How to make it formally correct?

Question

Can someone help me formalizing this statement:

$$ z= x^0 +ix^1 $$ And therefore $$ \frac{\partial}{\partial z} = \frac{\partial}{\partial (x^0 +ix^1)} = \frac{\partial}{\partial x^0} + \frac{1}{i} \frac{\partial}{\partial x^1} $$

My problem is with the last equality, I see it's right, but I'm not sure I'm allowed to do it in that way. Is there a way to procede more formally?

Answer 1

Your formula is only correct for holomorphic functions.

Usually one defines: $$ df(z)=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial z^*}dz^* $$ where $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial z^*}$ are called Wirtinger derivatives, defined as follows: $$ \frac{\partial}{\partial z}:=\frac{1}{2}\left(\frac{\partial}{\partial x^0}-i\frac{\partial}{\partial x^1}\right) $$ $$ \frac{\partial}{\partial z^*}:=\frac{1}{2}\left(\frac{\partial}{\partial x^0}+i\frac{\partial}{\partial x^1}\right) $$

Compared to your proposal, there is a 1/2 missing factor.

• this is ok for holomorphic functions (thanks to Cauchy-Riemann, see below)
• but not in the general case

I can copy derivation from An Introduction to Complex Differentials and Complex Differentiability p8-9, however this tutorial is really well written and I think it is better to directly read it.

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From $F(x,y)=U(x,y)+iV(x,y)$ the bivariate function associated to $f(z)$, the total differential is: $$ dF=\frac{\partial}{\partial x}F(x,y)dx+\frac{\partial}{\partial y}F(x,y)dy $$ which can be rewritten as follows (have a look at the cited reference): $$ dF=\frac{1}{2}\left[\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y}+i\left(\frac{\partial V}{\partial x}-\frac{\partial U}{\partial y}\right)\right]dz+\frac{1}{2}\left[\frac{\partial U}{\partial x}-\frac{\partial V}{\partial y}+i\left(\frac{\partial V}{\partial x}+\frac{\partial U}{\partial y}\right)\right]dz^* $$ Assuming $f$ holomorphic and using Cauchy-Riemann conditions ($\frac{\partial U}{\partial x}=\frac{\partial V}{\partial y}$ and $\frac{\partial U}{\partial y}=-\frac{\partial V}{\partial x}$) we get your suggestion: $\frac{\partial}{\partial z}=\frac{\partial}{\partial x^0}-i\frac{\partial}{\partial x^1}$ and $\frac{\partial}{\partial z^*}=0$

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One of the big advantage of Wirtinger calculus is:

For the Wirtinger derivatives, the common rules for differentiation known from real-valued analysis concerning the sum, product, and composition of two functions hold as well...

in brief you can do differential calculus like in $\mathbb{R}^n$. This is very handy for automatic differentiation, gradient computation... etc.