The Problem:
Let $f(z)=f(x+iy)=u(x,y)+iv(x,y)$ be an entire function.
If $g(z)=au^2(x,y) – bv^2(x,y)$ find values for a and b so $g(z)$ is a harmonic function.
My attempt to find a solution:
Well since I know that $f(z)$ is entire, the Cauchy-Riemann equations must hold, then:
$U_x=U_y$ and $V_x=-V_y$.
If $f(z)$ is entire, then its components are harmonic, hence:
$U_{xx} = -U_{yy}$ and $V_{xx}=-V_{yy}$
With all this information, I procceed to derivate $g(z)$ (twice) in terms of x and then in terms of y. So I can fulfill Laplace's equation and make $g(z)$ harmonic:
$g_{xx} + g_{yy}=0$.
I do this and end up with an expression that equals 0. The problem is, I don't know how to continue after this. I don't know if I lack another condition, or how to apply the information provided by the fact that $f(z)$ is entire correctly.
If I replace the conditions so everything is in terms of x (or y) I end up with an expression that equals 0 but contains both u and v, and I don't realize how to obtain values for a and b.
Any help is appreciated.
Best Answer
The product rule for computing the Laplacian of $g$ gives $$ \Delta g = 2 a(u u_{xx} + u_x^2 + u u_{yy} + u_y^2) - 2 b(v v_{xx} + v_x^2 + v v_{yy} + v_y^2) $$
Since $u$ and $v$ are harmonic and satisfy the Cauchy-Riemann equations $u_x = v_y$, $u_y = -v_x$, this simplifies to $$ \Delta g = 2 (a-b)(u_x^2 + v_x^2) = 2(a-b) |f'(z)|^2 $$ which means that $g$ is harmonic if $a=b$ or if $f$ is constant.
Remark: In the case $a=b$ we have $$ g(z) = a (u^2(z) - v^2(z)) = a \operatorname{Re}(f(z)^2) $$