How to Find the Conjugate of a Complex Exponential Function?

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Homework Help Overview

The discussion revolves around finding the conjugate of a complex exponential function, specifically \(\varphi=exp(-x^2/x_0^2)\). Participants are exploring the implications of the variables involved, particularly whether they are real or imaginary.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants attempt to determine the conjugate of the given function and question the conditions under which their expressions hold true, particularly focusing on the nature of \(x\) and \(x_0\).

Discussion Status

The discussion is active, with participants providing insights about the conditions under which their solutions apply. There is a recognition that the nature of \(x/x_0\) affects the outcome, and some participants affirm the correctness of their reasoning based on the assumption that \(x\) and \(x_0\) are real.

Contextual Notes

Participants are considering the implications of real versus imaginary values for the variables involved, which is central to understanding the conjugate in this context.

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Homework Statement



Find the conjugate of

\varphi=exp(-x^2/x_0^2)

Homework Equations





The Attempt at a Solution




Isn't the conjugate \varphi*=exp(x^2/x_0^2)
 
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noblegas said:

Homework Statement



Find the conjugate of

\varphi=exp(-x^2/x_0^2)

Homework Equations


The Attempt at a Solution

Isn't the conjugate \varphi*=exp(x^2/x_0^2)

Not if x and x0 are real, which I suspect they are. What is it in that case?
 
Last edited:
Dick said:
Not if x and x0 are real, which I suspect they are. What is it in that case?

oh ,my solution would only be correct if x/x0 is imaginary.would my expression
<br /> exp(-x^2/x_0^2)<br /> not change when taking its conjugate??
 
noblegas said:
oh ,my solution would only be correct if x/x0 is imaginary.would my expression
<br /> exp(-x^2/x_0^2)<br /> not change when taking its conjugate??

Right, sort of. If x is imaginary the conjugate(exp(x))=exp(-x). If x is real then conjugate(exp(x))=exp(x). But your solution is only correct if (x/x0)^2 is purely imaginary.
 
Dick said:
Right, sort of. If x is imaginary the conjugate(exp(x))=exp(-x). If x is real then conjugate(exp(x))=exp(x). But your solution is only correct if (x/x0)^2 is purely imaginary.

but x/x0 is not purely imaginary , but completely real. So my expression would remain the same when taking its conjugate
 
noblegas said:
but x/x0 is not purely imaginary , but completely real. So my expression would remain the same when taking its conjugate

Yesss.
 

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