$uv$ is harmonic if and only if $u+icv$ is analytic for some real c

Question

Let $u$ and $v$ be non constant harmonic functions on a complex domain. Prove that $uv$ is harmonic if and only if $u+icv$ is analytic for some real $c$.

I can prove the "if" part. I am having some trouble with the "only if" part.

My argument is : $uv$ is harmonic implies $u_xv_x+u_yv_y=0$. This means that

$<u_x,u_y>$ is perpendicular to $<v_x,v_y>$. This implies that $<v_x,v_y> = c<-u_y,u_x>$. This proves the result. My question is - does this sound rigorous enough?

Answer 1

$u_xv_x + u_yv_y = 0 \implies <u_x, u_y> = c(x,y) < v_y, -v_x>$

$ \implies u_x = c(x,y)v_y$ and $u_y = -c(x,y) v_x$.

Using $\triangle u =0$ and $\triangle v =0$ on this we get, $c_xv_y -c_yv_x =0$.

Again using $u_{xy} -u_{yx} =0$ we get $c_yv_y + c_xv_x=0$.

Eliminating $v_x$ and $v_y$ from the above equations leads to

$(c_x^2 +c_y^2)v_x =0$ and $(c_x^2 +c_y^2)v_y =0$ which means $c$ is a constant.

Hence $\exists c\in \mathbb{R}$ such that $u+icv$ is analytic.

Answer 2

As observed by the OP, if $u$, $v$ and $uv$ are harmonic in $\Omega\subset\mathbb R^2$ (open and connected), then $$ u_xv_x+u_yv_y=0, $$ and thus the there is a $c=c(x,y)$ such that $$ (v_y,-v_x)=c(x,y)(u_x,u_y), \qquad (\star) $$ with the above valid wherever $(u_x,u_y)\ne (0,0)$. But this can only happen in a subset of $\Omega$ without a limit in $\Omega$, otherwise the analytic function $f=u+iw$, where $w$ is a harmonic conjugate of $w$ would be constant.

$(\star)$ implies that \begin{align} 0=v_{xx}+v_{yy}=-c_xu_y -cu_{xy}+c_yu_x+cu_{xy}=-c_xu_y+c_yu_x, \\ 0=(v_y)_x-(v_x)_y=c_xu_x+cu_{xx}+c_yu_y+cu_{yy}=c_x u_x+c_yu_y, \end{align} and thus $(c_x,c_y)=(0,0)$, as it is perpendicular and parallel to a nonzero vector. Hence $c=c(x,y)$ is constant, and thus $cv$ is the harmonic conjugate of $u$.