Find the fixed points and draw the phase space

Question

\begin{align} x'&=x(2-x-y) \\ y'&=y(4x-x^2-3) \end{align}

• a) find the three fixed points and classify them
• b) sketch the phase portrait

I've found the first two fixed points $(0,0)$ and $(2,0)$ but I am confused to finding the third. I divided both equations by $x$ and $y$ respectively, and tried solving for $y$, but I am getting $\pm 2i$. I am not sure if this is correct. I understand how to find/classify them using the Jacobian matrix and the determinant/trace, but I do not understand how to draw the phase portrait (does that come from the eigenvalues/eigenvectors at each point?) $$ J = \pmatrix{ 2-2x-y & -x \\ 4y-2xy & 4x-x^2-3} $$

Answer 1

The singular points are solved as the solution for

$$ y(4x-x^2-3) = 0\\ x(2-x-y)=0 $$

or

$$ (1,1),(2,0),(3,-1),(0,0) $$

The Jacobian calculated at those points have respective eigenvalues

$$ (\frac{1}{2}(-1-i\sqrt 7),\frac{1}{2}(-1+i\sqrt 7)),(-2,1),(\frac{1}{2}(-3-i\sqrt{15}),\frac{1}{2}(-3+i\sqrt{15})),(-3,2) $$

so they can be classified as

sink, saddle, sink, saddle

Attached a plot showing the phase plane and the equilibrium points.