
#1
Jul2913, 02:53 AM

P: 149

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. 



#2
Jul2913, 03:46 AM

Sci Advisor
HW Helper
P: 4,301

Note that if g is analytic around z_{0}, you can write it as
$$g(z) = \sum_{n = 0}^\infty a_n (z  z_0)^n$$ for all z in a neighborhood of z_{0}, which is an easy expression to work with. 



#3
Jul2913, 04:02 AM

P: 149

Deleted.




#4
Jul2913, 04:34 AM

P: 149

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( zz_{0}\right) ^{n}= g(z) [/tex] Is this right? 



#5
Jul2913, 04:42 AM

Sci Advisor
HW Helper
P: 4,301

[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), a_{n} are real or complex numbers and the proof will be more or less straightforward. 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! 



#6
Jul2913, 05:07 AM

P: 149

Hi,
In my case I am now interested in real functions. Therefore, according to the link you gave coefficients a_{n} will be real. 1. (x + y)* = x* + y* 2. (xy)*= x*y* where x, y are complex numbers. But in my case how can I prove that z_{0} is real? 



#7
Aug213, 11:00 PM

P: 149

Could anyone help me next?




#8
Aug313, 04:16 AM

Sci Advisor
HW Helper
P: 4,301

##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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