We are asked to solve the equation $\sqrt{2+i} = x - iy$ for $x$ and $y$, where $i$ is the imaginary unit ($i^2 = -1$).

1. Problem Description

We are asked to solve the equation $\sqrt{2+i} = x - iy$ for $x$ and $y$, where $i$ is the imaginary unit ($i^2 = -1$).

2. Solution Steps

We square both sides of the equation:

$(\sqrt{2+i})^2 = (x-iy)^2$

$2+i = x^2 - 2ixy + (iy)^2$

$2+i = x^2 - 2ixy - y^2$

$2+i = (x^2 - y^2) - 2xyi$

We equate the real and imaginary parts:

Real part: $x^2 - y^2 = 2$

Imaginary part: $-2xy = 1$, so $xy = -\frac{1}{2}$

We have a system of two equations:

$x^2 - y^2 = 2$

$xy = -\frac{1}{2}$

From the second equation, we have $y = -\frac{1}{2x}$. Substituting this into the first equation, we get:

$x^2 - \left(-\frac{1}{2x}\right)^2 = 2$

$x^2 - \frac{1}{4x^2} = 2$

Multiplying by $4x^2$, we get:

$4x^4 - 1 = 8x^2$

$4x^4 - 8x^2 - 1 = 0$

Let $z = x^2$. Then $4z^2 - 8z - 1 = 0$.

Using the quadratic formula, we have:

$z = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(-1)}}{2(4)} = \frac{8 \pm \sqrt{64+16}}{8} = \frac{8 \pm \sqrt{80}}{8} = \frac{8 \pm 4\sqrt{5}}{8} = \frac{2 \pm \sqrt{5}}{2}$

Since $x^2 = z > 0$, we take the positive root:

$x^2 = \frac{2+\sqrt{5}}{2}$

$x = \pm \sqrt{\frac{2+\sqrt{5}}{2}}$

Since $xy = -\frac{1}{2}$, $x$ and $y$ have opposite signs.

If $x = \sqrt{\frac{2+\sqrt{5}}{2}}$, then $y = -\frac{1}{2x} = -\frac{1}{2\sqrt{\frac{2+\sqrt{5}}{2}}} = -\frac{1}{\sqrt{2(2+\sqrt{5})}} = -\frac{1}{\sqrt{4+2\sqrt{5}}}$

$y = -\frac{1}{\sqrt{(\sqrt{5}+1)^2-2\sqrt{5} + 2\sqrt{5} -4} }= -\frac{1}{\sqrt{(\sqrt{5}+1)^2 }} = - \frac{\sqrt{2}}{2\sqrt{2+\sqrt{5}}} = - \sqrt{\frac{2}{4(2+\sqrt{5})}} = - \sqrt{\frac{1}{2(2+\sqrt{5})}} = - \sqrt{\frac{2-\sqrt{5}}{2(4-5)}} = \sqrt{\frac{2-\sqrt{5}}{2}}$ Note: $y = -\sqrt{\frac{\sqrt{5}-2}{2}}$

$\frac{1}{2x} = \sqrt{\frac{1}{4} * \frac{2}{2+\sqrt{5}}} = \sqrt{\frac{1}{2(2+\sqrt{5})}} = \sqrt{\frac{2-\sqrt{5}}{2(4-5)}} = \sqrt{\frac{\sqrt{5}-2}{2}}$

So $y = -\sqrt{\frac{\sqrt{5}-2}{2}}$

If $x = - \sqrt{\frac{2+\sqrt{5}}{2}}$, then $y = \sqrt{\frac{\sqrt{5}-2}{2}}$

Therefore, the solutions are:

$x = \sqrt{\frac{2+\sqrt{5}}{2}}$, $y = -\sqrt{\frac{\sqrt{5}-2}{2}}$

$x = -\sqrt{\frac{2+\sqrt{5}}{2}}$, $y = \sqrt{\frac{\sqrt{5}-2}{2}}$

3. Final Answer

$x = \sqrt{\frac{2+\sqrt{5}}{2}}$, $y = -\sqrt{\frac{\sqrt{5}-2}{2}}$ or

$x = -\sqrt{\frac{2+\sqrt{5}}{2}}$, $y = \sqrt{\frac{\sqrt{5}-2}{2}}$