Complex Conjugate of a Function

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Homework Help Overview

The discussion revolves around taking the complex conjugate of a specific complex function, expressed as \(\frac{1}{1-Ae^{i(a+b)}}\), where \(A\) is a real parameter and \(a\) and \(b\) are complex functions. Participants are exploring the validity of applying the complex conjugate operation to this function.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to determine if the complex conjugate can be expressed as \(\frac{1}{1-Ae^{-i(a^*+b^*)}}\) and questions the applicability of the rule \(f^*(x) = f(x^*)\). Some participants caution against this rule's general applicability, providing a counterexample to illustrate potential pitfalls. Others inquire whether the conclusion holds if \(a\) and \(b\) are simply complex numbers.

Discussion Status

The discussion is active, with participants providing insights and counterexamples. Some guidance has been offered regarding the conditions under which the original poster's assumption may hold true, particularly focusing on the nature of the functions \(a\) and \(b\).

Contextual Notes

There is an ongoing examination of the definitions and properties of complex functions, particularly in relation to complex conjugation. The implications of the parameters being complex or real are under consideration, which may affect the validity of the original poster's reasoning.

Niles
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Homework Statement


Hi

I have a complex function of the form
<br /> \frac{1}{1-Ae^{i(a+b)}}<br />
I want to take the complex conjugate of this: The parameters a and b are complex functions themselves, but A is real. Am I allowed to simply say
<br /> \frac{1}{1-Ae^{-i(a^*+b^*)}}<br />
where * denotes the c.c.? I seem to vaguely remember that f^*(x) = f(x^*).
 
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Be careful; the rule f(z^*)=f(z)^* doesn't in general work when z is a complex number.
Consider the following counterexample: take f(z)=i|z|, where |\cdot | is the absolute value. Then f(z^*)=i|z^*|=i|z|=f(z) but f(z)^*=(i|z|)^*=-i|z|=-f(z), so we have f(z^*)\neq f(z)^*.

In general, your function f can be very nice (the example above isn't complex-differentiable) but still fail to have this property.

For your problem, it's going to depend a lot on what your functions a,b are.
 
Thanks for helping out. If a and b are just complex numbers, then it should be correct, no?
 
Niles said:
Thanks for helping out. If a and b are just complex numbers, then it should be correct, no?

Yes, if a and b are just complex numbers, then it works out like that.
 

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