Complex Numbers with the prey-predator equilibrium?

Consider a model (very similar to Lotka-volterra prey-predator -model, exception $h$).

$$\frac{dx}{dt} = h+x(\alpha -\beta y)$$ $$\frac{dy}{dt} = -y(\gamma - \rho x).$$

Let's write this in matrix form:

$$\left(\begin{array}{c}\dot{x} \\ \dot{y} \end{array} \right) = \left( \begin{array}{cc} \alpha & -\beta x \\ \rho y & -\gamma \end{array} \right) \left(\begin{array}{c} x \\ y \end{array} \right) + \left(\begin{array}{c} h \\ 0 \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \end{array} \right)$$

where the final right hand-side is due to the definition of the equilibrium. So we have a system of the form $\bar{0} = A \bar{v} + \bar{h}$. The equation $\dot{f} = A \bar{v} + \bar{h}$ is non-linear ODE, more here, so I need to investigate the $J_{f} := \left[ \frac{\partial f_{i}}{\partial y_{j}} \right]$ where $f(\bar{x}, \bar{y}) = \frac{d \bar{v_{i}}}{d t}$ (read this link, clarifies things a lot!) So the Jacobian is needed for the stability more here. I am trying to solve the problem 5 here.

Clarifications needed, Some Helper Questions

1. Fix-points? Some material mention things called "fixpoints". $F(\bar{p}) =0$ and then checking some condition, thinking -- really, no need to check it like the history?

2. linearization? Confused (see the history).

First you have to find the equilibrium points of this system. One possibility is $x=\frac{-h}{\alpha}, y=0$ and another one is $x=\frac{\gamma}{\rho}, y = \frac{\alpha}{\beta}+\frac{\rho h}{\beta\gamma}$ (plug these values in to see it for yourself).

What the idea of linearization simply stating is: Take the 1st-order Taylor series approximation of the solutions. Around some(!) neighborhood of any point in the state space, the nonlinear system $\dot x = f(x)$ behaves like a linear system $\dot x = Ax$. For this you have to use the Jacobian (first term of Taylor approx.) evaluated at the point of linearization. If this point moreover, is an eq. point then the constant term vanishes and you are left with a linear system, otherwise a drift term stays. Thus the overall system describes $$\frac{d\Delta x}{dt} = A (x_0+\Delta x)$$ i.e. the small perturbation dynamics around the eq. point $x_0$. If $A$ is stable your system goes back to the eq. point or it goes to another eq. point or blows up.

Take the first eq. point as an example: $$\left.J\right|_{eq1} = \left.\pmatrix{\alpha-\beta y &-\beta x\\ \rho y &\rho x-\gamma}\right|_{x=\frac{-h}{\alpha}, y=0}=\pmatrix{\alpha & \frac{\beta h}{\alpha}\\ 0 &-\rho\frac{h}{\alpha}-\gamma }$$ Note that you have changed the sign of the variables so I can't directly tell you if this is a stable linearization, for example $\alpha = -\mu$. Check if it really is (question 3 asks this). Now you can plug in the second equilibrium point and you will get: $$\left.J\right|_{eq2} = \left.\pmatrix{\alpha-\beta y &-\beta x\\ \rho y &\rho x-\gamma}\right|_{x=\frac{\gamma}{\rho}, y = \frac{\alpha}{\beta}+\frac{\rho h}{\beta\gamma}}=\pmatrix{\frac{-\rho h}{\gamma} &-\frac{\beta\gamma}{\rho}\\ \frac{\rho\alpha}{\beta}+\frac{\rho^2 h}{\beta\gamma} &0 }$$ The rest is simple since the eigenvalues are the solutions of $$\lambda^2 + \frac{\rho h}{\gamma}\lambda + (\gamma\alpha +\rho h) = 0$$ From this you can compute the eigenvalues and see when these variables make the discriminant negative.