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?
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4$\begingroup$ No. $e^z$ is periodic with period $2\pi i$. $$e^0=e^{2\pi i}=e^{\pi i}\cdot e^{\pi i}=-e^{\pi i}.$$ $\endgroup$– Jyrki LahtonenMar 20, 2015 at 14:23
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$\begingroup$ 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 $\mathbb{C} \to \mathbb{C}$. $\endgroup$– anomalyMar 20, 2015 at 14:23
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1$\begingroup$ Would you conclude from $-(-1)^2=(-1)^3$ that $2=3$? $\endgroup$– egregMar 20, 2015 at 14:27
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1$\begingroup$ @Chou, what are you getting upset about? egreg is trying to help. (Unless comments have been deleted...) $\endgroup$– TonyKMar 20, 2015 at 14:37
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2$\begingroup$ @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. $\endgroup$– egregMar 20, 2015 at 16:06
2 Answers
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 $z=w$.
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 $i\pi = 0$.