# What's wrong with this proof that $e^{i\theta} = e^{-i\theta}$?

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"?

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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? – Herman Chau Mar 18 '12 at 8:18
You actually want $\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}$. – Qiaochu Yuan Mar 18 '12 at 8:19
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? – Herman Chau Mar 18 '12 at 8:23

## 1 Answer

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

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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. – Marc van Leeuwen Mar 18 '12 at 10:05