This is a function that I have spent a lot of time on, attacking it from several directions, to no fruition I really need some help with this one, the problem is:
Prove that: ($\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}})(log(|f'(z)|)) = 0$
For Analytic $f$
I tried to proceed by factorising the Laplacian into Wirtinger Derivatives as so:
$\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} = 4.\frac{\partial}{\partial z}.\frac{\partial}{\partial \bar{z}}$
And then proceeding from there, by writing the function $f$ with its Real and Imaginary parts as: $f(z) = u(x,y) + iv(x,y) $, and then reverting the Wirtinger derivatives to their definition in 'x' and 'y' ( e.g.: $\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} + \frac{1}{i}.\frac{\partial}{\partial y})$ ),during the long and arduous calculation I also had to use the fact that the real and imaginary parts of an analytic function are Harmonic ( i.e. satisfy $\nabla u = \nabla v = 0 )$, and also the Cauchy-Riemann equations, but I couldn't reach a satisfactory conclusion to the problem.
On a side note there is some ambiguity if in the problem the correct expression involved $f'(z)$ or $f(z)$, I solved the problem both ways and in my $f(z)$ version of the solution I arrived at the conclusion that this reduces to:
$\frac{|f'(z)|^{2}}{|f(z)|^{2}}$, which is clearly not equal to zero.
assuming $f'(z) \ne 0$, then $f'(w) \ne 0$ on a small disc around $z$, hence there is an analytic function $g(w)=\ln f'(w)$ there. But then $\ln |f'(w)| =\Re g(w)$ is harmonic hence it satisfies laplace on the full neighborhood, hence in particularly at $z$. Done!
Same proof applies to $\ln |f|$ wherever $f(z) \ne 0$
(at the zeroes $\ln |f| =-\infty$ is only subharmonic)