13. Write the complex number \( z \) in \( x+\mathrm{i} y \) form, where \( x, y \in \mathbb{Z} \) if \( z \) satisfies the equation \( \sqrt{z}=\frac{2}{1-\mathrm{i}}+1-4 \mathrm{i} \) 15. Find the real numbers \( p \) and \( q \) that satisfy the equation \( 2(p+\mathrm{i} q)=q-\mathrm{i} p-2(1-\mathrm{i}) \)

To solve the first problem, we start by simplifying the right-hand side of the equation \( \sqrt{z}=\frac{2}{1-\mathrm{i}}+1-4 \mathrm{i} \). 1. First, we multiply the numerator and the denominator of \( \frac{2}{1-\mathrm{i}} \) by the conjugate of the denominator: \[ \frac{2(1+\mathrm{i})}{(1-\mathrm{i})(1+\mathrm{i})} = \frac{2(1+\mathrm{i})}{1^2 + 1^2} = \frac{2(1+\mathrm{i})}{2} = 1 + \mathrm{i} \] 2. Next, we substitute this back: \[ \sqrt{z} = (1 + \mathrm{i}) + (1 - 4\mathrm{i}) = 2 - 3\mathrm{i} \] 3. To find \( z \), we square both sides: \[ z = (2 - 3\mathrm{i})^2 = 4 - 12\mathrm{i} - 9(-1) = 4 - 12\mathrm{i} + 9 = 13 - 12\mathrm{i} \] Thus, the complex number \( z \) in \( x+\mathrm{i}y \) form is \( 13 - 12\mathrm{i} \). --- For the second problem, let's solve \( 2(p+\mathrm{i} q)=q-\mathrm{i} p-2(1-\mathrm{i}) \). 1. First, simplify the right-hand side: \[ = q - \mathrm{i} p - 2 + 2\mathrm{i} = q - 2 + (2 - p)\mathrm{i} \] 2. Now equate the real and imaginary parts on both sides: Real part: \[ 2p = q - 2 \] Imaginary part: \[ 2q = 2 - p \] 3. We can express \( q \) from the first equation: \[ q = 2p + 2 \] 4. Substitute \( q \) into the second equation: \[ 2(2p + 2) = 2 - p \rightarrow 4p + 4 = 2 - p \rightarrow 5p = -2 \rightarrow p = -\frac{2}{5} \] 5. Substitute \( p \) back to find \( q \): \[ q = 2(-\frac{2}{5}) + 2 = -\frac{4}{5} + 2 = -\frac{4}{5} + \frac{10}{5} = \frac{6}{5} \] Thus, the real numbers are \( p = -\frac{2}{5} \) and \( q = \frac{6}{5} \).