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Since $e^{i\pi} = \cos \pi + i\sin \pi = -1,$ a suspicious argument is to proceed to conclude that $$-e^{i\pi} = 1.$$ However, this leads to $$-e^{i\pi} = e^{0}.$$ Is the above reasoning wrong?

Yes
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    No. $e^z$ is periodic with period $2\pi i$. $$e^0=e^{2\pi i}=e^{\pi i}\cdot e^{\pi i}=-e^{\pi i}.$$ – Jyrki Lahtonen Mar 20 '15 at 14:23
  • Why is it suspicious to conclude from $e^{\pi i} = -1$ that $-e^{\pi i} = 1$? In any case, $z \to e^z$ is not injective as a map $\mathbb{C} \to \mathbb{C}$. – anomaly Mar 20 '15 at 14:23
  • Thanks. You mean that argument is admissible? – Yes Mar 20 '15 at 14:23
  • @anomaly: Thanks; but that is the question – Yes Mar 20 '15 at 14:24
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    Would you conclude from $-(-1)^2=(-1)^3$ that $2=3$? – egreg Mar 20 '15 at 14:27
  • Not in provoking tone, etiquette. According to this site's law, being a dumb question is not a condition for downvoting. – Yes Mar 20 '15 at 14:31
  • @egreg: I am telling you; Do not talk like it is too trivial to bother you. Nobody forces you to look at this question. – Yes Mar 20 '15 at 14:34
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    @Chou, what are you getting upset about? egreg is trying to help. (Unless comments have been deleted...) – TonyK Mar 20 '15 at 14:37
  • @TonyK: No upset, never mind. I think the definition of trying to help varies from person to person. The comments you are interested are unimportant in the sense that they are not relevant to the question itself. – Yes Mar 20 '15 at 14:39
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    @Chou I'm sorry if you found my comment offensive, I surely didn't mean it. Nor I wanted to say your question is dumb or too trivial and in fact I didn't downvote it. I was pointing out that you neglected the minus sign. – egreg Mar 20 '15 at 16:06

2 Answers2

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That's correct. There's nothing wrong with the above reasoning. Is it equally wrong that $e^{2\pi n i}=1$ for all $n\in\mathbb{Z}$? The Euler formula you quoted shows that the exponential function, as a complex function, is periodic. Namely, it is non-injective, or $e^z=e^w$ does not imply $z=w$.

Moya
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There's nothing suspicious if you remember that $$e^{i\pi} = -1 \iff -(e^{i\pi}) = 1$$ and being clear that this is not to say $$(-e)^{i\pi} = 1$$

So there is nothing wrong with $$-(e^{i\pi}) = e^{0}.$$ But that is not to say that $i\pi = 0$.

amWhy
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