Harmonic function and harmonic conjugate Let $u:G\subset\mathbb{R} \rightarrow \mathbb{R}$ a harmonic function $v:G\rightarrow \mathbb{R}$ the harmonic conjugate function, with $G$ a domain. Prove that $u^2-v^2$ and $uv$ are harmonic without derivatives.
Before this, I proved that $u^2$ is harmonic, if $u$ is an harmonic function. Then, I thought that $u^2$ and $v^2$  are harmonic functions, and I wanted to conclude that $u^2-v^2$ is a harmonic function.
Nonetheless, this interpretation is wrong.
Thanks in advance
 A: Since $v$ is a harmonic conjugate to $u$, this means that the function $f=u+iv$ is holomorphic in the domain $G$ specified. However it is easy to show that if $f$ is holomorphic, then $f^2$ is holomorphic and by the Cauchy-Riemann equations, it's real and it's imaginary part are harmonic functions. Expanding shows that
$$f^2=(u+iv)^2=u^2-v^2+i(2uv)\equiv \Re f+ i \Im f$$
and therefore the functions $\Re f=u^2-v^2$ and $\Im f=2uv$ and are harmonic (obviously if $2uv$ is harmonic then also $uv$ is).
A: If $v$ is the harmonic conjugate of $u$, then
$f(z) = u(z) + iv(z) \tag 1$
is holomorphic; thus so is $(f(z))^2$; now,
$(f(z))^2 = (u(z) + iv(z))^2 = u^2(z) - v^2(z) + 2iu(z)v(z); \tag 2$
since $u^2 - v^2$ and $2uv$ are the real and imaginary parts of $f^2(z)$, they are harmonic; $2uv$ harmonic implies that $uv$ is.
Nota Bene:  if $u$ is harmonic, then $u^2$ is only harmonic if and only if $\nabla u = 0$, for we have
$\nabla u^2 = \nabla \cdot (\nabla u^2) = \nabla \cdot (2u\nabla u)$
$= 2\nabla u \cdot \nabla u + 2u\nabla \cdot \nabla u = 2\nabla u \cdot \nabla u + 2u\nabla^2 u = 2\nabla u \cdot \nabla u, \tag 3$
since
$\nabla^2 u = 0; \tag 4$
we are left with
$\nabla u^2 = 2\nabla u \cdot \nabla u; \tag 5$
thus for harmonic $u$, $u^2$ harmonic is equivalent to
$\nabla u = 0, \tag 6$
that is, $u$ is constant on connected components of $G$.
In this argument, I have used the well-known identity
$\nabla \cdot (u \nabla u) = \nabla u \cdot \nabla u + u \nabla \cdot \nabla u, \tag 7$
which is found in many sources on the gradient operator $\nabla$ and vector calculus identities; the reader my check out wikipedia or simply google around for more.  End of Note.
