# Complex function derivative.

Spent 2 hours solving this, but still no results. Let's define $$\frac{\partial}{\partial \overline{z}} = \frac12 \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right), \\ \frac{\partial}{\partial z} = \frac12 \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right).$$ Proof, that $$\frac{\partial^2 |f(z)|^p}{\partial z \partial \overline{z}} = \frac{p^2}4 |f(z)|^{p-2}|f'(z)|^2.$$ Made lots of calculations. Is it true, that $$\partial_z \partial_{\overline{z}} = \frac14 (\partial_{xx} + \partial_{yy})?$$

We can write $\lvert f(z) \rvert^p = \left(f(z)\overline{f(z)}\right)^{p/2}$, by the definition of the modulus. Now, if we define $\bar{f}(z) = \overline{f(\bar{z})}$, then $\overline{f(z)} = \bar{f}(\bar{z})$. In particular, $\bar{f}$ is analytic if and only if $f$ is (by expanding in a power series, for example). Now, $$\left(f(z)\bar{f}(\bar{z})\right)^{p/2} = \exp{\left(\frac{p}{2}\log{\left(f(z)\bar{f}(\bar{z})\right)}\right)}$$ by definition. Doing the first derivative gives $$\frac{\partial}{\partial \bar{z}} \exp{\left(\frac{p}{2}\log{\left(f(z)\bar{f}(\bar{z})\right)}\right)} = \frac{p}{2} \exp{\left(\frac{p}{2}\log{\left(f(z)\bar{f}(\bar{z})\right)}\right)} \frac{\partial_{\bar{z}} \bar{f}(\bar{z}) }{\bar{f}(\bar{z})},$$ since the $f(z)$ is a constant with respect to $\bar{z}$.
The last term contains no $z$s, so we can ignore it for the $z$ derivative. Differentiating again, we have $$\frac{\partial^2}{\partial z \partial \bar{z}} \exp{\left(\frac{p}{2}\log{\left(f(z)\bar{f}(\bar{z})\right)}\right)} = \frac{p^2}{4} \exp{\left(\frac{p}{2}\log{\left(f(z)\bar{f}(\bar{z})\right)}\right)} \frac{f'(z)}{f(z)}\frac{\partial_{\bar{z}} \bar{f}(\bar{z}) }{\bar{f}(\bar{z})}$$ It is clear that we will get what we want if we can show that $\partial_{\bar{z}} \bar{f}(\bar{z}) =\overline{f'(z)}$. This becomes clear when we unwind $\bar{f}$: $$\frac{\partial}{\partial \bar{z}} \overline{f(z)} = \overline{\frac{\partial}{\partial z} f(z)} = \overline{f'(z)},$$ since it is easy to see from the definition that $\overline{\partial_{z}}=\partial_{\bar{z}}$.
Indeed, $4\partial_z \partial_{\bar{z}} = \partial_x^2+\partial_y^2$ for twice-differentiable functions, since the mixed partial derivatives cancel in that case.