Let . Then is equal to: [2026]

Question

Let $S = {z \in C : z^{2} + 4 z + 16 = 0}$ . Then $\sum_{z \in S} | z + \sqrt{3} i |^{2}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$z^{2} + 4 z + 16 = 0$

$\Rightarrow (x + i y^{2}) + 4 (x + i y) + 16 = 0$

$\Rightarrow (x^{2} - y^{2} + 4 x + 16) + 2 i x y + 4 i y = 0 + i 0$

Comparing real and imaginary parts, we get

$2 y (x + 2) = 0 & x^{2} - y^{2} + 4 x + 16 = 0$

$\Rightarrow y = 0 or x = - 2$

If $y = 0, x^{2} + 4 x + 16 = 0, D = 16 - 4 \times 16 = - 48 < 0$

So, no real solution for x exists.

Now, if x = –2

$4 - y^{2} - 8 + 16 = 0 \Rightarrow y^{2} = 12$

$\Rightarrow y = \pm 2 \sqrt{3}$

So, $z = - 2 + 2 \sqrt{3} i and z = - 2 - 2 \sqrt{3} i$

$\sum_{z \in S} | z + \sqrt{3} i |^{2} = | - 2 + 2 \sqrt{3} i + \sqrt{3} i |^{2} + | - 2 - 2 \sqrt{3} i + \sqrt{3} i |^{2}$

$= | - 2 + 3 \sqrt{3} i |^{2} + | - 2 - \sqrt{3} i |^{2}$

$= 4 + 27 + 4 + 3 = 38$

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