We know $e^{2\pi i} = 1$, and that $(x^m)^n = x^{mn}$. This way, we can rewrite $e^{n}$ as some version of $(e^{2\pi i})^{\frac{n}{2\pi i}}$ for most n (right?).

But if this is true, then why isn't $e^3 = 1$, for example, if we can rewrite it as $(e^{2\pi i})^{\frac{3}{2\pi i}} = (1)^{\frac{3}{2\pi i}} = 1$ ? What am I missing here?

I just came upon this issue by accident while doing a problem, and I'm not sure how to best resolve it.

  • $\begingroup$ Have you seen a definition of exponentiation for complex numbers? There's a lot of insight in learning about that definition that a more brief answer is likely to miss. $\endgroup$ Commented Apr 8, 2019 at 3:40
  • $\begingroup$ @Milo Brandt I know Euler's formula, but that's about it... what are you referring to? $\endgroup$ Commented Apr 8, 2019 at 3:43
  • $\begingroup$ Usually, if you have complex numbers $x$ and $y$, you define $$x^y=e^{y\log(x)}$$ where $e^z=\sum_{i=0}^{\infty}\frac{z^i}{i!}$ and $\log$ is the natural logarithm. The thing to understand about this is that it's not clear how to define $\log$ - for instance, is $\log(1)=0$ or is $\log(1)=2\pi i$ since $e^0=e^{2\pi i}=1$? This breaks most hope of things working nicely, but it sort of needs a deeper understanding of what's going on. $\endgroup$ Commented Apr 8, 2019 at 3:50
  • $\begingroup$ Similar: math.stackexchange.com/questions/281528/… $\endgroup$ Commented Apr 8, 2019 at 5:01

1 Answer 1


Good question! The answer is that although the rule ${(x^b)^c} = x^{bc}$ holds when $b$ and $c$ are integers, it does not hold in general when they are not integers.

Consider the following simpler example. As you know, $(-1)^2 = 1$. Raising both sides to the power $\frac12$, we get $${((-1)^2)}^{1/2} = 1^{1/2},$$ which is still correct, but we cannot then apply the ${(x^b)^c} = x^{bc}$ rule to the left side to obtain $$(-1)^{2\cdot(1/2)} = (-1)^1 = -1 = 1.$$

  • $\begingroup$ This is much better than my (hypothetical) answer, much more elementary; I was kind of going on this rant about complex numbers and multivalued functions lol. $\endgroup$ Commented Apr 8, 2019 at 3:35
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    $\begingroup$ Also worth noting: It holds when $x$ is a non-negative real number and when $b$ and $c$ are both real. $\endgroup$ Commented Apr 8, 2019 at 3:35

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