# Show that if $g$ is nonconstant holomorphic and $f$ is harmonic such that $fg$ is harmonic, then $f$ is holomorphic.

Let $\Omega$ be an open and connected set in the complex plane and $g$ be a nonconstant holomorphic function on $\Omega$. Show that if $f$ is harmonic on $\Omega$ such that $fg$ is also harmonic on $\Omega$, then $f$ is in fact holomorphic.

I tried showing the Cauchy-Riemann equations hold for $f$ by writing out the Laplacian for $fg$. This was quite messy, and I was unable to solve the problem, either due to a computational error or a failure to see the next step. For what it's worth, setting $f=u+iv$ and $g=a+ib$, I arrived at

$b_y(u_x-v_y)+a_y(u_y+v_x)=0$

and

$a_y(-u_x+v_y)+b_y(u_y+v_x)=0$.

Any help is appreciated. Thank you.

The slick way to do this is with the operators \eqalign{\partial &= \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} - i \dfrac{\partial}{\partial y}\right)\cr \overline{\partial} &= \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\right)\cr}
Note that $4 \partial \overline{\partial} = \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2}$, so a function $f$ is harmonic iff $\partial \overline{\partial} f = 0$, while $f$ is holomorphic iff $\overline{\partial} f = 0$, in which case $\partial f = f'$. These operators obey Leibniz's rule, so since $g$ is holomorphic
$$\overline{\partial} (fg) = (\overline{\partial} f) g + f (\overline{\partial} g) = (\overline{\partial} f) g$$ and since $f$ and $fg$ are harmonic
$$0 = \partial \overline{\partial} (fg) = \partial \left(( \overline{\partial} f)g\right) = (\partial \overline{\partial} f) g + ( \overline{\partial} f) (\partial g) = ( \overline{\partial} f) g'$$ Thus $\overline{\partial} f = 0$ except perhaps on a discrete set of points of $\Omega$ where $g' = 0$. But since $f$ is harmonic, its partial derivatives are continuous, so in fact $\overline{\partial} f = 0$ everywhere in $\Omega$, and thus $f$ is holomorphic.