What is the derivative of the real part of a complex variable?

Question

If I have the complex variable $z=x+iy$ and the function $f(z)=z$, is it possible to calculate $\frac{d\Re{f(z)}}{dz}$, or in this particular case $\frac{dx}{dz}$? It should be equal to $\frac{1}{2}$, but I don't know why.

Answer 1

Thing is, you have the two Wirtinger operators $$\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\qquad\mbox{and}\qquad\frac{\partial}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right),$$which are useful for the following reason: if $f:U\subseteq \Bbb C \to \Bbb C$ is continuous and has partial derivatives with respect to $x$ and $y$ at all points, $f$ is holomorphic if and only if $\partial f/\partial \overline{z}=0$, in which case we have ${\rm d}f/{\rm d}z = \partial f/\partial z$. The former (with straight ${\rm d}$) is only defined for points on which $f$ is $\Bbb C$-differentiable. So ${\rm d}({\rm Re})/{\rm d}z$ does not exist, since $$\frac{\partial ({\rm Re})}{\partial \overline{z}}(z) =\frac{1}{2}\neq 0,\quad\mbox{but}\quad\frac{\partial ({\rm Re})}{\partial z}(z) =\frac{1}{2}.$$