Proving that $\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) \lvert f(z) \rvert^2 = 4 \lvert f'(z) \rvert^2$ Given $f$ entire show that
$$
\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) \lvert f(z) \rvert^2 = 4 \lvert f'(z) \rvert^2
$$
I've come close to getting the exact answer by writing $f(z)$ as $u(x,y)+iv(x,y)$ and realizing the Laplacian of f is equal to zero. This leads to a lot of cancellations but I'm still making mistakes in my computations and I can't figure out where. Any help is appreciated.
 A: From $f=u+iv$ we obtain $|f|^2=u^2+v^2$, which implies
$${\partial\over\partial x}\bigl(|f|^2\bigr)=2uu_x+2vv_x$$
and then
$${\partial^2\over\partial x^2}\bigl(|f|^2\bigr)=2(u_x^2+v_x^2)+2uu_{xx}+2vv_{xx}\ .$$
Since $u$ and $v$ both are harmonic (a consequence of the CR equations) we therefore obtain
$$\Delta\bigl(|f|^2\bigr)=2(u_x^2+v_x^2+u_y^2+v_y^2)\ .\tag{1}$$
On the other hand we have
$$u_x(p)+iv_x(p)=\lim_{h\to 0,\ {\rm real}}{f(p+h)-f(p)\over h}=f'(p)$$
for each fixed $p$, so that using the CR equations again we obtain
$$u_y^2+v_y^2=u_x^2+v_x^2=|f'|^2\ .$$
This together with $(1)$ proves the claim.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Note, with $\ds{z = x + y\ic\ \mbox{and}\ \overline{z} = x - y\ic}$, that

$$
\partiald{}{x}=\partiald{}{z} + \partiald{}{\overline{z}}\,,\qquad
\partiald{}{y}=\ic\,\partiald{}{z} - \ic\,\partiald{}{\overline{z}}
\quad\imp\quad
\left\{\begin{array}{lcl}
\partiald{}{z} & = & \half\pars{\partiald{}{x} - \ic\,\partiald{}{y}} 
\\[2mm]
\partiald{}{\overline{z}} & = & \half\pars{\partiald{}{x} + \ic\,\partiald{}{y}} 
\end{array}\right.
$$
  such that $\ds{\partiald{}{\overline{z}} = \overline{\pars{\partiald{}{z}}}}$

$$
\partiald[2]{}{x}
=\partiald[2]{}{z} + \partiald[2]{}{\overline{z}}
+ 2\,{\partial^{2} \over \partial z\,\partial\overline{z}}\,,\qquad
\partiald[2]{}{y}
=-\,\partiald[2]{}{z} - \partiald[2]{}{\overline{z}}
+ 2\,{\partial^{2} \over \partial z\,\partial\overline{z}}
$$

$$
\partiald[2]{}{x} + \partiald[2]{}{y}
=4\,{\partial^{2} \over \partial z\,\partial\overline{z}}
$$

\begin{align}&\color{#66f}{\large%
\pars{\partiald[2]{}{x} + \partiald[2]{}{y}}\verts{\fermi\pars{z}}^{2}}
=4\,{\partial^{2}\bracks{\fermi\pars{z}\overline{\fermi\pars{z}}} \over \partial z\,\partial\overline{z}}
=\dsc{4\,\partiald{\fermi\pars{z}}{z}\,\partiald{\overline{\fermi\pars{z}}}{\overline{z}}}
\\[5mm]&=4\,\partiald{\fermi\pars{z}}{z}\,
\overline{\bracks{\partiald{\fermi\pars{z}}{z}}}
=4\fermi'\pars{z}\overline{\fermi'\pars{z}}
=\color{#66f}{\large 4\verts{\fermi'\pars{z}}^{2}}
\end{align}
