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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Complex Number

Let z ∈ C be such that (z^2+3 i)/(z-2+i)=2+3 i . Then the sum of...

Let z∈Cz \in C be such that z2+3iz−2+i=2+3i\frac{z^2+3 i}{z-2+i}=2+3 i. Then the sum of all possible values of z2z^2 isLet z ∈ C be such that (z^2+3 i)/(z-2+i)=2+3 i . Then the sum of all possible values of z^2 is

Options

  1. A19−2i19-2 \mathrm{i}
  2. B−19−2i-19-2 \mathrm{i}
  3. C19+2i19+2 i
  4. D−19+2i-19+2 i

Correct answer

B. −19−2i-19-2 \mathrm{i}

Step-by-step solution

z2+3i=z(2+3i)−7−4iz^2+3 i=z(2+3 i)-7-4 i
z2−z(2+3i)+7+7i=0\mathrm{z}^2-\mathrm{z}(2+3 \mathrm{i})+7+7 \mathrm{i}=0Let z ∈ C be such that (z^2+3 i)/(z-2+i)=2+3 i . Then the sum of all possible values of z^2 is
<br/>z12+z22=(z1+z2)2−2z1z2=4−9+12i−14−14i=−19−2i<br/>\begin{aligned}<br/>& z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =4-9+12 i-14-14 i \\ & =-19-2 i<br/>\end{aligned}
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