Solution of this non-linear PDE What is the general solution $h: \mathbb{R}^2 \rightarrow \mathbb{C}$ (or maybe $h: \mathbb{C}^2 \rightarrow \mathbb{C}$ if necessary), $(x, y) \mapsto h(x, y)$ to the non-linear partial differential equation
$$
h_x^2 + h_y^2 = -1
$$
where the subscript denotes a partial derivative. I tried to approach the problem by looking at limiting cases and using ansatz, but nothing interesting came out.
 A: Changing coords 
$$
x\to iv\\
y\to iu
$$
We find
$$
\partial_x = -i\partial_v\\
\partial_y = -i\partial_u
$$
Thus you get
$$
\left(-i\partial_vh\right)^2 + \left(-i\partial_uh\right)^2 = -\left(h_v^2 + h_u^2\right) = -1
$$
Thus you get 
$$
h_v^2 + h_u^2 = 1
$$
Solutipns of the above have the form given by
$$
h^2 = (v-c_1)^2 + (u-c_2)^2
$$
where
$$
c_1^2 +c_2^2 = 1
$$
Leading to
$$
h(x,y) =\sqrt{ (-ix-c_1)^2 + (-iy-c_2)^2}\\
h(x,y) =\pm i\sqrt{(x+ic_1)^2 +(y+ic_2)^2}\\
$$
A: $h_x^2+h_y^2=-1$
$h_x^2=-h_y^2-1$
$h_x=\pm i\sqrt{h_y^2+1}$
$h_{xy}=\pm\dfrac{ih_yh_{yy}}{\sqrt{h_y^2+1}}$
Let $u=h_y$ ,
Then $u_x=\pm\dfrac{iuu_y}{\sqrt{u^2+1}}$
$u_x+\mp\dfrac{iuu_y}{\sqrt{u^2+1}}=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dy}{dt}=\mp\dfrac{iu}{\sqrt{u^2+1}}=\mp\dfrac{iu_0}{\sqrt{u_0^2+1}}$ , letting $y(0)=f(u_0)$ , we have $y=\mp\dfrac{iu_0t}{\sqrt{u_0^2+1}}+f(u_0)=\mp\dfrac{iux}{\sqrt{u^2+1}}+f(u)$ , i.e. $u=F\left(y\pm\dfrac{iux}{\sqrt{u^2+1}}\right)$
