No point in two "pictures"
$\renewcommand{\Re}{\operatorname{Re}}$
Characteristic equation
$$
\left|\begin{matrix}
1-k & -2\\ 2 &1-k\end{matrix}\right|=(k-1)^2+4\implies k_{1,2}=1\pm 2i.
$$
Finding eigenvectors
$$
k_1=1+2i\implies
\begin{pmatrix}
-2i & -2\\ 2 &-2i
\end{pmatrix}
\begin{pmatrix}
\alpha\\ \beta
\end{pmatrix}=
\begin{pmatrix}
0\\ 0
\end{pmatrix}\implies \beta=-i\alpha\implies \mathbf{e}=\begin{pmatrix}
1\\ -i
\end{pmatrix}
$$
and $k_2$ and $\mathbf{e}_2$ are complex conjugate. Then
\begin{align*}
\mathbf{x}=\Re \Bigl[(C_1+iC_2)\begin{pmatrix}
1\\ -i
\end{pmatrix}
e^{(1+2i)t}\Bigr]=
&e^{t}\Re \Bigl[(C_1+iC_2)\begin{pmatrix}
1\\ -i
\end{pmatrix}
(\cos(2t)+i\sin(3t))\Bigr]=\\
&e^{t} \begin{pmatrix}
C_1\cos(2t)-C_2\sin(2t)\\
C_1\sin(2t)+C_2\cos (2t)
\end{pmatrix}
\end{align*}
