Is $h(x,y,z) = \sin(x)\cos(z)$ harmonic?

Question

I am given the definition of a harmonic function in terms of differential forms. That is, for a function $f$,

$d*df = 0 $

where $*$ is the Hodge-star.

I have to show $f(x,y,z) = \sin(x)\cos(z)$. However as I calculated,

$ df = \cos(x)\cos(z)\,dx - \sin(x)\sin(z) $

$ *df = \cos(x)\cos(z)\,dy \wedge dz -\sin(x)\sin(z)\,dx \wedge dz$

$ d*df = d(\cos(x)\cos(z))\,dy \wedge dz -d(\sin(x)\sin(z))\,dx \wedge dz$

$ = -2\sin(x)\cos(z)\,dx \wedge dy \wedge dz$

I do not think the resultant 3-form is equal to 0. Did I make any mistake?

Answer 1

This $f$ is certainly not harmonic. Perhaps it was supposed to be $\sin(x) \cosh(z)$ or $\sinh(x) \cos(z)$?