My problem is:
Suppose $u$, $v$ are harmonic in region $\Omega$, and $\nabla u$, $\nabla v$ never vanish in $\Omega$. The level curves of $u$ and $v$ are perpendicular throughout $\Omega$. Moreover, assume $|\nabla u(z_0)| = |\nabla v(z_0)|$ at some specific point $z_0$. Prove that either $u+iv$ or $u-iv$ is conformal in $\Omega$.
Could anyone outline the ideas for the proof?
An alternative but closely related problem is: $u, w$ are harmonic and non-degenerate (gradients never vanish) in $\Omega$, and level curves of $u$ and $w$ coincide everywhere. What can you say about $u$ and $w$?
My conjecture is that $u$ and $w$ are proportional. If so, the problem above can be solved by taking the harmonic conjugate of $u$, and deduce that it equals $\pm v$ from the $|\nabla u(z_0)| = |\nabla v(z_0)|$ condition.
In Problem 3.4 (a), write $$ w_x = \lambda v_x, \; \; w_y = \lambda v_y \; . $$
Calculate $w_{xx}$ and $w_{yy}$ and add.
Calculate $w_{xy}$ two ways and subtract.