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