Question: I have two questions about a specific system of differential equations. First, if a complex number can be an equilibrium point. Second, and related with the first question, how can I verify that $(0,0)$ is the only solution that satisfy $X'=0$ and $Y'=0.$ The system is as follows:
$X'=xy^3-xy^2$
$Y'=-y^3-3x^4$

Thanks a lot!

Answer: 
Normally, you only consider real equilibrium points in a first course on differential equations.

$X′=xy^3−xy^2=0$ gives $xy^3=xy^2$ so either (a) $x=0$ (b) $y=0$ or (c) $y=1$

$Y′=−y^3−3x^4=0$ gives $−y^3=3x^4$. So now all we have to do is to plug (a-c) above in to this equation to see what equilibrium points we get.

(a) $x=0$ then $−y^3=3x^4=0$ so $y=0$.  

(b) $y=0$ similarly this gives $x=0$.   

(c) $y=1$ gives $−1=3x^4$ so you could take the complex cube roots of $-1/3$, but as I said this would be unusual for a first course in differential equations to my knowledge.

So your only equilibrium point is $(0,0).$
We have two coupled polynomial ODEs

$$\dot x = xy^3 - xy^2 \qquad \qquad \dot y = -y^3 - 3x^4$$

To find the equilibria, solve the system of polynomial equations

$$0 = xy^3 - xy^2 \qquad \qquad 0 = -y^3 - 3x^4$$

Using [SymPy](http://live.sympy.org/),

```
>>> from sympy import roots, solve_poly_system
>>> x, y = symbols('x y')
>>> p1 = x * y**3 - x * y**2
>>> p2 = -y**3 - 3 * x**4
>>> solve_poly_system([p1, p2], x,y)

```

which outputs

\begin{equation\*}\left [ \left ( 0, \quad 0\right ), \quad \left ( - \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} - \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right ), \quad \left ( - \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} + \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right ), \quad \left ( \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} - \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right ), \quad \left ( \frac{\sqrt{2}}{6} 3^{\frac{3}{4}} + \frac{\sqrt{2} i}{6} 3^{\frac{3}{4}}, \quad 1\right )\right ]\end{equation\*}

Hence, the origin is the only (real) equilibrium point.