May i ask you for a little help about the following problem:

Prove that the Wirtinger derivatives $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \bar{z}}$ are:

1) $\mathbb{C}$-linear.

2) The Leibniz rule for derivatives of higher order is valide for them.

We know that the Wirtinger derivatives are defined so $\frac{\partial }{\partial z}= \frac{1}{2} \left ( \frac{\partial }{\partial x} - i\frac{\partial }{\partial y}\right )$ and $\frac{\partial }{\partial \bar{z}}= \frac{1}{2} \left ( \frac{\partial }{\partial x} + i\frac{\partial }{\partial y}\right )$.

Thank you in advance!

Since $\partial_x (cf) = c\partial_x f$ and $\partial_y(cf) = c\partial_y f$ for all differentiable $f: dom(f) \subseteq \mathbb{C} \rightarrow \mathbb{C}$ and $c \in \mathbb{C}$ the $\mathbb{C}$-linearity of the Wirtinger derivatives follows by simply taking the appropriate complex-linear combinations of the identities I claim above. Likewise, the usual Leibniz rule for products is true for the real partial derivatives hence you can add those rules to obtain the Wirtinger Leibniz rules. For example:

$$ \partial_x(fg) = \partial_x(f)g+ f\partial_x(g) $$ $$ i\partial_y(fg) = i[\partial_y(f)g+ f\partial_y(g) ]$$

Then divide by two and add to obtain: $$ \partial_z(fg) = \partial_z(f)g+ f\partial_z(g) $$ Similar comments apply to the $\bar{z}$-derivatives.

It is interesting to note that for $f: \mathbb{C} \rightarrow \mathbb{C}$ the differential $df = \partial_x f dx+ \partial_yf dy$ can be written as $df = \partial_z f dz+ \partial_{\bar{z}}f d\bar{z}$. The condition $df(ch) = cdf(h)$ is equivalent to the condition of complex-differentiability. You can think of $ch$ being fed into the $dz$ or $d\bar{z}$. The $c$ filters out of the $dz$ but it comes out as $\bar{c}$ if we feed it to $d\bar{z}$. Thus the condition that a function on the complex plane be antiholomorphic (which means it's just a function of $\bar{z}$) is that the $\partial_z f=0$. In contrast, we get holomorphic (just a function of $z$) functions if $\partial_{\bar{z}}f=0$. Of course, if you're only interested in calculations then these comments about the differential maybe tangential to your interests.

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