Linear autonomous systems in the plane: When do phase curves rotate clockwise? For a linear autonomous system in the plane
$$ \mathbf{\dot{x}} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mathbf{x} \qquad (a,b,c,d \in \mathbb{R})$$
with determinant $D = ad - bc$ and trace $T = a + d$ we have the characteristic polynomial
$$ \chi(\lambda) = \lambda^2 - T\lambda + D$$
and the eigenvalues
$$ \lambda_{1, 2} = \frac{T \pm \sqrt{\Delta}}{2}, \qquad \Delta = T^2 - 4D $$
We know that the phase curves will rotate around the origin iff $\Delta < 0$.
Question: How can one determine using $a, b, c, d$, whether the phase curves turn clockwise or counter-clockwise?
Counterexample: The System
$$ \mathbf{\dot{x}} = \begin{pmatrix} 1 & s\\ -s & 1 \end{pmatrix}\mathbf{x} \qquad (s \in \mathbb{R} \setminus \{ 0 \})$$
will rotate clockwise for $s > 0$ and counter-clockwise for $s<0$, but in both cases we have
$$ D = 1 + s^2 > 0, \qquad T = 2, \qquad \Delta = -4s^2 < 0 $$
so these quantities will not suffice to determine the orientation.
 A: Inspired by amd's help I did the math:
The original system
$$ \mathbf{\dot{x}} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mathbf{x}, \qquad D = ad - bc, \qquad T = a + d $$
has the eigenvalues
$$ \lambda_{\pm} = \frac{T \pm i\sqrt{|\Delta|}}{2}, \qquad |\Delta| = 4D - T^2 > 0 $$
with corresponding eigenvectors
$$ v_{\pm} = \begin{pmatrix} -a+d \mp i\sqrt{|\Delta|}\\ -2c \end{pmatrix} $$
We want to relate it to the system
$$ \mathbf{\dot{w}} = \begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix}\mathbf{w}, \qquad \alpha = \Re\left(\lambda_{+}\right) = \frac{T}{2}, \qquad \beta = \Im\left(\lambda_{+}\right) = \frac{\sqrt{|\Delta|}}{2} > 0 $$
which is spinning counter-clockwise because of $\beta > 0$ (as pointed out by amd) and has the same eigenvalues, but now with the corresponding eigenvectors
$$ w_{\pm} = \begin{pmatrix} \pm i\\ 1 \end{pmatrix} $$
We are looking for a Matrix $B$ such that
$$ \begin{pmatrix} a & b\\ c & d \end{pmatrix} = B\begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix}B^{-1}, \qquad \mathbf{x} = B\mathbf{w} $$
Such a matrix is uniquely defined by the condition
$$ (v_{+} \mid v_{-} ) = B(w_{+} \mid w_{-} ) = B\begin{pmatrix} i & -i\\ 1 & 1\end{pmatrix} $$
and hence computes to
$$ B = \begin{pmatrix} -a+d - i\sqrt{|\Delta|} & -a+d + i\sqrt{|\Delta|}\\ -2c & -2c \end{pmatrix} \frac{1}{2i} \begin{pmatrix} 1 & i\\ -1 & i \end{pmatrix} = \begin{pmatrix} -\sqrt{|\Delta|} & d - a\\ 0 & -2c\end{pmatrix} $$
with determinant $\det B = 2c\sqrt{|\Delta|}$ yielding the result

  
*
  
*$c > 0$: The matrix $B$ preserves orientation as $\det B > 0$, hence $\mathbf{x} = B\mathbf{w}$ rotates counter-clockwise just like $\mathbb{w}$.
  
*$c < 0$: The matrix $B$ inverts orientation as $\det B < 0$, hence $\mathbf{x} = B\mathbf{w}$ rotates clockwise.
  
*$c = 0$: Leads to real eigenvalues $a$ and $d$, so the system does not rotate.
  

Should you be surprised that the orientation only depends on $c$, then please note: If
$$ B = \begin{pmatrix} e & f\\ g & h \end{pmatrix} \in GL(2,\mathbb{R}) $$
is an invertible real matrix, then we have $g \neq 0$ or $h \neq 0$ and
$$ B\begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix}B^{-1} = \begin{pmatrix} ? & -\beta\frac{e^2 + f^2}{\det B}\\ \beta\frac{g^2 + h^2}{\det B} & ? \end{pmatrix} $$
and since $g^2 + h^2 >0$ the sign of the "lower left coefficient" only changes if $\det B < 0$.
A: You answered your question yourself, but here is somewhat easier answer. 
First note that to have the origin with complex eigenvalues if is necessary to have $b,c\neq 0$ and $bc<0$. Moreover $(a-d)^2+4bc<0$.
Now consider the expression
$$
\frac{d}{dt}\arctan\frac{x_2(t)}{x_1(t)}=\frac{\dot x_2 x_1-\dot x_1x_2}{x_1^2+x_2^2}=\frac{cx_1^2+(d-a)x_1x_2-bx_2^2}{x_1^2+x_2^2}\,.
$$
Assuming that $b>0$ or (equivalently, $c<0$), one should show that the numerator is a negative definite form, which means that
$$
\dot \theta(t)<0,
$$
or that the rotation is clockwise. Similarly for the case $b<0,c>0$, which, of course, coincides with your work.
A: In short, you’ve more or less answered your own question: it depends on the sign of $s$.  
If the matrix of the equation is of the form $\pmatrix{\alpha&-\beta\\ \beta&\alpha}$, with $\beta$ positive, the phase curves are counterclockwise spirals with the direction of motion dependent on $\alpha$: When $\alpha>0$, the point moves outward as $t\to\infty$; when $\alpha<0$, the point moves inward instead.
In your specific example, when $s<0$, the solution is $$e^t\pmatrix{\cos\beta t&-\sin\beta t\\ \sin\beta t&\cos\beta t},$$ where $\beta=-s>0$. When $s>0$, this is the same as replacing $\beta$ with $-\beta$ in the solution, which will clearly spiral clockwise as a result.  
Another way to look at the $s>0$ case is to rewrite the matrix in the differential equation as $$\pmatrix{-1&0\\0&1}\pmatrix{1&-s\\ s&1}\pmatrix{-1&0\\0&1}^{-1}.
$$ This transformation flips the $x$-axis, turning the counterclockwise spiral into a clockwise one.
