Laplacian of $|f|^p,$ where $f$ is holomorphic I have to prove that if $f$ is a homolorphic function that doesn´t vanish on its domain then
$\triangle |f|^p=p^2 |f|^{p-2} |\frac{\partial f}{\partial z}|^2$ .
My attempt: I take $|f|^p=(z \bar{z})^{p/2}$ to use the alternative form of the Laplacian $\triangle u=\frac{\partial}{\partial z} \frac{\partial}{\partial \bar{z}}u$.
So I start computing as follows, using the chain rule:
$\triangle |f|^p = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \bar{z}} (z \bar{z})^{p/2}=4 \frac{\partial}{\partial z} \frac{p}{2} (z \bar{z})^{p/2 -1} (z) = 4 \frac{p}{2}(\bar{z})^{p/2-1}(\frac{p}{2})z^{p/2-1}=p^2|f|^{p-2}$.
So I´m clearly missing something, because the term $|\frac{\partial f}{\partial z}|^2$ doesn´t show off, I just don´t see what is it, as my differentiation skills are not quite good. Thanks.
 A: Let $f=u+iv$, where $u$ and $v$ satisfy the Cauchy-Riemann Equations and are harmonic.  Then, straightforward use of the chain rule exposes that
$$\nabla^2 |f|^p=\frac p2\left(\frac p2 -1\right)|f|^{p-4}\left(\left(\frac{\partial |f|^2}{\partial x}\right)^2+\left(\frac{\partial |f|^2}{\partial y}\right)^2\right)+\frac p2 |f|^{p-2}\nabla^2 |f|^2 \tag 1$$
Now, we evaluate the derivative terms on the right-hand side of $(1)$.  First, we have
$$\nabla^2 |f|^2=4\left|\frac{df}{dz}\right|^2 \tag 2$$
which follows directly.
Next, we evaluate the terms in parantheses in $(1)$.  There, we have 
$$\left(\frac{\partial |f|^2}{\partial x}\right)^2+\left(\frac{\partial |f|^2}{\partial y}\right)^2=4|f|^2\left|\frac{df}{dz}\right|^2 \tag 3$$
Using $(2)$ and $(3)$ in $(1)$ gives the desired result
$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 |f|^p=p^2\,|f|^{p-2}\left|\frac{df}{dz}\right|^2}$$
A: It goes like this:
$$\frac{\partial}{\partial z} (f^{p/2}(\bar f)^{p/2}) = \frac{p}{2} f' f^{p/2-1}(\bar f)^{p/2} $$
since the antiholomorphic term is treated as a constant. Then differentiate in $\bar z$ similarly, getting
$$ \frac{p}{2} f' f^{p/2-1}\frac{p}{2}\bar f' (\bar f)^{p/2-1}$$ 
Simplify to $ \dfrac{p^2}{4} |f'|^2 |f|^{p-2}$ and the result follows.

The above involves fractional powers of $f$ and $\bar f$. These are defined locally away from the zeros of $f$; since the computation is local, it is valid on $\{f\ne 0\}$. One has to be more careful at the zeros of $f$ where $|f|^p$ is not smooth in general. For $p>2$ the Laplacian of $|f|^p$ is  zero at the zeros of $f$, because all second derivatives vanish there. For $p<2$ it's not defined at the zeros of $f$.
