10 b.

The characteristic polynomial for the coefficient matrix is $\lambda^2 +2\lambda + 5$ . The roots of this are $\lambda = -1 \pm 2i$.  These are the eigenvalues of the coefficient matrix.

To find the eigenvectors, solve

with $\lambda=-1+2i$.

We get $(2-2i)a_1-4a_2=0$.  So, if $a_1=4$ then $a_2=2-2i$.  Hence an eigenvector is $\left[\begin{array}{c} 4 \\ 2-2i \end{array}\right]$

So a complex solution is

So
$x(t)=4e^{-t}(c_1\cos(2t)+c_2\sin(2t))$
and
$y(t)=2e^{-t}\big(c_1(\cos(2t)+\sin(2t))+c_2(-\cos(2t)+\sin(2t))\big)$
$y(t)$ could also be written as
$y(t)=2e^{-t}\big((c_1-c_2)\cos(2t)+(c_1+c_2)\sin(2t)\big)$

Here is a slope field with a solution curve through the point $(0,1)$ at time $t=0$ with a counter-clockwise direction: