Proof about complex conjugate of a function

In summary: Then, if ##z_0^\ast = z_0##, you can simplify ##(z^\ast - z_0^\ast)^n## to ##(z - z_0)^n##.In summary, in order to prove the complex conjugate of a function, you must first show that the coefficients are real if the function is analytic. Then, you can use the properties of complex conjugate to simplify the expression and show that z0 must be real for the equality to hold.
  • #1
anhnha
181
1
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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  • #2
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
Deleted.
 
Last edited:
  • #4
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?
 
  • #5
anhnha said:
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!
 
  • #6
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?
 
  • #7
Could anyone help me next?
 
  • #8
##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##.
 

1. What is the complex conjugate of a function?

The complex conjugate of a function is a function that has the same real part as the original function, but with the imaginary part multiplied by -1. In other words, it is a function that reflects the original function across the real axis on the complex plane.

2. Why is the complex conjugate of a function important?

The complex conjugate of a function is important because it allows us to simplify and manipulate complex expressions and equations. It also helps us to find the roots of complex polynomials and solve certain types of differential equations.

3. How do you find the complex conjugate of a function?

To find the complex conjugate of a function, you simply replace all instances of the imaginary variable, typically denoted as "i", with -i. For example, if the original function is f(z) = z^2 + 3iz - 2, its complex conjugate would be f*(z) = z^2 - 3iz - 2.

4. What is the relationship between a function and its complex conjugate?

The relationship between a function and its complex conjugate is that they are reflections of each other across the real axis. This means that if we graph both functions on the complex plane, they will be identical in the real part, but have opposite signs in the imaginary part.

5. Can the complex conjugate of a function be used to simplify expressions?

Yes, the complex conjugate of a function can be used to simplify expressions. This is because multiplying a complex number by its conjugate results in a real number, thus eliminating the imaginary part of the expression. This is particularly useful when dealing with complex numbers in the denominator of a fraction, as it allows us to rationalize the denominator.

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