Consider the two parameter family of linear systems.

$$y'_{1}(t)=ay_{1}(t)+by_{2}(t)$$ $$y'_{2}(t)=-by_{1}(t)+ay_{2}(t)$$

I know my determinant is $$a^2+b^2$$ and my trace is $$2a$$. To find my eigenvalues I think I look at $$a\pm bi$$.

In the $$ab$$ plane identify all the regions where this system possesses a saddle, a sink, a spiral sink.

Consider that $$(y_1+iy_2)'=(a-ib)(y_1+iy_2)$$ and $$(y_1-iy_2)'=(a+ib)(y_1-iy_2).$$

You might want to check your calculations, since the equation for eigenvalues, for a 2 by 2 matrix, in terms of its determinant and trace are given in this question as

$$\lambda = \frac12 \left(\text{tr} \pm \sqrt{\text{tr}^2-4\det}\right).$$

In general it might be easier to solve it with

$$\det\left(A - \lambda I\right)=0,$$

where $A$ is a matrix, such that $\dot{\vec{y}}=A\vec{y}$, $I$ the identity matrix of the same size as $A$ and $\lambda$ the eigenvalues, which have to be solved for such that the equation is true.

• Oh thanks! I figured out what I did wrong. I have that my eigenvalues values are a+_bi, does this mean that in my trace determinant graph, I got that spirals sink in the 2nd and 3rd quadrant and spiral sources in the 1st and 4th and center along the b'axis? Commented Oct 23, 2015 at 3:02

Here is another way of solving this problem, along the lines of my answer here: Trace Determinant Plane Differential Eqns. Let's assume that we are allowed use the trace-determinant plane and the different behavioral regions of it.

Let us first put $$A_{a,b}=\begin{pmatrix} a& b\\ -b & a\end{pmatrix}$$. Then like the OP states $$\operatorname{tr}(A_{a,b})=2a$$ and $$\det{A_{a,b}}=a^2+b^2$$. After the answer I referred to above, let us consider the coordinate change from the $$ab$$-plane to the $$td$$-plane

$$\Phi: \mathbb{R}^2\to \mathbb{R}^2,\,\, \begin{pmatrix} a\\b \end{pmatrix}\mapsto \begin{pmatrix} 2a\\a^2+b^2 \end{pmatrix}.$$

As in my previous answer, this coordinate change is natural as $$\Phi(a,b)=(\operatorname{tr}(A_{a,b}),\det(A_{a,b}))$$.

Next we need to optimize $$\Phi$$ a bit. First note that $$\Phi\begin{pmatrix}a\\-b\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}$$, so that $$\Phi$$ is two-to-one except when $$b=0$$, in which case it is one-to-one. Next note that $$\dfrac{(2a)^2}{4}=a^2\leq a^2+b^2$$, so that the image of $$\Phi$$ is precisely

$$R=\left\{(t,d)\in\mathbb{R}^2\,\left|\, \dfrac{t^2}{4}\leq d\right.\right\}.$$

Thus we have two inverses for $$\Phi$$, one with the upper half $$ab$$-plane as the image and another with the lower half $$ab$$-plane as the image:

\begin{align*} \Phi^{-1}_+:&R \to \mathbb{R}\times \mathbb{R}_{\geq0},\, \begin{pmatrix}t\\d\end{pmatrix}\mapsto \begin{pmatrix}\frac{t}{2}\\\sqrt{d-\frac{t^2}{4}}\end{pmatrix}\\ \Phi^{-1}_-:&R \to \mathbb{R}\times \mathbb{R}_{\leq0},\, \begin{pmatrix}t\\d\end{pmatrix}\mapsto \begin{pmatrix}\frac{t}{2}\\-\sqrt{d-\frac{t^2}{4}}\end{pmatrix} \end{align*}

Here is a humble caricature summarizing the situation (the trace-determinant plane is from Hirsch, Smale and Devaney's Differential Equations, Dynamical Systems, and an Introduction to Chaos (3e, p.64)):

This exactly matches with the OP's conclusion in a comment.

Finally note that in the critical case $$\dfrac{t^2}{4}=d$$, we have $$b=0$$, so that $$A_{a,b}=A_{\frac{t}{2},0}=\begin{pmatrix}\frac{t}{2}&0\\0&\frac{t}{2}\end{pmatrix}$$ is a diagonal matrix; hence accordingly we have either a star sink (if $$t<0$$), a star source (if $$t>0$$), or all equilibrium solutions (if $$t=0$$).