Let $S = \left\{ z \in \mathbb{C}:z^{2} + 4z + 16 = 0 \right\}$.
Then $\sum_{z \in S}\mspace{2mu}|z + \sqrt{3}i|^{2}$ is equal to :

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Let $S = \left\{ z \in \mathbb{C}:z^{2} + 4z + 16 = 0 \right\}$.Then $\sum_{z \in S}\mspace{2mu}|z + \sqrt{3}i|^{2}$ is equal to :

Accepted Answer

Sol. [Key Concept: Solve $z^{2} + 4z + 16 = 0$ for the two complex roots, then add their squared moduli $|z + \sqrt{3\ }{i|}^{2}$.]$${z^{2} + 4z + 16 = 0 }{\Rightarrow (z + 2)^{2} = - 12 }{\Rightarrow z = - 2 \pm 2\sqrt{3}i }{\Rightarrow S = \{ - 2 + 2\sqrt{3}i, - 2 - 2\sqrt{3}i\} }{	herefore\sum_{z \in S}\mspace{2mu}|z + \sqrt{3}i|^{2} }{= | - 2 + 2\sqrt{3}i + \sqrt{3}i|^{2} + | - 2 - 2\sqrt{3}i + \sqrt{3}i|^{2} }{= | - 2 + 3\sqrt{3}i|^{2} + | - 2 - \sqrt{3}i|^{2} }{= 38}$$

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