Let $f=f(z)$ and $g=g(w)$ be two complex valued functions which are differentiable in the real sense, $h(z)=g(f(z))$. Prove the complex chain rule. All partial derivatives: $$ \frac{\partial h}{\partial z} = \frac{\partial g}{\partial w}\frac{\partial f}{\partial z} + \frac{\partial g}{\partial \bar w}\frac{\partial \bar f}{\partial z} $$ and $$ \frac{\partial h}{\partial \bar z} = \frac{\partial g}{\partial w}\frac{\partial f}{\partial \bar z} + \frac{\partial g}{\partial \bar w}\frac{\partial\bar f}{\partial \bar z} $$ Are we supposed to arrive at this through Cauchy-Riemann?
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You can't use Cauchy-Riemann because that's a condition imposed on functions differentiable in the complex sense. Here, the functions are only differentiable in the real sense. – Vercassivelaunos Feb 25 '23 at 08:24
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This has nothing to do with Cauchy-Riemann's equations. Use the regular chain rule (for functions on $\mathbb{R}^2$) and the definition of the Wirtinger derivatives: $$ \frac{\partial}{\partial z} = \frac12 \left( \frac{\partial}{\partial x} - i\,\frac{\partial}{\partial y} \right) \qquad\text{and}\qquad \frac{\partial}{\partial \bar z} = \frac12 \left( \frac{\partial}{\partial x} + i\,\frac{\partial}{\partial y} \right) $$ It all boils down to a fairly long and tedious algebraic manipulation (See also: Wikipedia)
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\begin{gather*} \text{Let }f(z)=w=u(z)+iv(z)\\ \frac{\partial g}{\partial w} \frac{\partial f}{\partial z} +\frac{\partial g}{\partial \bar{w}} \frac{\partial \bar{f}}{\partial z}=\frac{1}{2}\left(\frac{\partial g}{\partial u}+\frac{\partial g}{i\partial v}\right)\left(\frac{\partial u}{\partial z}+\frac{i\partial v}{\partial z}\right) +\frac{1}{2}\left(\frac{\partial g}{\partial u}-\frac{\partial g}{i\partial v}\right)\left(\frac{\partial u}{\partial z}-\frac{i\partial v}{\partial z}\right)\\ =\frac{\partial g}{\partial u}\frac{\partial u}{\partial z}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial z}=\frac{\partial g}{\partial z} \end{gather*} Proceed similarly for the second chain rule
