# Zero sets of harmonic fucntions

Can a two variable Harmonic function f(x,y) be zero on a curve with a cusp?

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No. A two variable harmonic function is the real part of an analytic function. Near a zero, an analytic function behaves like a power of $z-z_0$.
Let u be your harmonic function, and suppose u(0,0)=0. Let v be the harmonic conjugate, and w.l.o.g. let v(0,0)=0. Then u+iv is an analytic function, which has a Taylor series $$u+iv=a_n z^n +O(z^{n+1}).$$ We can write this in the form $w(z)^n$, where $w$ is an analytic function of $z$ that is locally invertible. Now $u=0$ is equivalent to $Re(w^n)=0$. In the $w$ plane, this is given by $n$ straight lines intersecting at the origin. These lines map to smooth curves in the $z$ plane. – Michael Renardy Jun 16 '11 at 14:53
Perhaps study this one: $$f(x,y) = \frac{2 \sqrt{x^{2} + y^{2}} - \sqrt{2 \sqrt{x^{2} + y^{2}} + 2 x}}{2 \sqrt{x^{2} + y^{2}} - 2 \sqrt{2 \sqrt{x^{2} + y^{2}} + 2 x} + 2}$$ which vanishes on a cardioid with a cusp at the origin. Harmonic except possibly at the one point $(0,0)$.