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Proof about complex conjugate of a function

by anhnha
Tags: complex, conjugate, function, proof
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anhnha
#1
Jul29-13, 02:53 AM
P: 170
Hi,
I need to understand the proof about complex conjugate of a function.
g(z) = g*(z*)
I don't know what it it called in English and can't search for it.
If anyone knows where can I get the proof, please let me know.
Thanks for help.
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CompuChip
#2
Jul29-13, 03:46 AM
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Note that if g is analytic around z0, you can write it as
$$g(z) = \sum_{n = 0}^\infty a_n (z - z_0)^n$$
for all z in a neighborhood of z0, which is an easy expression to work with.
anhnha
#3
Jul29-13, 04:02 AM
P: 170
Deleted.

anhnha
#4
Jul29-13, 04:34 AM
P: 170
Proof about complex conjugate of a function

Thanks for the idea!
[tex]g\left( z^{\ast }\right) =\sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}[/tex]
[tex]g^{\ast }\left( z^{\ast }\right) = \left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }= \sum _{n=0}^{\infty }\left( z-z_{0}\right) ^{n}= g(z) [/tex]
Is this right?
CompuChip
#5
Jul29-13, 04:42 AM
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Quote Quote by anhnha View Post
Thanks for the idea!
[tex]g\left( z^{\ast }\right) =\sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}[/tex]
This is actually a very specific function. In general, you will have coefficients an
[tex]g\left( z \right) =\sum _{n=0}^{\infty }a_n \left( z -z_{0}\right) ^{n}[/tex]
Depending on whether you look at real or complex analytic functions (also see this Wikipedia page and related pages), an are real or complex numbers and the proof will be more or less straightforward.

[tex]g^{\ast }\left( z^{\ast }\right) = \left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }= \sum _{n=0}^{\infty }\left( z-z_{0}\right) ^{n}= g(z) [/tex]
Is this right?
You are pretty quick jumping from ## \left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }## to ##\sum _{n=0}^{\infty }\left( z-z_{0}\right) ^{n}##. You are trying to prove whether this equality holds, and it almost seems like you are assuming it now. At the moment, your 'proof' basically says "this theorem is true because it's true.". To convince a mathematician, you will need some more intermediate steps.

For example: is the conjugate of the sum the sum of the conjugate? If so, you can write
$$\left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast } = \sum _{n=0}^{\infty } \left(\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }$$
Then, how do you get from ##\left( (z^\ast - z_0)^n \right)^\ast## to ##(z - z_0)^n##?

You will find that you actually need some restrictions!
anhnha
#6
Jul29-13, 05:07 AM
P: 170
Hi,
In my case I am now interested in real functions. Therefore, according to the link you gave coefficients an will be real.
You are pretty quick jumping from ## \left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }## to ##\sum _{n=0}^{\infty }\left( z-z_{0}\right) ^{n}##. You are trying to prove whether this equality holds, and it almost seems like you are assuming it now. At the moment, your 'proof' basically says "this theorem is true because it's true.". To convince a mathematician, you will need some more intermediate steps.

For example: is the conjugate of the sum the sum of the conjugate? If so, you can write
$$\left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast } = \sum _{n=0}^{\infty } \left(\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }$$
Then, how do you get from ##\left( (z^\ast - z_0)^n \right)^\ast## to ##(z - z_0)^n##?

You will find that you actually need some restrictions!
Yes, that is my mistake. I use two properties of complex conjugate.
1. (x + y)* = x* + y*
2. (xy)*= x*y*
where x, y are complex numbers.
Then, how do you get from ##\left( (z^\ast - z_0)^n \right)^\ast## to ##(z - z_0)^n##?

You will find that you actually need some restrictions!
I think that z0 has to be real, right?
But in my case how can I prove that z0 is real?
anhnha
#7
Aug2-13, 11:00 PM
P: 170
Could anyone help me next?
CompuChip
#8
Aug3-13, 04:16 AM
Sci Advisor
HW Helper
P: 4,300
##z_0## is real if ##(z_0)^\ast = z_0##.

You can use the two properties you mentioned to show that
##\left( (z^\ast - z_0)^n \right)^\ast = (z^\ast - z_0^\ast)^n##.


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