# Why isn't $e^n$ equal to 1?

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.

• 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. Commented Apr 8, 2019 at 3:40
• @Milo Brandt I know Euler's formula, but that's about it... what are you referring to? Commented Apr 8, 2019 at 3:43
• 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. Commented Apr 8, 2019 at 3:50
• Commented Apr 8, 2019 at 5:01

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 $$(-1)^{2\cdot(1/2)} = (-1)^1 = -1 = 1.$$
• Also worth noting: It holds when $x$ is a non-negative real number and when $b$ and $c$ are both real. Commented Apr 8, 2019 at 3:35