COMPASSLearning

Talk to us Register Login

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Complex Numbers

Let

z_{1}, z_{2} \in C

be the distinct solutions of the equation

z^{2} + 4 z - (1 + 12 i) = 0

. Then

∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}

is equal to :

(A)18
(B)22
(C)29
(D)34✓ Correct

Answer: (D)

Solution

By the sum of roots

z_{1} + z_{2} = - 4

. Solving

z^{2} + 4 z - (1 + 12 i) = 0

,

z = - 2 \pm 5 + 12 i

, and since

(3 + 2 i)^{2} = 5 + 12 i

, the roots are

z_{1} = 1 + 2 i

and

z_{2} = - 5 - 2 i

. Then

∣ z_{1} ∣^{2} = 1^{2} + 2^{2} = 5

and

∣ z_{2} ∣^{2} = 5^{2} + 2^{2} = 29

, so

∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} = 34

.

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Complex Numbers →

More JEE Main questions on Complex Numbers

If $x^{2} + x + 1 = 0$ , then the value of $(x + \frac{1}{x})^{4} + (x^{2} + \frac{1}{x ^{2}})^{4} + (x^{3} + \frac{1}{x ^{3}})^{4} + \dots + (x^{25} + \frac{1}{x ^{25}})^{4}$ is :
JEE Main 2026View solution →
Let z be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument.
Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to:
JEE Main 2026View solution →
Let $α = \frac{- 1 + i 3}{2}$ and $β = \frac{- 1 - i 3}{2}, i = - 1$ . If $(7 - 7 α + 9 β)^{20} + (9 + 7 α - 7 β)^{20} + (- 7 + 9 α + 7 β)^{20} + (14 + 7 α + 7 β)^{20} = m^{10}$ , then m is
JEE Main 2026View solution →
Let $S = {z \in C : 4 z^{2} + \overline{z} = 0}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ is equal to :
JEE Main 2026View solution →
Let $S = {z : 3 \leq ∣2 z - 3 (1 + i) ∣ \leq 7}$ be a set of complex numbers. Then $Min_{z \in S} (z + \frac{1}{2} (5 + 3 i))$ is equal to:
JEE Main 2026View solution →
If $z = \frac{3}{2} + \frac{i}{2}, i = - 1$ , then $(z^{201} - i)^{8}$ is equal to
JEE Main 2026View solution →