COMPASSLearning

Talk to us Register Login

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Complex Numbers

Let

A = {z \in C : ∣ z - 2 - i ∣ = 3}

,

B = {z \in C : Re (z - i z) = 2}

and

S = A \cap B

. Then

\sum_{z \in S} ∣ z ∣^{2}

is equal to

Answer: 22

Solution

Writing

z = x + i y

, the circle

∣ z - 2 - i ∣ = 3

is

(x - 2)^{2} + (y - 1)^{2} = 9

. Since

z - i z = (x + y) + i (y - x)

, the condition

Re (z - i z) = 2

is

x + y = 2

. Substituting

y = 2 - x

into the circle gives

x^{2} - 3 x - 2 = 0

, whose roots satisfy

x_{1} + x_{2} = 3

and

x_{1} x_{2} = - 2

. For each point

∣ z ∣^{2} = x^{2} + (2 - x)^{2} = 2 x^{2} - 4 x + 4

, so the sum over both points is

2 (x_{1}^{2} + x_{2}^{2}) - 4 (x_{1} + x_{2}) + 8

. With

x_{1}^{2} + x_{2}^{2} = 9 - 2 (- 2) = 13

, this is

2 (13) - 4 (3) + 8 = 22

.

JEE Main · 2025 · 04 Apr 2025 · 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 →