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
0
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!
 
Physics news on Phys.org
  • #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?
 
Last edited:
  • #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!
 
Last edited:
  • #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).
 
  • Like
Likes 1 person
  • #9
Thanks a lot, it's clear now!
 

1. What is the derivative of the complex conjugate of z with respect to z?

The derivative of the complex conjugate of z with respect to z is equal to the complex conjugate of the derivative of z with respect to z. In other words, if z = x + yi, then the derivative of z with respect to z is equal to 1, and the derivative of the complex conjugate of z with respect to z is equal to 1.

2. How do you find the derivative of the complex conjugate of z with respect to z?

To find the derivative of the complex conjugate of z with respect to z, you can use the chain rule. First, you need to express the complex conjugate of z as a function of x and y. Then, you can take the derivative of this function with respect to x and y, and use the chain rule to find the derivative with respect to z. Alternatively, you can use the Cauchy-Riemann equations to directly find the derivative.

3. Why is the derivative of the complex conjugate of z with respect to z equal to the complex conjugate of the derivative of z with respect to z?

This is because the complex conjugate of a complex function is equal to the complex conjugate of each term in the function. Therefore, when taking the derivative of the complex conjugate of z with respect to z, the derivative of each term will also be complex conjugated. This results in the complex conjugate of the derivative of z with respect to z.

4. Can the derivative of the complex conjugate of z with respect to z be simplified?

Yes, the derivative of the complex conjugate of z with respect to z can be simplified using the Cauchy-Riemann equations. These equations relate the partial derivatives of a complex function to each other, allowing you to express the derivative in terms of only one of the partial derivatives. This simplification can be useful for solving complex integration problems.

5. What is the geometric interpretation of the derivative of the complex conjugate of z with respect to z?

The geometric interpretation of the derivative of the complex conjugate of z with respect to z is that it represents the rate of change of the complex conjugate of z as z changes. In other words, it measures how much the complex conjugate of z changes for a small change in z. This can be visualized as the slope of a tangent line to the complex conjugate of z on the complex plane.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
981
  • Calculus and Beyond Homework Help
Replies
3
Views
772
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Replies
1
Views
609
  • Calculus and Beyond Homework Help
Replies
8
Views
471
Replies
2
Views
897
  • Calculus and Beyond Homework Help
Replies
3
Views
278
  • Calculus and Beyond Homework Help
Replies
5
Views
997
Back
Top