JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

Let $z$ be a complex number such that $∣ z - 6∣ = 5$ and $∣ z + 2 - 6 i ∣ = 5$ . Then the value of $z^{3} + 3 z^{2} - 15 z + 141$ is equal to

A
42
B
37
C
50
D
61

Solution

The two equations describe circles in the complex plane centred at $C_{1} = (6, 0)$ and $C_{2} = (- 2, 6)$ , both of radius 5.

Distance between centres: $8^{2} + 6^{2} = 10 = r_{1} + r_{2}$ . The circles touch externally, and the unique common point is the midpoint of $C_{1} C_{2}$ :

$z = (\frac{6 - 2}{2}, \frac{0 + 6}{2}) = (2, 3) = 2 + 3 i$ .

Compute $z^{2} = (2 + 3 i)^{2} = 4 + 12 i + 9 i^{2} = - 5 + 12 i = 4 z - 13$ (since $4 z = 8 + 12 i$ , so $4 z - 13 = - 5 + 12 i$ . ✓)

Then $z^{3} = z \cdot z^{2} = z (4 z - 13) = 4 z^{2} - 13 z = 4 (4 z - 13) - 13 z = 3 z - 52$ .

$z^{3} + 3 z^{2} - 15 z + 141 = (3 z - 52) + 3 (4 z - 13) - 15 z + 141 = (3 + 12 - 15) z + (- 52 - 39 + 141) = 0 + 50 = 50$ .

Option (3).

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