= Solution
Write the <holomorphic function> as $f=u+iv$. Analyticity gives smooth real and imaginary parts satisfying the <Cauchy-Riemann equations> $u_x=v_y$, $u_y=-v_x$. Differentiating these identities shows
$$
\Delta u=v_{yx}-v_{xy}=0,\qquad \Delta v=-u_{yx}+u_{xy}=0.
$$
Thus both parts are <harmonic functions>. The <product rule> then gives
$$
\Delta|f|^2=\Delta(u^2+v^2)
=2(u_x^2+u_y^2+v_x^2+v_y^2)+2u\Delta u+2v\Delta v
=4(u_x^2+v_x^2).
$$
Since $f'=u_x+iv_x$, this proves the <Laplacian identity for the squared modulus of a holomorphic function>,
$$
\boxed{\Delta|f(z)|^2=4|f'(z)|^2.}
$$
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