Cauchy-Riemann conditions implied by harmonicity and orthogonal gradients? If I have two functions $\eta:\mathbb{R}^2\to\mathbb{R}$ and $\xi:\mathbb{R}^2\to\mathbb{R}$ for which the following relations hold:
$$
\begin{align}
\left( \frac{\partial\xi}{\partial x} \right)^2 + \left( \frac{\partial\xi}{\partial y} \right)^2 &= \left( \frac{\partial\eta}{\partial x} \right)^2 + \left( \frac{\partial\eta}{\partial y}\right)^2 \neq 0\,,\\
\frac{\partial^2\xi}{\partial x^2} + \frac{\partial^2\xi}{\partial y^2} &= \frac{\partial^2\eta}{\partial x^2} + \frac{\partial^2\eta}{\partial y^2} = 0\,,\\
\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial x} + \frac{\partial\xi}{\partial y}\frac{\partial\eta}{\partial y} &= 0\,.
\end{align}$$
Do these conditions imply that these two functions satisfy Cauchy-Riemann conditions?
$$
\begin{align}
\frac{\partial\xi}{\partial x} &= \frac{\partial\eta}{\partial y}\,,\\
\frac{\partial\xi}{\partial y} &= -\frac{\partial\eta}{\partial x}\,.
\end{align}
$$
If so, how to show that? If not, is there some simple condition that would need to be added?
 A: A proof for the statement that it $u,w$ are real harmonic and satisfy the two conditions above:
$u_x^2+u_y^2=w_x^2+w_y^2$ (and not identically zero) and $u_xw_x+u_yw_y=0$ then if $v$ is a harmonic conjugate of $u$ we have $w=v+C$ or $w=-v+C$ and conversely those functions work, so one needs an extra condition on $w$ to ensure it is a harmonic conjugate and not a negative of such, for the C-R to hold wr $u,w$.
First, we note that by C-R if $w= \pm v+C$ the identities work.
Then taking $v$ a conjugate of $u$ and substituting the C-R equations in the orthogonality relation, we get $w_xv_y=v_xw_y$ while also $v_x^2+v_y^2=w_x^2+w_y^2$
But now assume at a point $w_y(z) \ne 0$ then it follows that $v_y(z) \ne 0$ (otherwise we must have $v_x(z)=0$ but then $w_x^2(z)+w_y^2(z)=0$ contradicting $w_y(z) \ne 0$) so we get at $z$:
$w_x^2/w_y^2=v_x^2/v_y^2$ or adding $1$ to each side $(w_x^2+w_y^2)/w_y^2=(v_x^2+v_y^2)/v_y^2$ and since the numerators are equal and non zero, we get $w_y^2(z)=v_y^2(z)$ so $w_y = \pm v_y$
If $w_y(z)=0$ then automatically $v_y(z)=0$ (since if it would not, then $w_x(z)=0$ and then the sum of squares is zero etc) so the equality holds everywhere.
By the same reasoning $w_x= \pm v_x$ everywhere and of course, if we are at a non-zero value of both $v_x, v_y$ the signs must coincide which means that $w_y=v_y$ or $w_y=-v_y$ on any connected component of the complement of $v_x=0$ (set being same with $w_x=0$ by the above, while the common zeroes of $v_x,v_y$ are discrete of course as they are zeroes of a non-constant analytic function $(u+iv)'$); but now $w_y+v_y$ is a real harmonic function and same with $w_y-v_y$ so if either is zero on any open set, it is zero in the plane, hence we must have the same sign everywhere and then the same holds for $w_x,v_x$ and we are done as we get that either $w=v+C$ or $w=-v+C$
