I have been proposed this enigma, but can't solve it. So here it is:

$$\begin{align} e^{2 \pi i n} &= 1 \quad \forall n \in \mathbb{N} && (\times e) \tag{0} \\ e^{2 \pi i n + 1} &= e &&(^{1 + 2 \pi i n})\ \text{(raising both sides to the $2\pi in+1$ power)} \tag{1} \\ e^{(2 \pi i n + 1)(2 \pi i n + 1)} &= e^{(2 \pi i n + 1)} = e &&(\text{because of (1)}) \tag{2} \\ e^{1 + 4 \pi i n - 4 \pi^2 n^2} &= e && (\div e) \tag{3} \\ e^{4 \pi i n - 4 \pi^2 n^2} &= 1 &&(n \rightarrow +\infty) \tag{4} \\ 0 &= 1 &&(?) \tag{5} \end{align}$$

So the question is: where is the error?

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    $\begingroup$ The other problem, which the three current answers do not mention, is that $\hspace{1.88 in}$ e-to-the does not have a limit at complex infinity. $\:$ $\endgroup$
    – user57159
    Sep 11, 2014 at 8:51
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    $\begingroup$ In all rigor, there is a step missing: from [1] you can establish $(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}$, which is indeed true; but as pointed by several answerers, this cannot be turned to [2] because of the properties of the complex power ($(a^b)^c$ may not be replaced by $a^{(bc)}$). $\endgroup$
    – user65203
    Sep 11, 2014 at 9:14
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    $\begingroup$ @Ricky: I am not sure that $4\pi in-4\pi^2n^2$ be a case of complex infinity, it has a well defined argument ($\pi$). In addition the exponential equals $e^{-4\pi^2n^2}$, and clearly tends to $0$. $\endgroup$
    – user65203
    Sep 11, 2014 at 10:58
  • $\begingroup$ @RickyDemer: Since $n\in\Bbb N$ (clearly stated) the limit $n\to\infty$ is just one over the natural numbers (and it converges); no complex infinity is involved. $\endgroup$ Sep 12, 2014 at 5:52
  • $\begingroup$ @MarcvanLeeuwen : $\;\;\;$ $4\pi in-4\pi^2n^2$ is a sequence of complex numbers $\hspace{1.89 in}$ whose absolute values converge to infinity. $\;\;\;\;\;\;\;$ $\endgroup$
    – user57159
    Sep 12, 2014 at 6:20

2 Answers 2


The problem is that the power rule

$$ (a^b)^c = a^{bc}$$

only holds when $a$ and $b$ are positive real numbers. In that derivation the crucially wrong step is

$$ (e^{2 \pi i n + 1})^{2 \pi i n + 1} = e^{(2\pi i n + 1)(2 \pi i n+1)}.$$

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    $\begingroup$ $b$ doesn't have to be positive if $a$ is (strictly) positive. $\endgroup$
    – TonyK
    Sep 11, 2014 at 9:42
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    $\begingroup$ Yes, of course, since in that case $a^b = (1/a)^{-b}$. Thanks for pointing it out. $\endgroup$
    – Zavosh
    Sep 11, 2014 at 9:43
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    $\begingroup$ "only holds" is just too strong a claim in general. One can come up with various counterexamples. But I think it gets the point across that one should never assume the property holds without reason, and the only version following from "basic algebra" rules learned in school is that $(a^b)^c=a^{bc}$ if $a>0$ and $b,c\in\mathbb R$ (even though irrational exponents are hardly understood at that level). Like Halmos said, "You are allowed to lie a little, but you must never mislead." $\endgroup$ Sep 11, 2014 at 13:04
  • $\begingroup$ @Jonas: I originally wrote "only holds in general...", but thought the extra words would be distracting. Anyway, I apologize if it sounds misleading. $\endgroup$
    – Zavosh
    Sep 11, 2014 at 15:10
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    $\begingroup$ Prometheus: No, I'm trying to say that you lied a little but didn't mislead :) Hence the "gets the point across". I like it. $\endgroup$ Sep 11, 2014 at 15:17

In complex numbers, $e^a=e^b$ does not imply that $a=b$. For instance, $e^{2\pi in+1}=e$ does not imply that $2\pi in+1=1$.

For the same reason, $\log e^a$ is not the same as $a$, and $(e^a)^a:=e^{(\log e^a)a}$ is not the same as $e^{a^2}$ (instead it is $e^{(a+2\pi ik)a}$, for some $k$).

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    $\begingroup$ Ah, and this $\log$ introduces a branch cut that effectively removes $2i\pi n$ from the exponent, which it is difficult notice when you just write it on the form $(e^{2i\pi n +1})^{2i\pi n +1}$. Nice explanation! $\endgroup$ Sep 11, 2014 at 21:24
  • $\begingroup$ Right, the $\log$ works "modulo" $2i\pi$ so that exponentiation behaves unlike the real counterpart. $\endgroup$
    – user65203
    Sep 11, 2014 at 23:49
  • $\begingroup$ @YvesDaoust I think your explanation is more exhaustive, but Prometheus' is simpler and more straightforward, hence I accepted that answer. Thanks $\endgroup$
    – astabada
    Sep 15, 2014 at 12:54
  • $\begingroup$ @astabada: thanks for the comment, you can't accept all. $\endgroup$
    – user65203
    Sep 15, 2014 at 13:08

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