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We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$

Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?

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marked as duplicate by 23rd, Brandon Carter, Micah, Amzoti, Asaf Karagila May 21 '13 at 8:19

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17  
The exponential map on the complex plane is not injective. You're right to say that both $e^{2\pi i}=1$ and $e^0=1$, but without injectivity, we cannot conclude that $2\pi i=0$. –  Jared May 21 '13 at 1:16
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It's the same as why $-1\neq 1$, despite the fact that $(-1)^2=1^2$. –  Glen O May 21 '13 at 1:17
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I'm 70 inches tall and so is Joe Smith. Since we're both the same height, I must be Joe Smith, right? No. Just because $f(x) = f(y)$ doesn't mean that $x=y$. –  MJD May 21 '13 at 1:19
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Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm. –  mathguy May 21 '13 at 1:27
    
sin(0) = sin(2pi) so 0 = 2pi, how does that work? –  imranfat May 21 '13 at 3:24

4 Answers 4

up vote 26 down vote accepted

You have shown that $e^{2\pi i} = e^0$. This does not imply $2\pi i = 0$, because $e^z$ is not injective. You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

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1  
I think the thing about real functions may be a red herring. As Glen O pointed out in comments, there are plenty of noninjective functions. For example, $(-1)^2 = 1^2$, and yet $-1\ne 1$. Or even $3\cdot 0 = 4\cdot 0$, and yet $3\ne 4$. –  MJD May 21 '13 at 1:20
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Sure, but I don't think that's where the OP is confused. I think they understand that $x^2$ and $0$ are noninjective functions, but they assume without thinking about it that $e^x$ is injective, because in their experience (with real $e^x$) it is. –  Goos May 21 '13 at 1:24
    
+1 for understanding what was really confusing OP. –  Raskolnikov May 21 '13 at 8:06

The $\log$ function is multi-valued on $\mathbb{C}^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because $$e^{2\pi i}=e^0$$ does not imply that $2\pi i=0$.

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I did not figure such a big change occurs when moving over to the complex plane and figured it would be as intuitive as with real numbers. Thank you (and other answerers/commenters). –  mathguy May 21 '13 at 1:24

It's like saying that, because $\sin{\pi} = \sin{0}$, that $\pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^{2\pi i} = e^0 \implies \ln{e^{2\pi i}} = \ln{e^0} \implies 2 \pi i = 0$. The error here is that $\forall x \in \mathbb{C} \ \ln{e^x} = x$ is not true! $ln$ isn't even a function, just like the naive version of $\arcsin(x)$. You have to make a choice of range, which is usually $\Im(x) \in (-\pi, \pi]$.

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Actually, the following is true:

$ e^{2n\pi i} = 1  \forall  n \in \mathbb{Z} $

As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.

You could as well assume that $2\pi = 0$ because of

$\sin 2\pi = \sin 0 \land \cos 2\pi = \cos 0$.

As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.

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