Here is the problem:
If f= u+ iv is analytic in a region, show that uv is harmonic in the region but that u^2 need not be harmonic.
Thanks
$\displaystyle f(x+iy)=u+iv$
The function is analyic, hence homolorphic.
It satisfies the Cauchy-Riemann Equations.
$\displaystyle u_x=v_y$
$\displaystyle u_y=-v_x$
Thus,
$\displaystyle u_xv_x+u_yv_y=0$
We need to show,
$\displaystyle uv$
Solves the PDE,
$\displaystyle f_{xx}+f_{yy}=0$.
Now,
$\displaystyle (uv)_{xx}=(u_xv+uv_x)_x=u_{xx}v+2u_xv_x+uv_{xx}$
$\displaystyle (uv)_{yy}=u_{yy}v+2u_yv_y+uv_{yy}$
Add them,
$\displaystyle (uv)_{xx}+(uv)_{yy}=u_{xx}v+2(u_xv_x+u_yv_y)+uv_{y y}$
But from Cauchy-Riemann Equations,
$\displaystyle (uv)_{xx}+(uv)_{yy}=u_{xx}v+uv_{yy}$
Maybe there is some theorem that tell the right hand side is zero. I do not know of one.
The function is analyic, hence both u and v are harmonic.
It satisfies the Cauchy-Riemann Equations.
$\displaystyle u_x=v_y$ & $\displaystyle u_y=-v_x$
Now,
$\displaystyle (uv)_{xx}=(u_xv+uv_x)_x=u_{xx}v+2u_xv_x+uv_{xx}$
$\displaystyle (uv)_{yy}=u_{yy}v+2u_yv_y+uv_{yy}$
Add them,
$\displaystyle (uv)_{xx}+(uv)_{yy}=2(u_xv_x+u_yv_y) $ because
$\displaystyle u_{xx}+ u_{yy}=0$ & $\displaystyle v_{xx}+ v_{yy}=0.$
But $\displaystyle u_x v_x + u_y v_y = u_x \left( { - u_y } \right) + u_y \left( { u_x } \right) = 0$
For the complex analyst the definition is: a function $\displaystyle f$ said to be analytic at $\displaystyle z_0$ if its derivative exists at each point in some neighborhood of $\displaystyle z_0$.
Moreover, a function $\displaystyle u$ is said to be harmonic if $\displaystyle u_{xx} + u_{yy} = 0$, that is the Laplace equation as you have noted.
Theorem: If $\displaystyle f(z) = u(x,y) + iv(x,y)$ is analytic is the domain $\displaystyle D$, then both $\displaystyle u(x,y)$ and $\displaystyle v(x,y)$ are harmonic in $\displaystyle D$.