Let z1, z2 C be the distinct solutions of the equation z2+4z-(1+12i)=0. Then |z1|2+|z2|2 is equal to :
- 18
- 22
- 29
- 34
Answer
(4) 34
Solution
z2+4z-(1+12i)=0
(z+2)2=4+1+12i=5+12i
(3+2i)2=9-4+12i=5+12i
z+2= (3+2i)
z1=1+2i, z2=-5-2i
|z1|2+|z2|2=(1+4)+(25+4)=34
Answer: (4)