JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

Let $S = {z \in C : z^{2} + 4 z + 16 = 0}$ .

Then $z \in S \sum ∣ z + 3 i ∣^{2}$ is equal to:

Solution

Solve $z^{2} + 4 z + 16 = 0$ : completing the square, $(z + 2)^{2} = - 12$ , so $z = - 2 \pm 23 i$ .

Compute $∣ z + 3 i ∣^{2}$ for each root.

For $z = - 2 + 23 i$ : $z + 3 i = - 2 + 33 i$ , so $∣ z + 3 i ∣^{2} = 4 + 27 = 31$ .

For $z = - 2 - 23 i$ : $z + 3 i = - 2 - 3 i$ , so $∣ z + 3 i ∣^{2} = 4 + 3 = 7$ .

Sum $= 31 + 7 = 38$ .

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