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.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.