Demostration with the laplacian Suppose $f(z)=u(x,y)+iv(x,y)$ is an integer function and that $u (x,y)$ and $v(x, y)$ are of class $C^2$ in $\Bbb R^2$
Show that

I've done:
Since $f(z)=u(x,y)+iv(x,y)$ is an integer function, $f$ is differenciable
$$f'(z) = (∂u/∂x) + i (∂v/∂x)$$ 
$$4|f'(z)| = 4((∂u/∂x)^2+(∂v/∂x)^2)$$
And  
$$Δ|f(z)|^2 = Δ(u^2+v^2) =  (∂^2u/∂x^2)+(∂^2v/∂x^2) =                   (∂u/∂x)^2+(∂v/∂x)^2$$
This expressions are obviouly not igual... Could someone say me what i've done wrong?
 A: According to a commet of our OP aluno20000 to the question itself, we are to take the words "$f(z)$ is integral" to mean $f(z)$ is differentialble with respect to $z \in \Bbb C$; that is, $f(z)$ is holomorphic on $\Bbb C$ in addition to being $C^2$ on $\Bbb R^2$.
We want to show that
$\nabla^2 \vert f(z) \vert^2 = 4 \vert f'(z) \vert^2; \tag 0$
we have
$f(z) = u(z) + iv(z); \tag{0.1}$
thus,
$\vert f(z) \vert^2 = u^2(z) + v^2(z), \tag{0.2}$
and
$\nabla^2 \vert f(z) \vert^2 = \nabla^2(u^2 + v^2) = \nabla \cdot \nabla (u^2 + v^2)$
$= \nabla \cdot (2u\nabla u + 2v \nabla v) = 2\nabla u \cdot \nabla u + 2u\nabla^2 u + 2\nabla v \cdot \nabla v + 2v\nabla^2 v; \tag 1$
here we have used the well-known identity from vector calculus,
$\nabla \cdot (\psi \mathbf A) = \nabla \psi \cdot \mathbf A + \psi \nabla \cdot \mathbf A, \tag{1.5}$
taking
$\psi = u, v; \; \mathbf A = \nabla u, \nabla v, \; \text{respectively}; \tag{1.6}$
with $f(z)$ holomorphic, $u$ and $v$ are harmonic:
$\nabla^2 u = \nabla^2 v = 0; \tag 2$
(1) then becomes
$\nabla^2 \vert f(z) \vert^2 = \nabla^2(u^2 + v^2) = 2\nabla u \cdot \nabla u + 2\nabla v \cdot \nabla v; \tag 3$
now,
$\nabla u = (u_x, u_y), \tag 4$
$\nabla v = (v_x, v_y); \tag 5$
since $f(z) = u(z) + iv(z)$, Cauchy-Riemann asserts
$u_x = v_y, \tag 6$
$u_y = -v_x, \tag 7$
whence
$\nabla u \cdot \nabla u = u_x^2 + u_y^2 = u_x^2 + v_x^2 = \vert f'(z) \vert^2; \tag 8$
$\nabla v \cdot \nabla v = v_x^2 + v_y^2 = u_y^2 + v_y^2 = \vert f'(z) \vert^2; \tag 9$
thus,
$ 2\nabla u \cdot \nabla u + 2\nabla v \cdot \nabla v = 4\vert f'(z) \vert^2; \tag {10}$
substitution of this into (3) yields (0).  $OE\Delta$.
