Question

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

Answer 1

Answer:

$\displaystyle x_1 = 2-\sqrt{3} \text{ and } y_1 = \frac{\sqrt{3}-1}{2}$

Or:

$\displaystyle x _ 2 = 2 + \sqrt{3} \text{ and } y _ 2 = -\frac{1+\sqrt{3}}{2}$

Step-by-step 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:

$\displaystyle \begin{aligned} y &= \frac{-(2)\pm\sqrt{(2)^2 - 4(2)(-1)}}{2(2)} \\ \\ &= \frac{-2\pm\sqrt{12}}{4} \\ \\ &= \frac{-2\pm2\sqrt{3}}{4}\\ \\ &= \frac{-1\pm\sqrt{3}}{2} \end{aligned}$

Hence:

$\displaystyle y_1 = \frac{-1+\sqrt{3}}{2} \text{ or } y_2 = \frac{-1-\sqrt{3}}{2}$

Then:

$\displaystyle x _ 1 = 1 - 2\left(\frac{-1+\sqrt{3}}{2}\right) = 1 + (1 - \sqrt{3}) = 2 - \sqrt{3}$

And:

$\displaystyle x _ 2 = 1 - 2\left(\frac{-1-\sqrt{3}}{2}\right) = 1 + (1 + \sqrt{3}) = 2 + \sqrt{3}$

In conclusion, the values of x and y are:

$\displaystyle x_1 = 2-\sqrt{3} \text{ and } y_1 = \frac{\sqrt{3}-1}{2}$

Or:

$\displaystyle x _ 2 = 2 + \sqrt{3} \text{ and } y _ 2 = -\frac{1+\sqrt{3}}{2}$