Functions whose derivatives cycle through the cube roots of unity Does anyone know if there exists three complex-valued functions, $f, g, h$ such that
$f' = e^{\frac{1}{3}i\pi}g$
$g' = e^{\frac{2}{3}i\pi}h$
$h' = f$  
and are they unique?
Thanks.
 A: The general solution $(f,g,h)$ is given by
$$
f(z)=\sum_\omega c_\omega\mathrm{e}^{\omega z},\quad
g(z)=\mathrm{e}^{-i\pi/3}\sum_\omega \omega c_\omega\mathrm{e}^{\omega z},\quad
h(z)=-\sum_\omega\omega^2c_\omega\mathrm{e}^{\omega z},
$$
where the sums are over the three cubic roots $\omega$ of $-1$, that is $\omega\in\{\mathrm{e}^{i\pi/3},-1,\mathrm{e}^{-i\pi/3}\}$, and the three coefficients $c_\omega$ are any complex numbers. 
The proof is simple: Start from the observation that if $(f,g,h)$ is a solution then the third derivative of $f$ must be $-f$. Deduce from this observation that $f$ must be of the form above. Then assume $f$ is any function of the form above and is part of a solution $(f,g,h)$, deduce the value of $g$ by the first differential equation, then the value of $h$ by the second differential equation. Finally, check that these $f$ and $h$ solve the third differential equation. They do, hence you are done.
A: For an example, how about
$f(z) = e^{\zeta_6 z}$ where $\zeta_6$ = $e^{2\pi i/6}.$ 
As for finding all the solutions,
any $f$ with the properties you desire must satisfy the differential equation $D^3(f) + f = 0.$ Thus, the space of solutions for this equation is 3-dimensional generated and by the functions $e^{-z},e^{\zeta_6 z},e^{\zeta_6^5 z}.$
