Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The conjugate of $e^{-iwt}$ is $e^{iwt}$.

Then, what would be the conjugate of $e^{iwt}$? Would it be $e^{-iwt}$?

Also, for $|e^{iwt}|^2$, what would the value look like?

share|improve this question
    
Your first formula is correct only if $w t$ is real... –  Morgan Sherman Aug 9 '12 at 21:09
    
What are $\,w,t\,$? Real, complex...? –  DonAntonio Aug 9 '12 at 21:10
    
Also, "linear-algebra" is not a good tag for this question (try instead something like complex-variables). –  Morgan Sherman Aug 9 '12 at 21:12
    
Given the first, sure. –  André Nicolas Aug 9 '12 at 21:13
    
In general $\overline{e^{z}} = e^{\overline{z}}$ for all $z \in \mathbb{C}$. –  Mikko Korhonen Aug 9 '12 at 21:28

1 Answer 1

up vote 4 down vote accepted

Complex conjugation is an automorphism of order 2, meaning $\,\overline{\overline z}=z\,\,,\,\,\forall\,z\in\Bbb C\,$ , so if the conjugate of $\,e^{-iwt}\,$ is $\,e^{iwt}\,$ , then the conjugate of the latter is the former.

Also, writing the trigonometric version of $\,e^{ix}\,\,,\,x\in\Bbb R\,$ , you can check at once that $\,|e^{ix}|=1\,\,\,,\,\,\forall x\in\Bbb R\,$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.