Let z_1, z_2 ∈ mathbbC be the distinct solutions of the equation z^2 + 4z - (1 + 12i) = 0. Then |z_1|^2 + |z_2|^2 is equal to :

Given equation is $z^2 + 4z - (1 + 12i) = 0$z2+4z−(1+12i)=0

Let the roots be $z_1$z1​ and $z_2$z2​.

Sum of roots: $z_1 + z_2 = -4$z1​+z2​=−4

Product of roots: $z_1 z_2 = -(1 + 12i)$z1​z2​=−(1+12i)

We know that $(z_1 - z_2)^2 = (z_1 + z_2)^2 - 4z_1 z_2$(z1​−z2​)2=(z1​+z2​)2−4z1​z2​

$$(z_1 - z_2)^2 = (-4)^2 - 4(-(1 + 12i)) = 16 + 4 + 48i = 20 + 48i$$(z1​−z2​)2=(−4)2−4(−(1+12i))=16+4+48i=20+48i

Taking modulus on both sides:

$$|z_1 - z_2|^2 = |20 + 48i| = \sqrt{20^2 + 48^2} = \sqrt{400 + 2304} = \sqrt{2704} = 52$$∣z1​−z2​∣2=∣20+48i∣=202+482​=400+2304​=2704​=52

Using the parallelogram law for complex numbers:

$$|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$$∣z1​+z2​∣2+∣z1​−z2​∣2=2(∣z1​∣2+∣z2​∣2)

$$|-4|^2 + 52 = 2(|z_1|^2 + |z_2|^2)$$∣−4∣2+52=2(∣z1​∣2+∣z2​∣2)

$$16 + 52 = 2(|z_1|^2 + |z_2|^2)$$16+52=2(∣z1​∣2+∣z2​∣2)

$$68 = 2(|z_1|^2 + |z_2|^2)$$68=2(∣z1​∣2+∣z2​∣2)

$$|z_1|^2 + |z_2|^2 = 34$$∣z1​∣2+∣z2​∣2=34

Answer: $34$34