Let z C be such that z2+3iz-2+i = 2+3i. Then the sum of all possible values of z2 is :
- -19+2i
- -19-2i
- 19-2i
- 19+2i
Answer
(2) -19-2i
Solution
z2 + 3i = (2+3i)(z-2+i)
z2 + 3i = (2+3i)z + (2+3i)(-2+i)
(2+3i)(-2+i) = -4 + 2i - 6i + 3i2 = -7 - 4i
z2 - (2+3i)z + (7+7i) = 0
Its roots z1, z2 (neither equals 2-i, so both are admissible):
z1 + z2 = 2+3i, z1z2 = 7+7i
z12 + z22 = (z1+z2)2 - 2z1z2
= (2+3i)2 - 2(7+7i)
= (4 - 9 + 12i) - 14 - 14i
= -19 - 2i
Answer: (2)