{|}^{2}=0+1+\frac{1}{2}+\frac{5}{2}=4$∑∣z∣2=0+1+21​+25​=4