Complex Eigenvalues

Sorry about being a little misleading with the complex eigenvalues.

Let me try again:

Let's say we have eigenvalue $\lambda=a+ib$ and eigenvector$\xi=\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)+i\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)$.

Then the solution $\xi e^{\lambda t}=e^{at}\left[\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\cos bt-\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\sin bt\right]+i\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt+\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]\right]$

Thus we get the two real solutions: $x_{1}(t)=\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\cos bt-\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\sin bt\right]$

and $x_{2}(t)=e^{at}\left[\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt+\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]$.

You don't need to memorize these formulas, I'm just doing the problem
abstractly.

To determine the direction of the rotation, let's look at $x_{1}$.

$x_{1}(0)=\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)$

Now let's look at the derivative: $x_{1}^{\prime}(t)=\left[-b\left(\begin{array}{c}
u_{1}\\
u_{2}\end{array}\right)\sin bt-b\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)\cos bt\right]$

$x_{1}^{\prime}(0)=-b\left(\begin{array}{c}
v_{1}\\
v_{2}\end{array}\right)$. The spiral should be tangent to this vector at $x_{1}(0)$ and be
the direction goes in the direction of this vector.