Let S = {z in mathbb{C} : z^2 + 4z + 16 = 0}. | vedclass

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(D) The roots of the quadratic equation $z^2 + 4z + 16 = 0$ are found using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Substitutin

Vedclass · Accepted Answer

$38$

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(D) The roots of the quadratic equation $z^2 + 4z + 16 = 0$ are found using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Substituting the values $a = 1, b = 4, c = 16$, we get $z = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} = -2 \pm i 2\sqrt{3}$.Let the two roots be $z_1 = -2 + 2\sqrt{3}i$ and $z_2 = -2 - 2\sqrt{3}i$.We need to calculate the sum $S = |z_1 + \sqrt{3}i|^2 + |z_2 + \sqrt{3}i|^2$.First, calculate $|z_1 + \sqrt{3}i|^2 = |-2 + 2\sqrt{3}i + \sqrt{3}i|^2 = |-2 + 3\sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + 27 = 31$.Next, calculate $|z_2 + \sqrt{3}i|^2 = |-2 - 2\sqrt{3}i + \sqrt{3}i|^2 = |-2 - \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7$.Finally, the sum is $31 + 7 = 38$.

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

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