Let z = x + iy where x and y are real
x - iy = (x + iy)³
x - iy = x³ + 3x²(iy) + 3x(iy)² + (iy)³
x - iy = x³ - 3xy² + i(3x²y - y³)
Equate real and imaginary parts:
(1) x = x³ - 3xy²
(2) - y = 3x²y - y³
From (1):
x(x² - 3y² - 1) = 0
x = 0 OR x² = 3y² + 1
When x = 0:
Sub x = 0 into (2):
y³ = y
y(y² - 1) = 0
y = 0, 1, - 1, when x = 0
So complex number solutions are: 0, i, - i
When x² = 3y² + 1:
Sub x² = 3y² + 1 into (2)
- y = 3y(3y² + 1) - y³
8y³ + 4y = 0
y(2y² + 1) = 0
y = 0 is the only real solution (since y is real)
Sub y = 0 into (1)
x = x³
x(x² - 1) = 0
x = 0, 1, -1, when x = 0
So complex number solutions are: 0, 1, - 1
Therefore total solutions are: 0, 1, - 1, i, - i