Question

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

Answer 1

`\displaystyle x_1 = 2-√(3) \text{ and } y_1 = (√(3)-1)/(2)`

Or:

`\displaystyle x _ 2 = 2 + √(3) \text{ and } y _ 2 = -(1+√(3))/(2)`

Explanation:

We are given the equation:

`\displaystyle (x^2 + 2xy) + i(y-1) = (x^2 -2x + 2y) - i(x +y)`

And we want to find the values of x and y such that the equation is true.

First, distribute:

`\displaystyle (x^2 + 2xy) + i(y-1) = (x^2 -2x + 2y) +i(-x -y)`

If two complex numbers are equivalent, their real and imaginary parts are equivalent. Hence:

`\displaystyle x^2 + 2xy = x^2 - 2x +2y \text{ and } y - 1 = -x -y`

Simplify:

`\displaystyle 2xy = -2x +2y \text{ and }x = 1 - 2y`

Substitute:

`\displaystyle 2(1-2y)y = -2(1-2y) + 2y`

Solve for y:

`\displaystyle \begin{aligned} 2(y - 2y^2) &= (-2 + 4y) + 2y \\ 2y - 4y^2 &= 6y -2\\ 4y^2 + 4y - 2& = 0 \\ 2y^2 + 2y - 1 &= 0 \\ \end{aligned}`

From the quadratic formula:

`\displaystyle \begin{aligned} y &= (-(2)\pm√((2)^2 - 4(2)(-1)))/(2(2)) \\ \\ &= (-2\pm√(12))/(4) \\ \\ &= (-2\pm2√(3))/(4)\\ \\ &= (-1\pm√(3))/(2) \end{aligned}`

Hence:

`\displaystyle y_1 = (-1+√(3))/(2) \text{ or } y_2 = (-1-√(3))/(2)`

Then:

`\displaystyle x _ 1 = 1 - 2\left((-1+√(3))/(2)\right) = 1 + (1 - √(3)) = 2 - √(3)`

And:

`\displaystyle x _ 2 = 1 - 2\left((-1-√(3))/(2)\right) = 1 + (1 + √(3)) = 2 + √(3)`

In conclusion, the values of x and y are:

`\displaystyle x_1 = 2-√(3) \text{ and } y_1 = (√(3)-1)/(2)`

Or:

`\displaystyle x _ 2 = 2 + √(3) \text{ and } y _ 2 = -(1+√(3))/(2)`