Given the equation z2+4z+16=0, we find its roots using the quadratic formula:
z=2−4±16−64=2−4±−48=−2±23i
The set S contains two complex numbers: z1=−2+23i and z2=−2−23i.
We need to evaluate the sum z∈S∑∣z+3i∣2, which is ∣z1+3i∣2+∣z2+3i∣2.
For z1=−2+23i:
∣z1+3i∣2=∣−2+23i+3i∣2=∣−2+33i∣2
∣−2+33i∣2=(−2)2+(33)2=4+27=31
For z2=−2−23i:
∣z2+3i∣2=∣−2−23i+3i∣2=∣−2−3i∣2
∣−2−3i∣2=(−2)2+(−3)2=4+3=7
Adding these values together:
z∈S∑∣z+3i∣2=31+7=38
Answer: 38