# Solve for z = a + bi, where a and b are real numbers. (z minus conjugate.z) ^ 2 = 2 z + 6i?

### 2 Answers

- 1 decade agoFavorite Answer
(z - z̅ )² = 2z + 6i

Remember that the conjugate of the complex number

a + bi

Is

a - bi

So, if z = a + bi

Then z̅ = a - bi

And, z - z̅ = (a + bi) - (a - bi)

= a + bi - a + bi

= 2bi

So, we have

(2bi)² = (z - z̅ )² = 2z + 6i

But, 2z + 6i

= 2(a + bi) + 6i

= 2a + 2bi + 6i

And,

(2bi)² = 4b²i² = -4b²

So, we must have

-4b² = 2a + 2bi + 6i = 2a + (2b + 6)i

The left hand side is real, so the right hand side must be as well.

In other words, the imaginary part on the right hand side,

(2b + 6)i, must be 0

So, we have

2b + 6 = 0

⇒ 2b = -6

⇒ b = -3

Substituting this into the equation

-4b² = 2a + (2b + 6)i, we get

-4(-3)² = 2a

⇒ -4(9) = 2a

⇒ -36 = 2a

⇒ a = -18

So we have

z = a + bi

= -18 - 3i

- Jake.Lv 51 decade ago
(a + bi - a + bi)^2 = 2z + 6i

(2bi)^2 = 2z + 6i

4b^2i^2 = 2z + 6i

-4b^2 = 2z + 6i

2z = -4b^2 - 6i

z = -2b^2 - 3i

b = -3

therefore

z = -18 - 3i