Do there exist Jordan analytic curves $J$ in the complex plane $C$, other than circles, with the following property:


There exists a harmonic function $u$ in the unbounded component of $C\backslash J$, 
$u(z)=0$ and $|\mathrm{grad}\, u(z)|=1$ for $z\in J$.


If we impose mild restrictions on $u$ near $\infty$, for example that $u$ is bounded
from below or from above, then such things do not exist.



A free boundary problem
Sorry, I solved it few hours after posting.


Consider two functions
$$\phi(z)=z+\frac{2a^2}{z}-\frac{a^4}{3z^3},$$
and
$$v(z)=(1+a^4)\log|z|-\frac{a^2}{2}\Re(z^2-z^{-2}).$$
They have the following properties.


(i) When $a>0$ is sufficiently small, $\phi$ is
univalent in $\Delta=\{ z:|z|\geq 1\}$, and $\phi'(z)\neq 0$ in $\Delta$.


(ii) $v$ is harmonic in $\Delta$ and $v(z)=0$ on the unit circle.


(iii) For $z$ on the unit circle, we have $|\mathrm{grad}\, v(z)|=|\phi'(z)|$.


First two properties are evident, and (iii) is verified by
direct calculation.


We choose $J=\phi(\{ z:|z|=1\})$, and $u=v\circ\phi^{-1}$
in the outer component of $J$.
Evidently
So $u$ is harmonic
and $u(z)=0$ on $J$. Finally for $z\in J$ we have
$$|\mathrm{grad}\, u|=|\mathrm{grad}\, v||(\phi^{-1})'|=1,$$
in view of (iii).


