Verified Solution • 2025–26 Session moderate

QJEEAdv23Ma30cc: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Ma30cc Solution: Let be a complex number satisfying , where denotes the complex…

Let zz be a complex number satisfying ∣z∣3+2z2+4zˉ−8=0|z|^3+2 z^2+4 \bar{z}-8=0 , where zˉ\bar{z} denotes the complex conjugate of zz . Let the imaginary part of zz be nonzero.

Match each entry in List-I to the correct entries in List-II.

List - I List - II
(P) ∣z∣2|z|^2 is equal to (1) 12
(Q) ∣z−zˉ∣2|z-\bar{z}|^2 is equal to (2) 4
(R) ∣z∣2+∣z+zˉ∣2|z|^2+|z+\bar{z}|^2 is equal to (3) 8
(S) ∣z+1∣2|z+1|^2 is equal to (4) 10
(5) 7

The correct option is:

Multiple Choice Options

  • A (P)→(1)(Q)→(3)(R)→(5)(S)→(4)(P) \rightarrow(1) \quad(Q) \rightarrow(3) \quad(R) \rightarrow(5) \quad(S) \rightarrow(4)
  • B (P)→(2)(Q)→(1)(R)→(3)(S)→(5)(P) \rightarrow(2) \quad(Q) \rightarrow(1) \quad(R) \rightarrow(3) \quad(S) \rightarrow(5)
  • C (P)→(2)(Q)→(4)(R)→(5)(S)→(1)(P) \rightarrow(2) \quad(Q) \rightarrow(4) \quad(R) \rightarrow(5) \quad(S) \rightarrow(1)
  • D (P)→(2)(Q)→(3)(R)→(5)(S)→(4)(P) \rightarrow(2) \quad(Q) \rightarrow(3) \quad(R) \rightarrow(5) \quad(S) \rightarrow(4)

Step-by-Step Verified Solution

∵∣z∣3+2z2+4zˉ−8=0\because|z|^3+2 z^2+4 \bar{z}-8=0 .........(1)

Take conjugate both sides

⇒∣z∣3+2z‾2+4z−8=0\Rightarrow|\mathrm{z}|^3+2 \overline{\mathrm{z}}^2+4 \mathrm{z}-8=0 ........(2)

amp; By (1)−(2)amp;⇒2(z2−zˉ2)+4(zˉ−z)=0amp;⇒z+z‾=2amp;⇒∣z+z‾∣=2\begin{aligned}& \text { By }(1)-(2) \\\\& \Rightarrow 2\left(\mathrm{z}^2-\bar{z}^2\right)+4(\bar{z}-\mathrm{z})=0 \\\\& \Rightarrow \mathrm{z}+\overline{\mathrm{z}}=2 \\\\& \Rightarrow|\mathrm{z}+\overline{\mathrm{z}}|=2\end{aligned}

Let z=x+iyz=x+i y

∴x=1∴z=1+iy\therefore \mathrm{x}=1 \quad \therefore \mathrm{z}=1+\mathrm{iy}

Put in (1)

amp;⇒(1+y2)3/2+2(1−y2+2iy)+4(1−iy)−8=0amp;⇒(1+y2)3/2=2(1+y2)amp;⇒1+y2=2=∣z∣\begin{aligned}& \Rightarrow\left(1+y^2\right)^{3 / 2}+2\left(1-y^2+2 i y\right)+4(1-i y)-8=0 \\\\& \Rightarrow\left(1+y^2\right)^{3 / 2}=2\left(1+y^2\right) \\\\& \Rightarrow \sqrt{1+y^2}=2=|z|\end{aligned}

Also y=±3\mathrm{y}= \pm \sqrt{3}

amp;∴z=1±i3amp;⇒z−z‾=±2i3amp;⇒∣z−z‾∣=23amp;⇒∣z−z‾∣2=12amp; Now z+1=2+i3amp;∣z+1∣2=4+3=7amp;∴P→2;Q→1;R→3;S→5amp;∴ Option [B] is correct. \begin{aligned}& \therefore \mathrm{z}=1 \pm \mathrm{i} \sqrt{3} \\\\& \Rightarrow \mathrm{z}-\overline{\mathrm{z}}= \pm 2 \mathrm{i} \sqrt{3} \\\\& \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|=2 \sqrt{3} \\\\& \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|^2=12 \\\\& \text { Now } \mathrm{z}+1=2+\mathrm{i} \sqrt{3} \\\\& |\mathrm{z}+1|^2=4+3=7 \\\\& \therefore \mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5 \\\\& \therefore \text { Option }[\mathrm{B}] \text { is correct. }\end{aligned}

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