Let S = {z in mathbb{C} : z^2 + sqrt{6}iz - 3 | vedclass

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(A) Given the quadratic equation $z^2 + \sqrt{6}iz - 3 = 0$.Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:$z = \frac{-\

Vedclass · Accepted Answer

$162$

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(A) Given the quadratic equation $z^2 + \sqrt{6}iz - 3 = 0$.Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:$z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{-\sqrt{6}i \pm \sqrt{-6 + 12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$.Thus, the roots are $z_1 = \frac{\sqrt{6}}{2}(1 - i)$ and $z_2 = \frac{\sqrt{6}}{2}(-1 - i)$.Now, calculate $z^2$ for each root:$z_1^2 = \left(\frac{\sqrt{6}}{2}(1 - i)\right)^2 = \frac{6}{4}(1 - 2i + i^2) = \frac{3}{2}(1 - 2i - 1) = -3i$.$z_2^2 = \left(\frac{\sqrt{6}}{2}(-1 - i)\right)^2 = \frac{6}{4}(1 + 2i + i^2) = \frac{3}{2}(1 + 2i - 1) = 3i$.Now, calculate $z^8 = (z^2)^4$:$z_1^8 = (-3i)^4 = (-3)^4 \cdot i^4 = 81 \cdot 1 = 81$.$z_2^8 = (3i)^4 = 3^4 \cdot i^4 = 81 \cdot 1 = 81$.Therefore, $\sum_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$.

Let $S = \{z \in \mathbb{C} : z^2 + \sqrt{6}iz - 3 = 0\}$. Then $\sum_{z \in S} z^8$ is equal to:

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