I have recently encountered a rather strange definition of a 'derivative' on functions with complex arguments. It was defined as $$d^* f(z) = \frac{\partial f}{\partial z} dz - \frac{\partial f}{\partial z^*} dz^*$$ It does follow the Leibniz rule so that $d^* (fg) = (d^*f)g + f(d^*g)$, so it seems to be a valid derivation. However, I have no idea about the context where this could be useful. Has anybody seen this derivation and can point me to relevant literature? Any mathematical insights are very welcome too.

Edit: Please specifically note the negative sign in the definition. I've found references to a similar expression, but with a + sign. How are they related?

Edit2: Ok, so with user44874's hint I've performed the calculation explicitly in complex cartesian coordinates (x,y) and I get $$ d^* f(x+iy) = i \left( \frac{\partial f}{\partial x} dy - \frac{\partial f}{\partial y} dx\right)$$ which is a symplectic "gradient" like the one used in Hamiltonian mechanics, up to a factor of $i$. Can anyone confirm this?


At "Notes on complex function theory" by Donald Sarason you can find this form of derivative under "III.16 COMPLEX PARTIAL DIFFERENTIAL OPERATORS".

Note that:

$$ \frac{\partial }{\partial z} = \frac{1}{2} (\frac{\partial }{\partial x}-i\frac{\partial }{\partial y} )$$

$$ \frac{\partial }{\partial z^*} = \frac{1}{2} (\frac{\partial }{\partial x}+i\frac{\partial }{\partial y} )$$

and that $f$ satisfies that Cauchy Reiman equation iff $\frac{\partial f }{\partial z^*}=0$ and thus holomorphic functions can be interpreted as those in the kernal of the operator $\frac{\partial f }{\partial z^*}$.

  • $\begingroup$ It seems you have a typo in the signs of your partial derivatives. Also, are you certain about the sign of the definition of what I wrote? I found the definition with a +, but not with a -. $\endgroup$ – Dirk Dinther Nov 2 '12 at 11:13
  • $\begingroup$ yes i had a typo, thanks. I am afraid I am not familiar with your notations, what do "dc" and "dc*" stand for? $\endgroup$ – user44874 Nov 2 '12 at 11:17
  • $\begingroup$ Argh, typo on my side. Sorry. dc and dc^* are really dz and dz^* $\endgroup$ – Dirk Dinther Nov 2 '12 at 11:18

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