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JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Complex Numbers

Let

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

.

Then

\sum_{z \in S} ∣ z + 3 i ∣^{2}

is equal to :

(A)42
(B)23
(C)27
(D)38✓ Correct

Answer: (D)

Solution

Completing the square,

z^{2} + 4 z + 16 = (z + 2)^{2} + 12 = 0

gives

(z + 2)^{2} = - 12

, so

z = - 2 \pm 23 i

and

S = {- 2 + 23 i, - 2 - 23 i}

. For the first root,

z + 3 i = - 2 + 33 i

, so

∣ z + 3 i ∣^{2} = 4 + 27 = 31

; for the second,

z + 3 i = - 2 - 3 i

, so

∣ z + 3 i ∣^{2} = 4 + 3 = 7

. Their sum is

31 + 7 = 38

.

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option

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