# Intersections of the level curves of two (conjugate) harmonic functions

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ with $f(x,y)=e^{-x}(x\sin y-y\cos y)$.

1 Let $g$ be one of the conjugate harmonics of $f$ on $\mathbb{R}^2$ and assume the level curves of $f$ and $g$ intersect.How do I show that the level curves intersect at right angles (by calculating)?
2 What is the conceptual explanation behind the right angle intersection?

What I know:
1 The harmonic conjugates are of the form $$g(x,y)=e^{-x}(x\cos y+y\sin y)-e^{-x_0}(x_0\cos y_0+y_0\sin y_0)$$ with $(x_0,y_0)\in\mathbb{R}^2.$
If I take $(x_0,y_0)=(0,0)$, I can take $g(x,y)=e^{-x}(x\cos y+y\sin y)=C$ and $f(x,y)=e^{-x}(x\sin y-y\cos y)=C$ as level curves. How would I then show that they intersect at right angles?
2 I know that being conjugate harmonic functions means that $f,g$ are the real and imaginary parts of a holomorphic function $\phi(z)$ with complex variable $z=x+iy$ and that $f,g$ satisfy the Cauchy-Riemann equations. But does this say anything about level curves intersecting at right angles?

If Cauchy-Riemann holds, then $$\langle \nabla f (x,y),\nabla g (x,y)\rangle = \frac{\partial f}{\partial x}(x,y)\frac{\partial g}{\partial x}(x,y) + \frac{\partial f}{\partial y}(x,y)\frac{\partial g}{\partial y}(x,y) = 0 .$$ So...
• The calculation needed for $1$ is essentialy what I did above. The point is that the gradients are orthogonal to the level curves. So, if the gradients are orthogonal to each other, it means that the level curves are orthogonal to each other too. Recall that if $\gamma$ parametrizes a level curve, then $f(\gamma(t)) = c$ for all $t$ implies $\langle \nabla f(\gamma(t)),\gamma'(t)\rangle = 0$. – Ivo Terek Mar 6 '16 at 16:24