# Proof about complex conjugate of a function

1. Jul 29, 2013

### anhnha

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. Jul 29, 2013

### CompuChip

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.

3. Jul 29, 2013

### anhnha

Deleted.

Last edited: Jul 29, 2013
4. Jul 29, 2013

### anhnha

Thanks for the idea!
$$g\left( z^{\ast }\right) =\sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}$$
$$g^{\ast }\left( z^{\ast }\right) = \left( \sum _{n=0}^{\infty }\left( z^{\ast }-z_{0}\right) ^{n}\right) ^{\ast }= \sum _{n=0}^{\infty }\left( z-z_{0}\right) ^{n}= g(z)$$
Is this right?

5. Jul 29, 2013

### CompuChip

This is actually a very specific function. In general, you will have coefficients an
$$g\left( z \right) =\sum _{n=0}^{\infty }a_n \left( z -z_{0}\right) ^{n}$$
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.

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!

6. Jul 29, 2013

### anhnha

Hi,
In my case I am now interested in real functions. Therefore, according to the link you gave coefficients an will be real.
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.
I think that z0 has to be real, right?
But in my case how can I prove that z0 is real?

7. Aug 2, 2013

### anhnha

Could anyone help me next?

8. Aug 3, 2013

### CompuChip

$z_0$ is real if $(z_0)^\ast = z_0$.

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