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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Complex Numbers

Let

S = {z \in C : z^{2} + 6 i z - 3 = 0}

. Then

\sum_{z \in S} z^{8}

is equal to :

(A)162✓ Correct
(B)184
(C)262
(D)324

Answer: (A)

Solution

For the roots,

α + β = - 6 i

and

α β = - 3

. Then

α^{2} + β^{2} = (α + β)^{2} - 2 α β = (- 6 i)^{2} + 6 = 0

,

α^{4} + β^{4} = (α^{2} + β^{2})^{2} - 2 (α β)^{2} = - 2 \cdot 9 = - 18

, and

α^{8} + β^{8} = (α^{4} + β^{4})^{2} - 2 (α β)^{4} = (- 18)^{2} - 2 \cdot 81 = 324 - 162 = 162

. Therefore

\sum_{z \in S} z^{8} = 162

.

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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