4
$\begingroup$

I recently learned that $\cos{\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2}$ Based on this, I managed to "prove" that: $$e^{i\theta} = e^{-i\theta}$$

Since $e^{i\theta} = \cos{\theta} + i\sin{\theta}$, we can substitute the above two identities to get: $$e^{i\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2} + i\frac{e^{i\theta} - e^{-i\theta}}{2}$$ Simplifying, I get $$(1-i)e^{i\theta} = (1-i)e^{-i\theta}$$ which implies $$e^{i\theta} = e^{-i\theta}$$ for all real $\theta$. Obviously, this isn't true in general, but I'm having a hard time seeing what's wrong. Can someone please point out the flaw in the above "proof"?

$\endgroup$
  • $\begingroup$ As a side note, I wasn't too sure what the relevant tags are for this question. I assume (fake-proofs) and (complex-numbers) are definitely relevant, but perhaps there are better tags? $\endgroup$ – Herman Chau Mar 18 '12 at 8:18
  • 4
    $\begingroup$ You actually want $\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}$. $\endgroup$ – Qiaochu Yuan Mar 18 '12 at 8:19
  • 3
    $\begingroup$ Oh, my textbook misprinted it as $\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2}$ and I just accepted it without bothering to check. -facepalm- So should I just delete this question? $\endgroup$ – Herman Chau Mar 18 '12 at 8:23
9
$\begingroup$

You don't have $\sin\theta = e^{i \theta}/2- e^{-i\theta} /2$. Actually, $$ \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2 i}.$$

$\endgroup$
  • 1
    $\begingroup$ One might add that the presence of $i$ in the formula for the sine is obviously necessary: $e^{i\theta}$ and $e^{-i\theta}$ are complex conjugates, so while their sum is real, their difference is purely imaginary. Since one knows that $\sin x$ is real for real $x$, one needs to divide out the $i$ to get from the imaginary to the real axis. This simple argument may help memorize the formulas. $\endgroup$ – Marc van Leeuwen Mar 18 '12 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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