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Mathematics · Complex Numbers

JEE Main 2025 — 3 April, Morning Shift — Question 33

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} is

  1. Option A:

    19−2i19-2 \mathrm{i}

  2. Option B:

    −19−2i-19-2 \mathrm{i}

    Correct
  3. Option C:

    19+2i19+2 \mathrm{i}

  4. Option D:

    −19+2i-19+2 \mathrm{i}

Answer: B

Step-by-step solution

z2+3i=z(2+3i)−7−4iz^2 + 3i = z(2 + 3i) - 7 - 4i z2−z(2+3i)+7+7i=0→z2z1z^2 - z(2 + 3i) + 7 + 7i = 0 \xrightarrow{z_2} z_1 z12+z22=(z1+z2)2−2z1z2z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2 =(2+3i)2−2(7+7i)= (2 + 3i)^2 - 2(7 + 7i) =(4+12i+(3i)2)−(14+14i)= (4 + 12i + (3i)^2) - (14 + 14i) =(4+12i−9)−(14+14i)= (4 + 12i - 9) - (14 + 14i) =−5+12i−14−14i= -5 + 12i - 14 - 14i =−19−2i= -19 - 2i

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Mathematics
Chapter
Complex Numbers
Topic
Conjugate of complex numbers & properties
Let z in C be such that frac z 2 +3 i z-2+i =2+3 i . Then the sum of… | JEE Main 2025 PYQ with Solution · DhiX AI