Proof that $\frac{u_x v_x - u_y v_y}{(u_x^2 + u_y^2)(v_x^2 + v_y^2)}$ is harmonic if $u$ and $v$ are

Question

Can any one help me with this question?

Let $u$ and $v$ be harmonic functions in a domain ($D$ is a simply connected space and open).

Assume $(u_x^2 + u_y^2)(v_x^2 + v_y^2) \ne 0$. Define $$\psi = \frac{u_x v_x - u_y v_y}{(u_x^2 + u_y^2)(v_x^2 + v_y^2)}.$$ Prove that $\psi$ is harmonic.

Answer 1

Recall that if $u$ harmonic then, $\phi=u_x+iu_y$ is analytic.

Also, recall that if $\phi$ analytic , then, $Re(\phi)$ is harmonic. (Use Cauchy-Riemann)

Now, $f=u_x+iu_y$ and $g=v_x+iv_y$ are both analytic. Then, $|fg|^2=(u_x^2+u_y^2)(v_x+iv_y)^2\not \equiv 0$. There exists some open subset of $D$ such that $fg(z)\neq 0$ for every $z$.

In particular, $\frac{1}{fg}$ is analytic in some open subset of $D$.

Then, $\displaystyle\frac{1}{fg}=\frac{(u_x-iu_y)(v_x-iv_y)}{(u_x^2+u_y^2)(v_x^2+v_y^2)}=\frac{(u_x v_x-u_yv_y)-i(u_x v_y+u_y v_x)}{(u_x^2+u_y^2)(v_x^2+v_y^2)}$. But note, that $Re(\frac{1}{fg})=\psi$