We are asked to solve for $x$ and $y$ in several equations involving complex numbers.

1. Problem Description

We are asked to solve for $x$ and $y$ in several equations involving complex numbers.

2. Solution Steps

(a) $(x + yi) = (3 + i)(2 - 3i)$

$x + yi = 3(2) + 3(-3i) + i(2) + i(-3i) = 6 - 9i + 2i - 3i^2 = 6 - 7i - 3(-1) = 6 - 7i + 3 = 9 - 7i$

Therefore, $x = 9$ and $y = -7$.

(b) $\frac{2 + 5i}{1 - i} = x + yi$

To simplify the left side, we multiply the numerator and denominator by the conjugate of the denominator:

$\frac{2 + 5i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(2 + 5i)(1 + i)}{(1 - i)(1 + i)} = \frac{2 + 2i + 5i + 5i^2}{1 - i^2} = \frac{2 + 7i - 5}{1 - (-1)} = \frac{-3 + 7i}{2} = -\frac{3}{2} + \frac{7}{2}i$

Thus, $x = -\frac{3}{2}$ and $y = \frac{7}{2}$.

(c) $3 + 4i = (x + yi)(1 + i)$

$3 + 4i = x(1) + x(i) + yi(1) + yi(i) = x + xi + yi + yi^2 = x + xi + yi - y = (x - y) + (x + y)i$

Equating the real and imaginary parts, we have:

$x - y = 3$ and $x + y = 4$.

Adding these two equations, we get $2x = 7$, so $x = \frac{7}{2}$.

Substituting this into $x + y = 4$, we have $\frac{7}{2} + y = 4$, so $y = 4 - \frac{7}{2} = \frac{8}{2} - \frac{7}{2} = \frac{1}{2}$.

Thus, $x = \frac{7}{2}$ and $y = \frac{1}{2}$.

(d) $(x + yi)^2 = 2$

$x^2 + 2xyi + (yi)^2 = 2$

$x^2 - y^2 + 2xyi = 2$

Equating the real and imaginary parts:

$x^2 - y^2 = 2$ and $2xy = 0$.

Since $2xy = 0$, either $x = 0$ or $y = 0$.

If $x = 0$, then $-y^2 = 2$, so $y^2 = -2$. Thus, $y = \pm i\sqrt{2}$, but we are looking for real values of $x$ and $y$, so we reject these solutions.

If $y = 0$, then $x^2 = 2$, so $x = \pm\sqrt{2}$.

Thus, $x = \pm\sqrt{2}$ and $y = 0$.

(e) $\frac{x + yi}{2 + i} = 5i$

$x + yi = 5i(2 + i) = 10i + 5i^2 = 10i - 5 = -5 + 10i$

Thus, $x = -5$ and $y = 10$.

(f) $\frac{1}{x + yi} + \frac{1}{x - yi} = -\frac{4}{5}i$

$\frac{(x - yi) + (x + yi)}{(x + yi)(x - yi)} = \frac{2x}{x^2 + y^2} = -\frac{4}{5}i$

Since the left side is a real number and the right side is purely imaginary, the only possible value for each side is

0. Therefore, $2x = 0$, which means $x = 0$. Also, $-\frac{4}{5}i = 0$ leads to a contradiction, so there are no solutions for this equation.

(g) $\frac{1}{x + iy} + \frac{1}{1 + 3i} = 1$

$\frac{1}{x + iy} = 1 - \frac{1}{1 + 3i} = \frac{1 + 3i - 1}{1 + 3i} = \frac{3i}{1 + 3i}$

$x + iy = \frac{1 + 3i}{3i} = \frac{(1 + 3i)(-3i)}{(3i)(-3i)} = \frac{-3i - 9i^2}{-9i^2} = \frac{-3i + 9}{9} = 1 - \frac{1}{3}i$

Thus, $x = 1$ and $y = -\frac{1}{3}$.

3. Final Answer

(a) $x = 9, y = -7$

(b) $x = -\frac{3}{2}, y = \frac{7}{2}$

(c) $x = \frac{7}{2}, y = \frac{1}{2}$

(d) $x = \pm\sqrt{2}, y = 0$

(e) $x = -5, y = 10$

(f) No solution

(g) $x = 1, y = -\frac{1}{3}$