In the study of differential equations, we often have to determine the equilibrium points of an ordinary differential equation (ODE). Let us assume we have the following differential equation
$$\dot{x} = x(x^2+1).$$
For an equilibrium point, we have to determine all the values $x_\text{eq}$ such that $\dot{x}_\text{eq}=0.$ The algebraic solutions for the given example are
$$x_{\text{eq};1} = 0$$ $$x_{\text{eq};2,3} = \pm i,$$
in which $i$ is the imaginary unit. From standard stability theory, we know that the first equilibrium is unstable. The other two equilibria are not investigated because they are not real. From a practical point of view, this makes sense.
But in the spirit of a quote from Einstein 'God does not play dice with the universe' I am not fully satisfied with the practical point of view.
I am interested in the theoretical meaning of these complex valued equilibria. I do not (want to) believe that there is absolutely no real relevance of these equilibria to the behaviour of the real system. I am looking forward to inspiring and enlightening answers from our clever MathStackExchange users :).
To answer in the context of your example, you can study your ODE either in the reals or in the complex plane.
In the first case, you are looking only at real equilibria so you only need to look for the real roots of $x(x^2+1)$. The local behaviour of this real equilibrium point is determined by... local data, namely the derivative of $x(x^2+1)$ at 0. No need to look further.
On the other hand, you can look at the complex valued ODE $\dot z= z(z^2+1)$. Then you will get the three equilibria you mentionned, including the complex ones $\pm i$. Since $z \mapsto z(z^2+1)$ is analytic, the solutions of the ODE will be analytic, so they can be viewed as holomorphic functions instead of just analytic functions from a real interval to $\mathbb C$. Then the complex derivative at the equilibria plays the same role as the real derivative for the local analysis.