Derivative of the complex conjugate of z with respect to z

In summary, the conversation discusses the use of Wirtinger derivatives in complex analysis and how they can be applied in a system of complex ordinary differential equations. The issue of finding the Jacobian of a function involving a complex conjugate is also brought up. It is determined that in this case, the Jacobian can be found using ordinary multivariate Jacobian techniques for real functions.
  • #1
fairy._.queen
47
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Hi all!

From Wirtinger derivatives, given [itex]z=x+iy[/itex] and indicating as [itex]\overline{z}[/itex] the complex conjugate, I get:
[itex]\frac{\partial\overline{z}}{\partial z}=\frac{1}{2}\left(\frac{\partial (x-iy)}{\partial x}-i\frac{\partial (x-iy)}{\partial y}\right)=0[/itex]

This puzzles me, because I cannot see why a number and its complex conjugate could be independent variables.

Thank you in advance!
 
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  • #2
fairy._.queen said:
Hi all!

From Wirtinger derivatives, given [itex]z=x+iy[/itex] and indicating as [itex]\overline{z}[/itex] the complex conjugate, I get:
[itex]\frac{\partial\overline{z}}{\partial z}=\frac{1}{2}\left(\frac{\partial (x-iy)}{\partial x}-i\frac{\partial (x-iy)}{\partial y}\right)=0[/itex]

This puzzles me, because I cannot see why a number and its complex conjugate could be independent variables.

Thank you in advance!

If ##\Delta z = h + i k## then ##\Delta \bar{z} = h - ik## (##h,k## real), so
[tex] \frac{\Delta \bar{z}}{\Delta z} =\left( \frac{h^2 - k^2}{h^2+k^2}\right)
- i \left( \frac{2kh}{h^2+k^2} \right) [/tex]
Since the limit as ##h, k \to 0## depends on the "direction", the function ##\bar{z}## is not differentiable in the complex-analysis sense.
 
  • #3
Thank you very much for your reply! Then, what about the following situation:

I have an ordinary differential equation system that depends on a number of variables, including [itex]z[/itex]. I have to compute the Jacobian of the function [itex]f[/itex] defining the system, so that I need [itex]\frac{\partial f}{\partial z}[/itex], but [itex]\overline{z}[/itex] is not a variable for which the algorithm solves.

To simplify things, let's assume I have
[itex]f(a) = cz + d\overline{z}[/itex]
and the integrator I'm using needs [itex]\frac{\partial f(a)}{\partial z}[/itex].

How can I compute that entry of the Jacobian?
 
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  • #4
fairy._.queen said:
Thank you very much for your reply! Then, what about the following situation:

I have an ordinary differential equation system that depends on a number of variables, including [itex]z[/itex]. I have to compute the Jacobian of the function [itex]f[/itex] defining the system, so that I need [itex]\frac{\partial f}{\partial z}[/itex], but [itex]\overline{z}[/itex] is not a variable for which the algorithm solves.

To simplify things, let's assume I have
[itex]f(a) = cz + d\overline{z}[/itex]
and the integrator I'm using needs [itex]\frac{\partial f(a)}{\partial z}[/itex].

How can I compute that entry of the Jacobian?

Sorry: I have no idea, and I am not sure the problem even makes sense.

I guess it all depends on whether z is some parameter and you need df/dz is the sense of complex analysis, or just in the sense that using z is a convenient way of encapsulating two real variables and you actually have real derivatives. I just cannot tell from the information you supply.
 
  • #5
I have a system of complex ODEs, with complex variables.

Let's assume that it looks like this (it's an oversimplification, but will do):
[itex]
z' = w+iz\\
w'=c\overline{z}+dz
[/itex]

My algorithm requires me to provide the Jacobian manually, so that I need [itex]\frac{\partial w'}{\partial z}[/itex].

What can I do?
 
  • #6
fairy._.queen said:
I have a system of complex ODEs, with complex variables.

Let's assume that it looks like this (it's an oversimplification, but will do):
[itex]
z' = w+iz\\
w'=c\overline{z}+dz
[/itex]

My algorithm requires me to provide the Jacobian manually, so that I need [itex]\frac{\partial w'}{\partial z}[/itex].

What can I do?

What does the "prime" stand for (in ##z^{\prime}## for example)?
 
  • #7
Derivative with respect to a time variable, i.e. [itex] z = z(t) [/itex], for instance. Thanks!
 
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  • #8
fairy._.queen said:
Derivative with respect to a time variable, i.e. [itex] z = z(t) [/itex], for instance. Thanks!

OK, that is sort of what I suspected.

Assuming t is real, that means that using complex quantities is really just a convenient way of encapsulating real quantities; that is, you can re-write the DE system as a system of coupled DEs for real quantities---similar to what we do in electronics/electrical engineering when we use complex voltages, currents, impedances, etc. You are not really doing "complex analysis", so you don't have to worry about Cauchy-Riemann equations and all that. The Jacobians you need are just ordinary multivariate Jacobians for real functions. Perhaps just re-writing everything in purely real terms would be the least confusing way to go---although it may be much longer (just as in electronics we can always re-write everything in purely real terms, but the expressions get much longer in many cases).
 
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  • #9
Thanks a lot, it's clear now!
 
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