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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?

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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...

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  • $\begingroup$ First of all, thank you for replying. Secondly, I don't see why that implies orthogonality of the level curves. And finally, how would I solve question 1? $\endgroup$ – Spider-Pig Mar 6 '16 at 16:11
  • $\begingroup$ 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 $\langle \nabla f(\gamma(t)),\gamma'(t)\rangle = 0$. $\endgroup$ – Ivo Terek Mar 6 '16 at 16:24
  • $\begingroup$ Thanks again! Different question: is there a way of showing orthogonality at 1 without using the gradients? $\endgroup$ – Spider-Pig Mar 6 '16 at 16:30
  • $\begingroup$ I think that the idea was to use precisely the gradients, in some sense. If you really want to avoid this, you can look at the tangent vectors to the level curves (which are the gradients with the coordinates and one sign swapped) $\endgroup$ – Ivo Terek Mar 6 '16 at 16:35

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