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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:

  1. A \(42\)
  2. B \(23\)
  3. C \(27\)
  4. D \(38\)
Verified Solution

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…