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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 $$ \partial_z \partial_{\overline{z}} = \frac14 (\partial_{xx} + \partial_{yy})? $$

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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.

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  • $\begingroup$ You're genius! Thanks! $\endgroup$ – Kamil Mar 26 '17 at 17:09
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Just derivate: \begin{eqnarray} \partial_{z\,\overline{z}} |f|^p &=& \partial_{z\,\overline{z}}\left(f^p\overline{f}^p \right)^\frac12\\ &=& \partial_{z}\left(\frac12\left(f^p\overline{f}^p \right)^{-\frac12} pf^p\overline{f}^{p-1} \frac{\partial\overline{f}}{\partial\overline{z}}\right)\\ &=& \partial_{z}\left(\frac{p}{2}f^{\frac{p}{2}}\overline{f}^{(\frac{p}{2}-1)} \frac{\partial\overline{f}}{\partial\overline{z}}\right)\\ &=& \frac{p}{2} \frac{p}{2} f^{\frac{p}{2}-1} \overline{f}^{(\frac{p}{2}-1)} \frac{\partial f}{\partial z}\frac{\partial\overline{f}}{\partial\overline{z}}\\ &=& \frac{p^2}4 |f(z)|^{p-2}|f'(z)|^2 \end{eqnarray}

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