[complex analysis] differentiation w.r.t. complex conjugate; does it make sense?


by nonequilibrium
Tags: analysis, complex, conjugate, differentiation, sense
nonequilibrium
nonequilibrium is offline
#1
Dec13-11, 09:38 PM
P: 1,406
Hello,

Differentiability of [itex]f : \mathbb C \to \mathbb C[/itex] is characterized as [itex]\frac{\partial f}{\partial z^*} = 0[/itex].

More exactly: [itex]\frac{\partial f(z,z^*)}{\partial z^*} := \frac{\partial f(z[x(z,z^*),y(z,z^*)])}{\partial z^*} = 0[/itex] where [itex]z(x,y) = x+iy[/itex] and [itex]x(z,z^*) = \frac{z+z^*}{2}[/itex] and analogously for y.

But anyway, does this make sense? More specifically, is it consistent? The thing I'm having trouble with is that it looks like that we can always make the differentation w.r.t. z* zero. For example, for clarity of my argument, define

[itex]g : \mathbb C \to \mathbb C: z \mapsto z^*[/itex]

then for the modulus function [itex] \frac{\partial |z|}{\partial z^*} = \frac{\partial \sqrt{z g(z)}}{\partial z^*} = 0[/itex] as g is a function of z and not z*.

Where is the mathematical error in this reasoning?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
nonequilibrium
nonequilibrium is offline
#2
Dec16-11, 05:01 PM
P: 1,406
One possible answer is that the technique uses the chain rule, which only works if the functions are differentiable.

Concretely in my example g is not differentiable, so the ``[itex]\frac{\partial f}{\partial z^*} = 0 \Leftrightarrow f[/itex] is differentiable'' technique doesn't strictly apply.

But if that is the explanation, then you can never use the ``[itex]\frac{\partial f}{\partial z^*} = 0 \Leftrightarrow f[/itex] is differentiable'' technique to prove differentiability (as is often done and as is the sole purpose of the characterization), as you need to assume the differentiability for the requirements... (because if you don't, then you can construct examples as I did in the OP)

Any thoughts?
zhentil
zhentil is offline
#3
Dec16-11, 05:52 PM
P: 491
I'm a bit confused when you say that g is a function of z and not its conjugate, hence the partial must be zero. Doesn't that only work when the variables are independent?

Mute
Mute is offline
#4
Dec17-11, 10:18 AM
HW Helper
P: 1,391

[complex analysis] differentiation w.r.t. complex conjugate; does it make sense?


Quote Quote by zhentil View Post
I'm a bit confused when you say that g is a function of z and not its conjugate, hence the partial must be zero. Doesn't that only work when the variables are independent?
The variables [itex]z[/itex] and [itex]\bar{z}[/itex] are linear combinations of x and y, so they can be treated as independent. This way of viewing things is called "Wirtinger calculus", and it is in this formalism in which the Cauchy-Riemann equations become [itex]\partial f(z,\bar{z})/\partial \bar{z} = 0[/itex]. See here.

However, I think that may have hit upon the error. The variables [itex]z[/itex] and [itex]\bar{z}[/itex] are independent, so the function g(z) is taking in z and returning the independent variable [itex]\bar{z}[/itex], which isn't possible.
nonequilibrium
nonequilibrium is offline
#5
Dec17-11, 10:25 AM
P: 1,406
Mute, I don't understand your exact objection. Are you saying that I'm not allowed to define such function g?
Mute
Mute is offline
#6
Dec17-11, 11:05 AM
HW Helper
P: 1,391
Quote Quote by mr. vodka View Post
Mute, I don't understand your exact objection. Are you saying that I'm not allowed to define such function g?
Yes. Your function takes in one independent variables and returns another independent variable. I don't think that's possible. If you want to treat z and z* as dependent, then the [itex]\partial f/\partial z^\ast[/itex] version of the Cauchy-Riemann equations just reduce to the usual form expressed in terms of x and y.
nonequilibrium
nonequilibrium is offline
#7
Dec17-11, 12:37 PM
P: 1,406
That doesn't really make sense though? How can there be a constriction on the functions that I'm allowed to define? So according to you, it's also illegal to use something like Re(z)? Where does it end?
Mute
Mute is offline
#8
Dec17-11, 07:36 PM
HW Helper
P: 1,391
Quote Quote by mr. vodka View Post
That doesn't really make sense though? How can there be a constriction on the functions that I'm allowed to define? So according to you, it's also illegal to use something like Re(z)? Where does it end?
The heart of the issue isn't even a complex-variables issue. Consider a function [itex]h: \mathbb{R}^2 \rightarrow \mathbb{R}[/itex]. If you have a function q = h(x,y), where q is the dependent variable and x and y are independent variables, how can you define a function p(x) that returns y (i.e., y = p(x))? If the variable y is independent from x, you can't - the two variables are independent, so you can choose the value of y regardless of the value of x, so x doesn't determine y at all, which contradicts the assumption that you can write y = p(x). That inconsistency is what prevents you from defining such a function.

When applied to the complex variable case, you have made a (formal) change of basis from x and y to z and z*, rendering z and z* independent variables. This is the Wirtinger calculus in which the Cauchy-Riemann equations take the form

[tex]\frac{\partial f(z,z^\ast)}{\partial z^\ast} = 0.[/tex]

If you are using this definition of the CR equations, then you are treating z and z* independently, which means you cannot define a function g(z) which returns z*. (A function like Re(z) could still be defined by (z+z*)/2).

Perhaps the confusion arises because we normally don't consider z and z* to really be independent, but when using the Wirtinger calculus, you are formally treating them that way. I myself am a bit foggy on why the Wirtinger calculus works, but if I recall correctly, it is in part because when you define a complex function as

[tex]f(z) = u(x,y) + iv(x,y),[/tex]

the functions u and v (presumably) have convergent taylor series for which you can formally let x and y themselves be complex variables, which allows you to treat z and z* independently. This is as far as my memory goes on this topic, though, so if you want exact details you will have to seek them out for yourself.
nonequilibrium
nonequilibrium is offline
#9
Dec18-11, 07:07 AM
P: 1,406
Okay I think I'm starting to see your point... It does seem really fishy to say that z and z* are independent variables, it doesn't make a lot of sense, but I suppose it's allowed if one can make it mathematically consistent...

But may I try to correct your point a bit more? You say I'm not allowed to define g as [itex]g : \mathbb C \to \mathbb C: z \mapsto z^*[/itex]? I don't think this can be correct, after all the CR characterization [itex]\frac{\partial f}{\partial z^*}=0[/itex] is used to show that functions such as g aren't complex differentiable. A more correct objection (correct me if you think I'm wrong) would seem to me to say that g as defined above is okay as a normal function, but in this Wirtinger calculus, complex functions now depend on both z and z*, and the corresponding function is now [itex]\tilde g : \mathbb C^2 \to \mathbb C: (z,z^*) \mapsto z^*[/itex] and for brevity one says, for obvious reasons, that [itex]g = \tilde g[/itex], which can be dangerous because you shouldn't apply the ``[itex]\frac{\partial f}{\partial z^*}=0[/itex]'' test to the LHS (i.e. g), but to the RHS (i.e. [itex]\tilde g[/itex]). Do you agree?


Register to reply

Related Discussions
[complex analysis] domain coloring applet? (visualizing complex functions) General Math 0
Complex Analysis Complex Integration Question Calculus & Beyond Homework 6
Complex analysis - graphing in complex plane Calculus & Beyond Homework 6
Partial differentiation & complex analysis Calculus & Beyond Homework 3
Advice on complex analysis, Riemann surface & complex mappings. Differential Geometry 3