Conjugate of a complex function

[Xerxes]

This should be laughably easy, but I am a little confused here.

 

Suppose that the function [math]f: \mathbb{C} \to \mathbb{C}[/math]. Suppose further that [math]z \in \mathbb{C} = x +iy,\,\, x,\, y \in \mathbb{R}[/math].

 

What is meant by the complex conjugate of this function?

 

My thoughts (such as they are!). Set [math]z = x +iy[/math], and set [math]f(z) = ax+iby[/math] and [math]\overline{f(z)}= ax-iby[/math]

 

Apparently this can be written as the identity [math]\overline{f}(z) = \overline{f(z)}= ax-iby[/math], which I don't quite get. Moreover,......

 

...... how does this differ from, say, [math] f(\overline{z}) =\overline{ax+iby}[/math]?

Edited April 7, 2011 by Xerxes

[DrRocket]

This should be laughably easy, but I am a little confused here.

 

Suppose that the function [math]f: \mathbb{C} \to \mathbb{C}[/math]. Suppose further that [math]z \in \mathbb{C} = x +iy,\,\, x,\, y \in \mathbb{R}[/math].

 

What is meant by the complex conjugate of this function?

 

My thoughts (such as they are!). Set [math]z = x +iy[/math], and set [math]f(z) = ax+iby[/math] and [math]\overline{f(z)}= ax-iby[/math]

 

Apparently this can be written as the identity [math]\overline{f}(z) = \overline{f(z)}= ax-iby[/math], which I don't quite get. Moreover,......

 

...... how does this differ from, say, [math] f(\overline{z}) =\overline{ax+iby}[/math]?

 

Just think of conjugation as a function then

 

[math] \overline {f}(z) = \overline {f(z)} = (conjugation \circ f)(z)[/math]

 

[math] f(\overline {z}) = (f \circ conjugation) (z) [/math]

[Xerxes]

I thank you for that. Without wishing to appear ungrateful, I had come to a similar conclusion: in [math]\overline{f}(z) \equiv \overline{f(z)}[/math] we are taking the conjugate of the image point in the codomain, whereas in [math]f(\overline{z})[/math] we are taking the conjugate of an element in the domain. Which is just wordy way of re-stating you point

 

So do me another kindness, and see if the following floats your goat:

 

We know from elementary analysis that [math]\overline{\sin(z)} = \sin (\overline{z})[/math] and is analytic (indeed entire) and also that where [math]f(z) = z^n[/math] for any nonzero real [math]n[/math] that [math]f(z)[/math] is analytic also. Using fingers and toes only, it appears that for small [math]n[/math] then [math]\overline{f(z)} \equiv \overline{f}(z) = f(\overline{z})[/math]. Likewise any other polynomial function.

 

I somewhat rashly propose the following generalization: if a function is analytic then [math]\overline{f}(z) = f(\overline{z})[/math] always. True or false?

 

I confess I am having trouble with the converse, namely that analycity is required for this equality to hold. I use as an example [math]f(z) = |z|^2 \equiv z\overline{z}[/math] which clearly not analytic, but where the equality seems to hold (unless I made a mistake).

 

Is this gibberish?

[DrRocket]

So do me another kindness, and see if the following floats your goat:

 

We know from elementary analysis that [math]\overline{\sin(z)} = \sin (\overline{z})[/math] and is analytic (indeed entire) and also that where [math]f(z) = z^n[/math] for any nonzero real [math]n[/math] that [math]f(z)[/math] is analytic also. Using fingers and toes only, it appears that for small [math]n[/math] then [math]\overline{f(z)} \equiv \overline{f}(z) = f(\overline{z})[/math]. Likewise any other polynomial function.

 

I somewhat rashly propose the following generalization: if a function is analytic then [math]\overline{f}(z) = f(\overline{z})[/math] always. True or false?

 

I don't follow your statement regarding sine as following from "elementary analysis" since extending the domain to complex numbers requires complex analysis, but the statement is true.

 

You have outlined the proof that [math]\overline{f}(z) = f(\overline{z})[/math]. To wit:

 

[math] \overline{z_1z_2} = \overline {z_1} \overline{z_2} [/math] clearly extends to show that [math]\overline{f}(z) = f(\overline{z})[/math] for any polynomial function [math]f[/math]. Since conjugation is continuous and since any analytic function is the uniform limit on compacta of polynomials the theorem now follows for entire functions. Since sine is entire, it follows for sine.

 

 

I confess I am having trouble with the converse, namely that analycity is required for this equality to hold. I use as an example [math]f(z) = |z|^2 \equiv z\overline{z}[/math] which clearly not analytic, but where the equality seems to hold (unless I made a mistake).

 

Analyticity is not required for equality. Your counterexample is valid. The only real-valued analytic functions are constant.

 

 

 

Is this gibberish?

no

[Xerxes]

Ha! My functional analysis text is a treasure!

 

It first states the Thm of Cauchy: If [math]f(z)[/math] is analytic within and on the closed contour [math]C[/math] and the derivtive [math]f'(z)[/math] is continuous throughout this region, then

 

[math]\oint_C f(z) \,dz = 0[/math]. (Parenthetically, given the amazing power of this thm., the proof is surprisingly straightforward. In fact I have seen two, one using Green's Thm and the other using the closely related Thm of Stokes. Um well - neither of these is so easily proved I guess)

 

It then points out that Goursat showed that the statement about continuity is superfluous, and re-states the Thm as Cauchy-Goursat.

 

It then comes out with this priceless gem:

 

"Some authors (never mathematicians!) define an analytic function as a differentiable function with continuous derivatives.......But this is a mathematical fraud of cosmic proportions"

 

No words minced there, then.

[DrRocket]

Ha! My functional analysis text is a treasure!

 

What is the book ? This material is usually treated in a text on complex analysis rather than one on functional analysis.

 

It first states the Thm of Cauchy: If [math]f(z)[/math] is analytic within and on the closed contour [math]C[/math] and the derivtive [math]f'(z)[/math] is continuous throughout this region, then

 

[math]\oint_C f(z) \,dz = 0[/math]. (Parenthetically, given the amazing power of this thm., the proof is surprisingly straightforward. In fact I have seen two, one using Green's Thm and the other using the closely related Thm of Stokes. Um well - neither of these is so easily proved I guess)

 

Those hypotheses are a bit strange. Once you know that f is analytic, you know that it has continuous derivatives of all orders. See below

 

But if the authopr is starting from scratch an intends to use Stokes Theorem then he needs the hypothesis of continuous derivatives. There are better ways to do this.

 

It then points out that Goursat showed that the statement about continuity is superfluous, and re-states the Thm as Cauchy-Goursat.

 

Yes, see above. Most treatments simply call the theorem Cauchy's Theorem. The usual proof is done first for triangles in a convex region and then generalized to arbitrary curves in a convex region.

 

It then comes out with this priceless gem:

 

"Some authors (never mathematicians!) define an analytic function as a differentiable function with continuous derivatives.......But this is a mathematical fraud of cosmic proportions"

 

No words minced there, then.

 

An analytic function is one that is locally representable by a power series. This is considerably more restrictive that even having continous derivatives of all orders ( [math]C^\infty[/math])

 

What is striking about complex-valued functions of a complex variable is that any differentiable function is automatically analytc. This makes complex analysis very different from real analysis. Analytic functions are, in a sense, "nice". But you give up existence of partitions of unity, so complex manifolds are so difficult that very little theory exists, and Kahler manifolds (about which I know very little) are what are studied.

[creyesk]

This is not true, for a polinomial to satisfy this it must have real coeficients, the same goes for the coeficients in the power series expansion of the analytic function.