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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?

Caleb Stanford
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Sidd Singal
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    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 01:16
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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 01:17
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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 01:19
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    Thanks everyone. I just did not realize that logarithm functions were not injective in the complex realm. – Sidd Singal May 21 '13 at 01:27
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    sin(0) = sin(2pi) so 0 = 2pi, how does that work? – imranfat May 21 '13 at 03:24
  • @imranfat: If $2\pi i=0$ then certainly $2\pi=0$, you just multiply by $-i$. :-) – Asaf Karagila May 21 '13 at 08:20
  • This really isn't a duplicate of the question it is linked to. They are about two separate issues with complex exponentation (this one how $e^z$ is not injective, and that one how raising to non-integer powers is not single-valued). The issues are related, but the relationship is not obvious and certainly seeing an answer to one of them does not make the answer to the other obvious for someone who is new to these ideas. – Eric Wofsey Aug 05 '18 at 19:18

4 Answers4

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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 $z$.

Caleb Stanford
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    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 01:20
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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. – Caleb Stanford May 21 '13 at 01:24
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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 $\Im(x) \in (-\pi, \pi]$.

Joe
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Henry Swanson
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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 $2\pi i=0$.

Zev Chonoles
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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). – Sidd Singal May 21 '13 at 01:24
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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.

Alfe
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