How to show that every non-trivial orbit of ODE is a periodic circle I am working with this system of first-order ODEs.
\begin{align}
\dot{x} & = y^2 - x^2 \ ,\\
\dot{y} & = - 2 x y\ .
\end{align}
We were told to convert the system using $z=x+iy$ where I found that $z'=-z^2$. From there we were asked to show that every other orbit is a circle. I am not sure where to begin. I also looked at the rest point which occurs at the origin and believe it to by hyperbolic.
 A: I will ask you to criticize my decision, because I just thought of it, there may be inaccuracies.
I assumed that
$$ t, x(t),y(t) \in \mathbb{R}$$
$$ z(t) = x(t) + y(t)i ,  $$
$$ \frac{dz}{dt} = \frac{dx}{dt} + \frac{dy}{dt}i$$
$$ -z^{2} = (y^2 - x^2) - 2x\cdot y \cdot i $$
$$ \frac{dz}{dt} = -z^2$$
$$ \frac{1}{z} = t + C$$
where $C$ is an arbitrary constant
$$ \frac{x-yi}{x^2+y^2} = \frac{x}{x^2+y^2} + \left[\frac{-y}{x^2+y^2}\right ] i =  t + C $$
we understand that $ t + C = t + (A + Bi) \in \mathbb{C}
\Rightarrow$
\begin{equation*}
 \begin{cases}
   \frac{x(t)}{x^{2}(t)+y^2(t)} = t + A, 
   \\
   \frac{-y(t)}{x^2(t)+y^2(t)} = B,
 \end{cases}
\end{equation*}
from the second equation we obtain the equation of a circle with a center at a point $(0 ; - \frac{1}{2B})$ and a radius $\frac{1}{2|B|}$
$$  x^2 + y^2 = \frac{-y}{B}$$
$$ x^2 + y^2 + \frac{2y}{2B} + \left(\frac{1}{2B}\right)^2 = \left(\frac{1}{2B}\right)^2 $$
$$ x^2 + \left(y + \frac{1}{2B} \right)^2 = \left(\frac{1}{2B}\right)^2 $$
If you decide to count this decision as correct, then I want to warn you that Lutz Lehmann helped me
