Question

find the real numbers x&y so that (x^2+2xy)+i(y-1) = (x^2-2x+2y) - i(x+y)​

Answer 1

In order for two complex numbers to be equal, their real and imaginary parts must match up.

So if

x ² + 2xy + i (y - 1) = x ² - 2x + 2y - i (x + y)

then

x ² + 2xy = x ² - 2x + 2y

and

y - 1 = -x - y

Solve for x and y : the first equation gives

x ² + 2xy = x ² - 2x + 2y

2xy = 2y - 2x

and the second gives

y - 1 = -x - y

2y = 1 - x

Substitute this into the first equation:

x (1 - x) = (1 - x) - 2x

x - x ² = 1 - 3x

x ² - 4x + 1 = 0

x = 2 ± √3

Then

x = 2 - √3   ==>   y = (-1 + √3)/2

x = 2 + √3   ==>   y = (-1 - √3)/2

both of which make both sides of the equation (2 - √3) + i (√3 - 3)/2.