Complex Equations

Relevant wiki: Solving Linear Systems Using Matrices

We can rewrite the system of equations as the product of matrices as shown below:

$\begin{bmatrix} 2-3i & 5+i \\ 1-i & 2+i \end{bmatrix} \begin{bmatrix} z \\ w \end{bmatrix} = \begin{bmatrix} 19-4i \\ 9-i \end{bmatrix}$

Let $A = \begin{bmatrix} 2-3i & 5+i \\ 1-i & 2+i \end{bmatrix}$ . Then $\det(A) = (2-3i)(2+i)-(5+i)(1-i) = (7-4i)-(6-4i) = 1$ .

Therefore, we can obtain the inverse matrix of $A$ :

$A^{-1} = \begin{bmatrix} 2+i & -5-i \\ -1+i & 2-3i \end{bmatrix}$ .

Hence, $A^{-1}\cdot A \begin{bmatrix} z \\ w \end{bmatrix} = \begin{bmatrix} z \\ w \end{bmatrix} = A^{-1}\begin{bmatrix} 19-4i \\ 9-i \end{bmatrix}$ .

$\begin{bmatrix} z \\ w \end{bmatrix} = \begin{bmatrix} 2+i & -5-i \\ -1+i & 2-3i \end{bmatrix} \begin{bmatrix} 19-4i \\ 9-i \end{bmatrix} = \begin{bmatrix} (2+i)(19-4i)+(-5-i)(9-i) \\ (-1+i)(19-4i)+(2-3i)(9-i) \end{bmatrix} = \begin{bmatrix} (42+11i)+(-46-4i) \\ (-15+23i)+(15-29i) \end{bmatrix} = \begin{bmatrix} -4+7i \\ -6i \end{bmatrix}$

Thus, $z+w = -4+7i -6i = -4+i$ .

Then, $|z+w|^2 = 4^2 + 1^2 = \boxed{17}$ .