# Understanding $\Delta(\vert f \vert ^p)$ when $f$ is holomorphic, $p>0.$

Let $\Delta$ denote the Laplacian. I am trying to prove that if $f=u+iv$ is holomorphic on an open set $U\subset \mathbb{C}$ and $f$ is nonvanishing, then $$\Delta (\vert f\vert^p)=p^2\vert f\vert^{p-2}\Bigg\vert \frac{\partial f}{\partial z} \Bigg\vert^2, \text{any}\,p>0.\,\,\,\,\,(*)$$ This is Exercise 44 of Ch.1 in Function Theory of one Complex Variable, by Greene & Krantz.

Using the fact that $f$ is holomorphic (i.e., that $\partial u/\partial x=\partial v/\partial y$ & $\partial v/\partial x = -\partial u/\partial y)$ , I can rewrite the right-hand side of $(*)$ as $$p^2\vert f\vert^{p-2}\Bigg\vert \frac{1}{2}(\partial u/\partial x+\partial v/\partial y)+\frac{i}{2}(\partial v/\partial x-\partial u/\partial y) \Bigg\vert^2= p^2\vert f\vert^{p-2}\Bigg\vert \partial u/\partial x-i\partial u/\partial y \Bigg\vert^2=\,\,\,\,\,\,\,\,\,p^2\vert f\vert^{p-2}\Big((\partial u/\partial x)^2+(\partial v/\partial y)\Big)^2$$

However, I am not clear on how to proceed with the left-hand side: here's what I have $\Delta(\vert f \vert ^p)=\Delta(\vert u+iv\vert^p)=\frac{\partial^2}{\partial x^2}\vert u+iv\vert^p+\frac{\partial^2}{\partial y^2}\vert u+iv\vert^p=\frac{\partial}{\partial x}\Big[p(u+iv)/\vert u+iv\vert)*\frac{\partial}{\partial x}\vert u+iv \vert^{p-1}\Big] +\frac{\partial}{\partial y}\Big[p(u+iv)/\vert u+iv\vert)*\frac{\partial}{\partial y}\vert u+iv \vert^{p-1} \Big]$

Can someone give a hint on how to proceed with the left hand side? My only idea is to use brute force and to continue to differentiate, but then I'd end up doing that indefinitely.

One way to do this without "brute force" is to write $\Delta = 4 \partial_z \partial_{\bar{z}}$, and use the fact that since $f$ is holomorphic, then $\partial_z \bar{f} = 0$ and $\partial_{\bar{z}} f = 0$.
Further hint: $|f|^2 = f \bar{f}$.
• I've added an additional hint to clarify. So, in essence, you act the holomorphic and anti-holomorphic derivatives on $f$ and $\bar{f}$ separately, and then combine at the end. This will avoid any need to "unravel" the expression in terms of real and imaginary parts. Commented Oct 6, 2014 at 8:52