Consider the linear system

$$\dot{\mathbf{z}}+A \mathbf{z}=\mathbf{h}$$

where

$$\mathbf{z}(t)=\left(\begin{array}{c} x(t) \\ y(t) \end{array}\right), \quad A=\left(\begin{array}{cc} 1+a & -2 \\ 1 & -1+a \end{array}\right), \quad \mathbf{h}(t)=\left(\begin{array}{c} 2 \cos t \\ \cos t-\sin t \end{array}\right)$$

where $\mathbf{z}(t)$ is real and $a$ is a real constant, $a \geq 0$.

Find a (complex) eigenvector, e, of $A$ and its corresponding (complex) eigenvalue, $l \underline{l}$. Show that the second eigenvector and corresponding eigenvalue are respectively $\overline{\mathbf{e}}$ and $\bar{l}$, where the bar over the symbols signifies complex conjugation. Hence explain how the general solution to $(*)$ can be written as

$$\mathbf{z}(t)=\alpha(t) \mathbf{e}+\bar{\alpha}(t) \overline{\mathbf{e}},$$

where $\alpha(t)$ is complex.

Write down a differential equation for $\alpha(t)$ and hence, for $a>0$, deduce the solution to $(*)$ which satisfies the initial condition $\mathbf{z}(0)=\underline{0}$.

Is the linear system resonant?

By taking the limit $a \rightarrow 0$ of the solution already found deduce the solution satisfying $\mathbf{z}(0)=0$ when $a=0$.