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I'm kind of stuck on this problem and been working on it for days and cannot come to the conclusion of the proof.

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I edited your post, make sure I did not change your question. – Mhenni Benghorbal Nov 29 '12 at 6:29

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.

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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

$$ (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}) |f|^2 = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \overline{z}} f \, \overline{f} = 4 \frac{\partial}{\partial z}(0 \cdot \overline{f} + f \, \overline{f'}) = 4(f' \, \overline{f'} + f \cdot 0) = 4|f'|^2. $$

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