How to find solutions to nonlinear ODE $\dot{z}=i|z|^2z$ I have the following ODE for complex $z$:
$$\dot{z}=i|z|^2z$$
and I would like to find the most general exact solution possible. It is easy to see that $z=0$ and $z=e^{it}$ are two solutions, but I am hoping for a way to see if these are the only ones or if more may be found.
Thanks
 A: The complex conjugate of your equation (which we denote by (1)) is
$$
\overline z'(t)=-i|z|^2\overline z(t),\qquad (2).
$$
Let's add $\overline z\cdot(1)$ to $z\cdot (2)$:
$$
\frac{d}{dt}\left(z(t)\overline z(t)\right)=0.
$$
This implies $|z(t)|=|z(0)|=:r$.  Then (1) reduces to
$$
z'(t)=ir^2 z(t)
$$
This means $z(t)=e^{ir^2t}z(0)$, so the general solution is
$$
z(t)=z(0)\exp(i|z(0)|^2 t)=:r e^{i(r^2t+\phi)}
$$
if $z(0)=r e^{i\phi}$.
A: Assume
$z \ne 0; \tag{1}$
then we may write
$z = r e^{i \theta}, \tag{2}$
so
$\dot z = \dot r e^{i \theta} + ir \dot \theta e^{i \theta}; \tag{3}$
$\vert z \vert^2 = r^2.  \tag{4}$
We thus assemble the equation
$\dot r e^{i \theta} + i r \dot \theta e^{i \theta} = ir^3 e^{i \theta}; \tag{5}$
$e^{i \theta}$ cancels:
$\dot r + ir \dot \theta = ir ^3; \tag{6}$
we equate real and imaginary parts, cancel a factor of $ir$:
$\dot r = 0, \tag{7}$
$\dot \theta = r^2.  \tag{8}$
We see that $r = r_0$ is constant and
$\theta = r_0^2 t + \theta_0; \tag{9}$
finally,
$z(t) = r_0 e^{i(r_0^2 t + \theta_0)}; \tag{10}$
we check:
$\dot z = ir_0^3 e^{i(r_0^2 t + \theta _0)} = i\vert z \vert^2 z.  \tag{11}$
Taking $r_0 = 1$, $\theta_0 = 0$ we obtain
$z = e^{it}; \tag{12}$
the remaining non-zero solutions are circles surrounding the origin; the rotation is counter'clockwise at a rate of $r_0^2$ rad/sec.
A: Let $z = x + iy$, where $x, y$ are real functions in $t$. Then the ODE
$$\dot{x} + i\dot{y} = i(x^2 + y^2)(x+iy)$$
In particular, $$\dot{x} = -y(x^2 + y^2), \dot{y} = x(x^2 + y^2)$$
Then, $$\frac{dy}{dx} = -\frac{x}{y} \implies \int ydy = -\int x dx$$
Solving, we find that any solution must have constant modulus, independent of time, i.e. $$x(t)^2 + y(t)^2 = C^2$$ where $C$ is a constant of integration.  Back to the ODE, it reduces to $\dot{z} = iC^2 z$. 
Now, in polar form $z = Ce^{i\theta(t)}$. Substituting in the ODE gives $$iC\dot{\theta(t)}e^{i\theta(t)}=iC^3e^{i\theta(t)}$$ That is, $\dot{\theta}(t) = C^2$. Therefore, $\theta(t) = C^2t + C_2$. And so $$z(t) = Ce^{i(C^2t+C_2)}$$ 
