Given the complex Z. = 3 + 2I, the complex Z satisfies Z. * z = 3Z + Z. then the complex Z is equal to?

Let z = a + bi
Then (3 + 2I) (a + bi) = 3 (a + bi) + 3 + 2I
That is, (3a-2b) + (2a + 3b) I = (3a + 3) + (3b + 2) I
So 3a-2b = 3A + 3, 2A + 3B = 3B + 2
So a = 1, B = - 3 / 2
So z = 1-3i / 2

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