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Complex Differentiation

by Prologue
Tags: complex, differentiation
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Prologue
#1
Nov13-08, 09:42 PM
P: 185
I saw the argument for complex differentiation today and I had a question about a 'well known' aspect of the argument. My professor said something like this (at least a version of it): Derivatives on complex variables are defined in the usual way. However, in the complex plane, delta(z) may approach zero from any direction.

Why???

For the complex derivative to exist, the limiting value must be independent of the direction the derivative is taken in the complex plane.

Why???

Ok, ok. I think that if the first 'Why???' is answered then the second one will be answered too. So what is the deal, what gives the complex plane this property? I suspect this may be anwered by another question, what is the interpretation of the derivative of a complex function?

Sorry, that's a lot of questions but I am completely baffled on this, thanks!



Here is a link to an argument for the differentiation. http://phyastweb.la.asu.edu/phy501-s...otes/lec29.pdf
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Hurkyl
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Nov13-08, 11:04 PM
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Quote Quote by Prologue View Post
Derivatives on complex variables are defined in the usual way. However, in the complex plane, delta(z) may approach zero from any direction.
The same is true when you're working only with real numbers. The only difference is that there are more 'directions' in the complex plane than in the real numbers.


You're taking the limit as [itex]\Delta z[/itex] approaches zero. If you really wanted the limit as [itex]\Delta z[/itex] approached zero along some explicit curve, then you would have said that instead.

And when you get to multivariable calclus, you'll see the same thing in that setting too.


Have you considered appealing to definitions? You remember epsilon-delta arguments, and all that stuff, right?
HallsofIvy
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Nov14-08, 06:22 AM
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Well, that's the definition of "limit" isn't it?

Saying that "[itex]\lim_{x\rightarrow a} f(x)= L[/itex]" means "Given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]||x- a||< \delta[/itex], then [itex]||f(x)- L||< \epsilon[/itex].

There, I intentionally use "|| ||" rather than "| |" to emphasize that I am not assuming real numbers here. "|| ||" is "norm" is some vector space. The point is that says if x and a are "close", then f(x) and L must be "close" for whatever definition of "close" we are using. In particular it says nothing about x being in a particular direction from a: We can "approach" a from whatever direction we choose but as soon as we are within distanc [itex]\delta[/itex] from a, f(x) must be within [itex]\epsilon[/itex] of L. If the limit exists, then we must get the same result no matter what path we take to reach a. Since that is true of a limit, and the derivative is defined in terms of a limit, it is true of the derivative.

You may be thinking of the "partial derivatives" of a function where we "approach" the point only parallel to the x and y axes and, typically, get different results.

Let me emphasize that the partial derivatives are NOT "the" derivative. And, in fact, it is possible for a function to have partial derivaties at a point where the function is NOT differentiable!

Prologue
#4
Nov14-08, 09:37 AM
P: 185
Complex Differentiation

I understand the limit of a function stuff, but it seems to change when you start throwing around derivatives (partial?), and isn't that simple anymore, at least to me. The argument that goes into the complex function is x + iy, and we just call it z, so z depends on x and y.

[tex]z=x+iy; W(z)=W(x+iy)[/tex]

[tex]dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial iy}d(iy)[/tex]

If

[tex] dz = d \sqrt{||z||} = d \sqrt{z*z}[/tex]

then I can understand why it is direction independent, but then I wouldn't understand why it is only the magnitude.



Quote Quote by HallsofIvy View Post
...

You may be thinking of the "partial derivatives" of a function where we "approach" the point only parallel to the x and y axes and, typically, get different results.

Let me emphasize that the partial derivatives are NOT "the" derivative. And, in fact, it is possible for a function to have partial derivaties at a point where the function is NOT differentiable!

This is exactly what I am thinking of. Could you expand on how partial derivatives are not 'the derivative'?

It is very confusing to be able to have partial derivatives that can be different in a real plane, but when we jump to the complex plane they have to be the same?

I am obviously missing some distinction here.
morphism
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Nov14-08, 09:44 AM
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The partials aren't "the derivative" by definition. The derivative of a function f:R^n-R^m is by definition a linear map going from R^n to R^m that satisfies certain properties; if it exists it's given by the Jacobian matrix. The buzzwords you should google for are "Jacobian matrix" and, maybe, "Frechet derivative."

When you think of a complex function f:C->C as a function mapping R^2 to R^2 you need to keep in mind that there is some substantial extra structure available to us here - namely, complex division. In R^2 we could not divide vectors - this is the main reason differentiation is kind of flaky in higher real dimensions. But in C, we can divide complex numbers, and so we can define the complex derivative of f in a manner analogous to how we did it for R.
HallsofIvy
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Nov14-08, 09:53 AM
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No one has said that partial derivatives have to be the same in the complex plane. I specifically said we are talking about the derivative, NOT the partial derivatives.

I'm don't see how I can explain why partial derivatives are not "the" derivative without know what YOU know about derivatives in two dimensions. Every textbook I have ever seen makes the point clearly. Most, without going into the details, at least say that a function of two variables is "differentiable at a point" if and only the graph of z= f(x,y) has a tangent plane at that point. In particular, The function defined by "f(x,y)= 1 if xy is not 0, 0 if xy= 0" has both partial derivatives at (0,0) but is not even continuous there so cannot be differentiable.

If a function is differentiable at a point, we can think of the gradient of f at that point as representing the derivative at that point.
Prologue
#7
Nov14-08, 10:05 AM
P: 185
Ok, I think I am catching on to what you are saying HallsofIvy.

For instance, if we think of a complex function w(z) as being represented in one real dimension, 1 imaginary dimension, and one complex dimension (w(z)) then we would have a surface in real/complex 3d.

If that is the case then I have the picture and I can see what you are saying. If you represent 'the derivative' as a tangent plane at a point on the surface, then 'the derivative' would be the same coming from all directions.

So, is my visualization correct?

The reason I am bringing up what I am calling the partials in the complex plane (perhaps wrongly) is as a result of having the Cauchy Riemann conditions for an analytic function thrown in there.


Edit: Let me tack on that yesterday was the first time I saw any of this so please bear with me.


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