The complex version of the chain rule

Question

I want to prove the following equality:

\begin{eqnarray} \frac{\partial}{\partial z} (g \circ f) = (\frac{\partial g}{\partial z} \frac{\partial f}{\partial z}) + (\frac{\partial g}{\partial \bar{z}} \frac{\partial \bar{f}}{\partial z}) \end{eqnarray}

So I decide to do the following:

\begin{eqnarray} \frac{\partial}{\partial z} (g \circ f) = \frac{1}{2}[(\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x}) + \frac{1}{i}(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial y})] \end{eqnarray}

but the thing is that I am doing something wrong here since I don't get any conjugate function and any derivative with respect to $\bar{z}$ so Can someone help me to see where I am wrong and fix it please?

In fact I don't see what to do next, so I appreciate your help.

Thanks a lot in advance.

Edition:

What I've got so far is the following:

$$\frac{1}{2}[(\frac{\partial g}{\partial x} \circ f + \frac{\partial g}{\partial y} \circ f)\frac{\partial f}{\partial z} ]$$

but I'm still stuck.

Answer 1

The question is taken in the context of Wirtinger Derivatives.

To that end, we let $g$ and $f$ be functions of both $z$ and $\bar z$. Then, the composite function $g\circ f$ can be expressed as

$$g\circ f=g(f(z,\bar z),\bar f(z,\bar z))$$

The partial derivative of $g\circ f$ with respect to $z$ is then given by

$$\begin{align} \frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\ &=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\ &=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z} \end{align}$$

Answer 2

You do some mistakes. Note that $$\frac{\partial}{\partial x} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\operatorname{Re}\frac{\partial f}{\partial x})+(\frac{\partial g}{\partial y} \circ f)(\operatorname{Im}\frac{\partial f}{\partial x})$$ and $$\frac{\partial}{\partial y} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\operatorname{Re}\frac{\partial f}{\partial y})+(\frac{\partial g}{\partial y} \circ f)(\operatorname{Im}\frac{\partial f}{\partial y}).$$ Now you should be able to finish your calculations.

Answer 3

Define $z=x+iy$, then$$\begin{cases} x=\frac{z+\overline{z}}{2}\\ y=\frac{z-\overline{z}}{2i} \end{cases}.$$ We consider $x$ and $y$ are functions of $z$ and $\overline{z}$, thus we have $$\frac{\partial }{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{1}{i}\frac{\partial}{\partial y} \right) ;\quad \frac{\partial }{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial}{\partial y} \right)$$which shows $z$ is exactly what we have already known. We consider $h$, $f$, $g$ all are the functions of $z$ and $\overline{z}$, then we have$$dh=\frac{\partial h}{\partial z}dz+\frac{\partial h}{\partial \overline{z}}d\overline{z}.$$Also we can express$g\circ f$ as $$g\circ f=g(f(z,\overline{z}),\overline{f}(z,\overline{z})).$$Then,$$d(g\circ f)=\left( \frac{\partial g}{\partial z}\frac{\partial f}{\partial z}+\frac{\partial g}{\partial\overline{z}}\frac{\partial \overline{f}}{\partial z}\right)dz+ \left( \frac{\partial g}{\partial z}\frac{\partial f}{\partial \overline{z}}+\frac{\partial g}{\partial\overline{z}}\frac{\partial \overline{f}}{\partial \overline{z}}\right)d\overline{z}. $$Compared with $dh$ and $d(g\circ f)$, the complex version of chain rule is proved.