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...