Sum of Squared Moduli of Roots of Complex Quadratic Equation

To find the sum of the squared moduli of the roots of the quadratic equation $z^2 + 4z - (1 + 12i) = 0$, we can complete the square for $z$:

$(z + 2)^2 - 4 - 1 - 12i = 0$ $(z + 2)^2 = 5 + 12i$

Let $w = z + 2 = x + iy$, where $x, y \in \mathbb{R}$. Then: $(x + iy)^2 = 5 + 12i$ $(x^2 - y^2) + 2ixy = 5 + 12i$

Equating the real and imaginary parts:

1. $x^2 - y^2 = 5$
2. $2xy = 12 \implies xy = 6$

Using the identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$, we get: $(x^2 + y^2)^2 = 5^2 + 12^2 = 25 + 144 = 169$ $x^2 + y^2 = 13$

Adding $x^2 - y^2 = 5$ and $x^2 + y^2 = 13$: $2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$

Subtracting $x^2 - y^2 = 5$ from $x^2 + y^2 = 13$: $2y^2 = 8 \implies y^2 = 4 \implies y = \pm 2$

Since $xy = 6 > 0$, $x$ and $y$ must have the same sign. Therefore, $w = z + 2 = \pm (3 + 2i)$.

This gives us the two roots $z_1$ and $z_2$: $z_1 = (3 + 2i) - 2 = 1 + 2i$ $z_2 = (-3 - 2i) - 2 = -5 - 2i$

Now, calculating the squared moduli of the roots: $|z_1|^2 = 1^2 + 2^2 = 1 + 4 = 5$ $|z_2|^2 = (-5)^2 + (-2)^2 = 25 + 4 = 29$

Sum of squared moduli: $|z_1|^2 + |z_2|^2 = 5 + 29 = 34$

Thus, the correct option is D.