JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(S = \{z \in \mathbb{C} : z^2 + \sqrt{6}\,iz - 3 = 0\}\). Then \(\sum\limits_{z \in S} z^8\) is equal to :
- A \(162\)
- B \(184\)
- C \(262\)
- D \(324\)
Answer & Solution
Correct Answer
(A) \(162\)
Step-by-step Solution
Detailed explanation
Given equation is \(z^2 + \sqrt{6}iz - 3 = 0\). Rearranging the terms, we get \(z^2 - 3 = -\sqrt{6}iz\). Squaring both sides, we obtain: \((z^2 - 3)^2 = (-\sqrt{6}iz)^2\) \(z^4 - 6z^2 + 9 = -6z^2\) \(z^4 + 9 = 0\) \(z^4 = -9\) Squaring again, we get: \((z^4)^2 = (-9)^2\)…
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