How to make it formally correct? 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?
 A: 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.

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  $\frac{\partial}{\partial z^*}=0$

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.
