Given equation is z2+4z−(1+12i)=0
Let the roots be z1 and z2.
Sum of roots: z1+z2=−4
Product of roots: z1z2=−(1+12i)
We know that (z1−z2)2=(z1+z2)2−4z1z2
(z1−z2)2=(−4)2−4(−(1+12i))=16+4+48i=20+48i
Taking modulus on both sides:
∣z1−z2∣2=∣20+48i∣=202+482=400+2304=2704=52
Using the parallelogram law for complex numbers:
∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)
∣−4∣2+52=2(∣z1∣2+∣z2∣2)
16+52=2(∣z1∣2+∣z2∣2)
68=2(∣z1∣2+∣z2∣2)
∣z1∣2+∣z2∣2=34
Answer: 34