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?

Complex exponentiation doesn't obey the same rules as real exponentiation — Stahl, Aug 19 '17 at 22:39
What's the contradiction? $(e^{i2\pi})^i$ can be equal to both of those. Why do you think that is a contradiction? — fleablood, Aug 19 '17 at 22:41
you should make sure that you understand the definition of $e^{ix}$ as it is not really $e$ to the power of $ix$... this "notation" comes from the Taylor series associated with $e^x$ and with $sin(x)$ and with $cos(x)$ — Yanko, Aug 19 '17 at 22:41
"however, if before we simplify e^(i*2π) to 1 we multiply the powers which we're allowed to do" Allowed to do? According to PEMDAS, we do parenthesis first. In general, it does not hold for non-positive $a,b,c$ that$$(a^b)^c=a^{bc}$$You don't even need to use complex numbers to see that this is false. For example,$$1=((-1)^2)^{1/2}\ne(-1)^{2/2}=-1$$ — Simply Beautiful Art, Aug 19 '17 at 22:45
You can find here many, many other questions where the answer is the same: $(a^b)^c = a^{bc}$ fails for complex numbers. — GEdgar, Aug 20 '17 at 00:08

score 1 · Answer 1 · answered Aug 19 '17 at 22:45

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.

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

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.

edited Jun 12 '20 at 10:38

Community

1

answered Aug 19 '17 at 23:10

Mr. Xcoder

1,163

You probably mean $$\forall a\in\mathbb R,\ a>0$$ since otherwise the statement has trivial counterexamples. – Did Aug 20 '17 at 08:05
Again incorrect: one must assume $a>0$, not only $a\ne0$. – Did Aug 20 '17 at 12:29
@Did Sorry. Now it's fixed. Was in a rush when I edited and didn't take the time to think about it. – Mr. Xcoder Aug 20 '17 at 12:43

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

2 Answers2

But why?

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