Verified Solution • 2025–26 Session moderate

QJEEM23Ma1471: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1471 Solution: Let . Then ∑ z ∈ S

Let S={z∈C:zˉ=i(z2+Re⁡(zˉ))}S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\} . Then ∑z∈S|z|2 is equal to :

Multiple Choice Options

  • A 72\frac{7}{2}
  • B 4
  • C 3
  • D 52\frac{5}{2}

Step-by-Step Verified Solution

Let z=x+iyz=x+i y

amp;zˉ=i(z2+Re⁡(zˉ))amp;⇒x−iy=i(x2−y2+2ixy+x)amp;x−iy=i(x2−y2+x)−2xyamp;x=−2xy⇒x(2y+1)=0amp;⇒x=0,y=−12........(i)amp;−y=x2−y2+x...........(ii)\begin{aligned}&\bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right) \\\\&\Rightarrow x-i y=i\left(x^2-y^2+2 i x y+x\right) \\\\& x-i y=i\left(x^2-y^2+x\right)-2 x y \\\\&x=-2 x y \Rightarrow x(2 y+1)=0 \\\\&\Rightarrow x=0, y=\frac{-1}{2} ........(i)\\\\&-y=x^2-y^2+x ...........(ii)\end{aligned}

Case (I) x=0x=0

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

∴\therefore z=0,iz=0, i

Case (II) y=−12y=\frac{-1}{2}

⇒12=x2−14+x⇒x2+x−34=0\Rightarrow \frac{1}{2}=x^2-\frac{1}{4}+x \Rightarrow x^2+x-\frac{3}{4}=0

⇒\Rightarrow 4x+4x−3=0⇒(2x−1)(2x+3)=04 x+4 x-3=0 \Rightarrow(2 x-1)(2 x+3)=0

⇒\Rightarrow x=12,−32x=\frac{1}{2}, \frac{-3}{2}

amp;z=12−12i,−32−12iamp;∑∣z∣2=0+1+12+52=4\begin{aligned}& z=\frac{1}{2}-\frac{1}{2} i, \frac{-3}{2}-\frac{1}{2} i \\\\& \sum|z|^2=0+1+\frac{1}{2}+\frac{5}{2}=4\end{aligned}

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