COMPASSLearning

Talk to us Register Login

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Complex Numbers

The sum of the square of the modulus of the elements in the set

{z = a + ib : a, b \in Z, z \in C, ∣ z - 1∣ \leq 1, ∣ z - 5∣ \leq ∣ z - 5 i ∣}

is ______.

Answer: 9

Solution

∣ z - 1∣ \leq 1

over the integer lattice gives

(0, 0), (1, 0), (2, 0), (1, 1), (1, - 1)

. The condition

∣ z - 5∣ \leq ∣ z - 5 i ∣

reduces to

a \geq b

, which all five satisfy. Sum of

∣ z ∣^{2} = 0 + 1 + 4 + 2 + 2 = 9

.

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Integer answer

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 →