JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(S=\{z \in \mathbb{C}: z^2+4z+16=0\}\). Then \(\sum_{z \in S}|z+\sqrt{3}i|^2\) is equal to:
- A \(42\)
- B \(23\)
- C \(27\)
- D \(38\)
Answer & Solution
Correct Answer
(D) \(38\)
Step-by-step Solution
Detailed explanation
Given the equation \(z^2 + 4z + 16 = 0\), we find its roots using the quadratic formula: \(z = \dfrac{-4 \pm \sqrt{16 - 64}}{2} = \dfrac{-4 \pm \sqrt{-48}}{2} = -2 \pm 2\sqrt{3}i\) The set \(S\) contains two complex numbers: \(z_1 = -2 + 2\sqrt{3}i\) and…
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