Two contradictory ways to calculate $(e^{2\pi i})^i$?

Question

As we know:

$$e^{i2π} = 1$$

so here's the first way that we can calculate the expression in the title:

$$(e^{i2π})^i = 1^i = 1$$

however, if before we simplify $e^{i2π}$ to $1$ we multiply the powers which we're allowed to do, we get a different result: $$(e^{i2π})^i = e^{ii2π} = e^{-2π} ≈ 0.18 $$

What's the explanation to this?

Answer 1

Complex number exponentiation doesn't follow the rules for rational exponentiation. It is true that: $$(\forall a,b,c\in\mathbb{R}, a> 0):(a^b)^c = a^{bc}$$

However, things change for $\mathbb{C}$: $$a,b,c\in\mathbb{C}, (a^b)^c\:\text{might not equal } a^{bc}$$

But why?

$\log(w^z) = z\log(w)$ doesn't hold for all $w,z\in\mathbb{C}$, regardless of how we treat $\log$. Very relevant information regarding this topic can be found here. That being said, going with your first method is the correct choice in this case.

Answer 2

You are assuming that$$(\forall a,b\in\mathbb{C}):(e^a)^b=e^{ab}.\tag{1}$$However, what is the meaning of $(e^a)^b$? Since $a$ is an arbitrary complex number, $e^a$ is an arbitrary non-zero complex number. So, what is $z^w$ when $z,w\in\mathbb C$ and $z\neq0$? It is any number of the form $e^{w\log z}$, where $\log z$ is some logarithm of $z$. So, the equality $(1)$ must be used with a lot of precautions.