Proving Complex Conjugate is Real: Euler's Identity

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Homework Help Overview

The discussion revolves around proving that the expression \(\alpha \exp(i\varphi) + \alpha^* \exp(-i\varphi)\) is real, utilizing concepts from complex numbers and Euler's Identity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the expansion of exponential functions into trigonometric forms, questioning if rewriting \(\alpha\) in terms of real and imaginary components aids in the proof.

Discussion Status

Some participants have shared their attempts and insights, with one noting a more tedious approach that aligns with earlier suggestions. There is an ongoing exploration of different methods, but no explicit consensus has been reached.

Contextual Notes

Participants are considering the implications of whether certain variables, like \(\omega\), are real or complex, which affects the validity of their reasoning.

JPBenowitz
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Homework Statement



So we are given \alphaexp(i\varpit) +\alpha*exp(-i\varpit) and are asked to prove the resulting equation is real.

Homework Equations



\alpha + \alpha* = 2Re(\alpha) and Euler's Identity

The Attempt at a Solution



I tried expanding out the exp's to cosines and isines but couldn't reach the solution.
 
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What if you also write ##\alpha = a + i b##, where a and b are both real numbers? Can you do the problem then?
 
Mute said:
What if you also write ##\alpha = a + i b##, where a and b are both real numbers? Can you do the problem then?

I figured it out but I didn't do it that way. That seems awfully more tedious than usual.
 
JPBenowitz said:
I figured it out but I didn't do it that way. That seems awfully more tedious than usual.

It's more tedious than the quick solution, yes, but it was more along the lines of the approach you had tried to take by expanding the exponentials into sines and cosines, so I opted to guide you along that direction, in case the problem wanted you to show it explicitly.
 
Why are you writing a bar over the omega? If omega is real then it's sort of obviously true. Because you are adding two complex conjugates. If omega isn't real then it's not even true.
 
Last edited:

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