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.

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    $\begingroup$ Let me clarify something, just in case, because the intention might be one, but the notation might be indicating another. If one wanted the Wirtinger derivative of $\Re(z)=\frac{z+\overline{z}}{2}$, then this is $\frac{1}{2}\left(\frac{\partial}{\partial z}-i\frac{\partial}{\partial\overline{z}}\right)\left(\frac{z+\overline{z}}{2}\right)=\frac{1-i}{4}$. However, if this is denoting the complex derivative, then it doesn't exists since $\lim_{h\to0}\frac{\frac{(z+h)+(\overline{z}+\overline{z})}{2}-\frac{z+\overline{z}}{2}}{h}=\lim_{h\to0}\frac{h+\overline{h}}{2h}$, which doesn't exist. $\endgroup$ – user647486 Apr 22 at 13:37
  • $\begingroup$ I have no idea of what a Wirtinger derivative is. $\endgroup$ – Mauro Giliberti Apr 22 at 13:42
  • $\begingroup$ Google. $\endgroup$ – user647486 Apr 22 at 13:44
  • $\begingroup$ Yes, I read that. I still don't know what that is or why my professor would have used that, so probably he meant the classic complex derivative. $\endgroup$ – Mauro Giliberti Apr 22 at 13:47
  • $\begingroup$ Then the second part of the comment is what you immediately need. $\endgroup$ – user647486 Apr 22 at 13:48

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 $$\frac{\partial ({\rm Re})}{\partial \overline{z}}(z) =\frac{1}{2}\neq 0,\quad\mbox{but}\quad\frac{\partial ({\rm Re})}{\partial z}(z) =\frac{1}{2}.$$

  • $\begingroup$ Are those Wirtinger operators normally used in the context of complex derivatives? I'm trying to prove an easy theorem and never heard of those. In particular, I'm trying to calculate $\frac{dF}{dz}=\frac{dx}{dz}\frac{dF}{dx}+\frac{dy}{dz}\frac{dF}{dy}$. Am I doing something wrong? $\endgroup$ – Mauro Giliberti Apr 22 at 13:52
  • $\begingroup$ Yes. Probably you don't have clear the distinction between ${\rm d}F/{\rm d}z$ and $\partial F/\partial z$. Again: the former is the classical derivative computed via limit, which happens to be equal to the second one (which always makes sense if $F$ has partial derivatives with respect to $x$ and $y$) for holomorphic functions. $\endgroup$ – Ivo Terek Apr 22 at 13:55
  • $\begingroup$ Probably what you want is $$\frac{\partial F}{\partial z} =\frac{\partial x}{\partial z}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial z}\frac{\partial F}{\partial y} =\frac{1}{2}\left(\frac{\partial F}{\partial x}-i\frac{\partial F}{\partial y}\right),$$which is nothing more than a proof that the Wirtinger operadors must be defined as above, using $\partial x/\partial z=1/2$ (did above) and $\partial y/\partial z=1/2$. $\endgroup$ – Ivo Terek Apr 22 at 14:00
  • $\begingroup$ Yes, sorry for not specifying! F is holomorphic, and his partial derivatives are $\frac{\partial{F}}{\partial{x}}=f(z)$, $\frac{\partial{F}}{\partial{y}}=if(z)$ where f is another (not important) function. $\endgroup$ – Mauro Giliberti Apr 22 at 14:02
  • $\begingroup$ You're insisting on the notational mistake, that's bound to cause you problems: it is $\partial F/\partial x$, not ${\rm d}F/{\rm d}x$, etc. You can produce the correct symbol with \partial. $\endgroup$ – Ivo Terek Apr 22 at 14:03

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