If $f$ is entire and $z=x+iy$, prove that for all $z$ that belongs to $C$, $\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}\right)|f(z)|^2= 4|f'(z)|^2$ I'm kind of stuck on this problem and been working on it for days and cannot come to the conclusion of the proof. 
 A: Hint: Write $f(z) = u(x,y) + iv(x,y)$, where $u$ and $v$ have real values. Then $|f(z)|^2 = u(x,y)^2 + v(x,y)^2$, and you can compute the left-hand side in terms of partial derivatives of $u$ and $v$. Using the Cauchy-Riemann equations, it should simplify nicely: for instance, you'll find that
$$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} = 0,$$
and a similar identity for $v$.
For the right-hand side, you can use the fact that
$$\frac{df}{dz} =
\frac12\left( \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right),$$
and write the right-hand side of that equation in terms of partial derivatives of $u$ and $v$, and use the Cauchy-Riemann equations again.
A: Let $\frac{\partial}{\partial z} = \tfrac{1}{2}(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \overline{z}} = \tfrac{1}{2}(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y})$.  Then $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \overline{z}}$ so
$$
(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}) |f|^2 = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \overline{z}} f \, \overline{f} = 4 \frac{\partial}{\partial z}(0 \cdot \overline{f} + f \, \overline{f'}) = 4(f' \, \overline{f'} + f \cdot 0) = 4|f'|^2.
$$
