7. Solve for \( x \) and \( y \) given that: a) \( \quad(x+i y)(4 i)=8 \) b) \( \quad-\frac{1}{x+i y}+\frac{1}{1+3 i}=1 \) c) \( \frac{x}{1+i}-\frac{y}{2-i}=\frac{1-5 i}{3-2 i} \)

Let's tackle each equation one by one. ### a) Solve \( (x + i y)(4 i) = 8 \) First, expand the expression: \[ (4i)(x + i y) = 4ix - 4y \] This gives us: \[ 4ix - 4y = 8 \] Now, equate the real and imaginary parts. On the left side, the real part is \(-4y\) and the imaginary part is \(4x\), while on the right side, the real part is \(8\) and there is no imaginary part (0): 1. \( -4y = 8 \) 2. \( 4x = 0 \) From equation 1: \[ y = -2 \] From equation 2: \[ x = 0 \] Thus, the solution for part (a) is: \[ (x, y) = (0, -2) \] ### b) Solve \( -\frac{1}{x + i y} + \frac{1}{1 + 3i} = 1 \) First, simplify \( \frac{1}{1 + 3i} \) by multiplying the numerator and denominator by the conjugate: \[ \frac{1}{1 + 3i} \cdot \frac{1 - 3i}{1 - 3i} = \frac{1 - 3i}{1 + 9} = \frac{1 - 3i}{10} = \frac{1}{10} - \frac{3}{10}i \] Now, substitute this back into the equation: \[ -\frac{1}{x + iy} + \left(\frac{1}{10} - \frac{3}{10}i\right) = 1 \] Rearranging it gives: \[ -\frac{1}{x + iy} = 1 - \frac{1}{10} + \frac{3}{10}i = \frac{9}{10} + \frac{3}{10}i \] Next, taking the negative and moving terms gives: \[ \frac{1}{x + iy} = -\left(\frac{9}{10} + \frac{3}{10}i\right) \] To find \( x + iy \), take the reciprocal: \[ x + iy = \frac{1}{-\frac{9}{10} - \frac{3}{10}i} \] Multiply by the conjugate: \[ x + iy = \frac{-\frac{9}{10} + \frac{3}{10}i}{\left(-\frac{9}{10}\right)^2 + \left(-\frac{3}{10}\right)^2} = \frac{-\frac{9}{10} + \frac{3}{10}i}{\frac{81}{100} + \frac{9}{100}} = \frac{-\frac{9}{10} + \frac{3}{10}i}{\frac{90}{100}} = \frac{-9 + 3i}{90} = -\frac{1}{10} + \frac{1}{30}i \] Thus, we have: \[ x = -\frac{1}{10}, \quad y = \frac{1}{30} \] So, the solution for part (b) is: \[ (x, y) = \left(-\frac{1}{10}, \frac{1}{30}\right) \] ### c) Solve \( \frac{x}{1+i} - \frac{y}{2-i} = \frac{1-5i}{3-2i} \) First, simplify the right side. Multiply the numerator and denominator by the conjugate: \[ \frac{1-5i}{3-2i} \cdot \frac{3 + 2i}{3 + 2i} = \frac{(1-5i)(3+2i)}{(3)^2 + (2)^2} = \frac{3 + 2i - 15i - 10}{9 + 4} = \frac{-7 - 13i}{13} = -\frac{7}{13} - i \] Next, solve the left side separately: \[ \frac{x(2-i)}{(1+i)(2-i)} - \frac{y(1+i)}{(2-i)(1+i)} = -\frac{7}{13} - i \] The denominator simplifies to: \[ (1+i)(2-i) =