Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $i := \sqrt{-1}$, $f$ be the frequency ($\frac1p$), and $\omega := 2 \pi f$.

From page 3 here, why does $\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2} (e^{-i \omega t} - e^{i\omega t})$?

I understand that the LHS's $\frac{1}{2i}$ is multiplied by $\frac ii$ to get the RHS's $\frac i2$, but I don't understand how the contents of the parentheses changes?

share|cite|improve this question
Because $$\frac1i=-i$$ – DonAntonio May 24 '13 at 13:57
I get it now thanks guys. – Jase May 24 '13 at 13:58

Just that $\frac{1}{i} = - i$, so by distributing the minus sign, $$\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = -\frac{i}{2}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2}(e^{-i\omega t} - e^{i\omega t}). $$

share|cite|improve this answer
Dear downvoter: Thank you; my rep is now a multiple of ten. That's been bugging me for a while. :D – Neal May 24 '13 at 16:13

Your Answer


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.