Let z_1, z_2 in mathbb{C} be the distinct sol | vedclass

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(D) Given the quadratic equation $z^2 + 4z - (1 + 12i) = 0$.Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1, b = 4,

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$34$

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(D) Given the quadratic equation $z^2 + 4z - (1 + 12i) = 0$.Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1, b = 4, c = -(1 + 12i)$:$z = \frac{-4 \pm \sqrt{16 - 4(1)(-(1 + 12i))}}{2} = \frac{-4 \pm \sqrt{16 + 4 + 48i}}{2} = \frac{-4 \pm \sqrt{20 + 48i}}{2} = -2 \pm \sqrt{5 + 12i}$.Let $\sqrt{5 + 12i} = x + iy$. Squaring both sides: $x^2 - y^2 + 2ixy = 5 + 12i$.Equating real and imaginary parts: $x^2 - y^2 = 5$ and $2xy = 12 \Rightarrow xy = 6$.Since $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 5^2 + 12^2 = 25 + 144 = 169$, we have $x^2 + y^2 = 13$.Solving $x^2 - y^2 = 5$ and $x^2 + y^2 = 13$, we get $2x^2 = 18 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$.If $x = 3, y = 2$; if $x = -3, y = -2$. Thus $\sqrt{5 + 12i} = \pm(3 + 2i)$.So, $z = -2 \pm (3 + 2i)$.$z_1 = -2 + 3 + 2i = 1 + 2i$ and $z_2 = -2 - 3 - 2i = -5 - 2i$.$|z_1|^2 = 1^2 + 2^2 = 1 + 4 = 5$.$|z_2|^2 = (-5)^2 + (-2)^2 = 25 + 4 = 29$.$|z_1|^2 + |z_2|^2 = 5 + 29 = 34$.

Let $z_1, z_2 \in \mathbb{C}$ be the distinct solutions of the equation $z^2 + 4z - (1 + 12i) = 0$. Then $|z_1|^2 + |z_2|^2$ is equal to:

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