B Z2 + (2 – 3I)Z – 1 – 3I = 0 (9)  = 4 – 12I – 9 + 4 + 12I = -1...

Câu 9.b z

²

+ (2 – 3i)z – 1 – 3i = 0 (9)  = 4 – 12i – 9 + 4 + 12i = -1 = i

²

i ii i     i   . 2 1 2 = -1 + i hay z = 2 3Do đó, (9)  z = 2 32