Question:
Let $\mu:\mathbb C\to \mathbb C$ be a $C^\infty$ function satisfying $|\mu|\le 1$.
Let us furthermore assume that the function $\mu$ never takes the value $-1$.
Does there exist a $C^\infty$ function
$f:\mathbb C\to \mathbb C$ such that $\frac{\,\partial f/\partial \bar z\,}{\partial f/\partial z}=\mu$ ?
(and $\scriptstyle\partial f/\partial z$ everywhere non-zero)
Note that $f$ is by no means expected to be a homeomorphism. For example, if $\mu\equiv 1$, then a general solution is of the form projection $\mathbb C\to\mathbb R:z\mapsto Re(z)$ followed by any embedding $\mathbb R\to\mathbb C$.
The special case $|\mu|<1$ is the usual Beltrami equation.
More generally, on which domains $U\subset\mathbb C$ can one guarantee that a solution exists?
[The exact problem I care about is defined on the strip $U=\{z\in\mathbb C:0\le Im(z)\le 1\}$, and the function $f$ is furthermore required to be periodic: $f(z)=f(z+1)$.]
Equivalent formulation:
Let $\nu:\mathbb C\to \mathbb C$ be a $C^\infty$ function satisfying $Im(\nu)\ge0$.
$\exists$? $f:\mathbb C\to \mathbb C$ such that $\frac{\,\partial f/\partial y\,}{\partial f/\partial x}=\nu$ ?
(and $\scriptstyle\partial f/\partial x$ everywhere non-zero)
(The two formulations are equivalent via the transformations $\mu=\frac{1+i\nu}{1-i\nu}$ and $\nu=i\frac{1-\mu}{1+\mu}$, which are a version of the Cayley transform. These transformations map $\nu$ in the closed upper half plane to $\mu$ in the closed unit disc minus $-1$.)
Partial result (positive solution in the completely degenerate case):
When $\nu$ is real, equivalently when $|\mu|=1$, then the problem admits a positive solution.
In that case, the vector field $\partial_y-\nu\partial_x$ defines a foliation of $\mathbb C$. That vector field is never horizontal, so all the leaves of the foliation are closed (diffeomorphic to $\mathbb R$). Any function $f:\mathbb C\to\mathbb C$ which is constant on the leaves (i.e., which factors through the 1-dimensional leaf space) is then a solution of the degenerate Beltrami equation $\frac{\,\partial f/\partial y\,}{\partial f/\partial x}=\nu$.
Note that, unlike the leaf space of a general foliation of $\mathbb C$, the leaf space of this specific foliation is a manifold (by which I mean a second-countable possibly non-Hausdorff manifold), and can thus be immersed into $\mathbb C$. That's where the crucial condition $\mu\not=1$ gets used.
