Is this the same as $e^x$?

Question

I was wondering if this is a correct way to apply "exponent rules",

$$e^{x}=\left(e^{2\pi i}\right)^{x/2\pi i}=1^{-ix/2\pi}$$

I thought it was safe, just found it a bit weird, but I wanted to check if I could plot this in Wolfram Alpha and get the usual graph for $e^{x}$ (WA allows to plot both the real and imaginary parts of a real, complex valued function).

To my surprise the result appears to be 1.

So, did I make a mistake in applying these rules? or is Wolfram Alpha making a mistake?

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EDIT: Here is an image (in case I'm right and it gets modified in the future).

Answer 1

The problem here is that the rule $$(a^m)^n=a^{mn},$$ which is true for positive $a\ne 1$ and real $m,n$ fails to be true when $m,n$ are nontrivially complex (that is, complex numbers with nonzero imaginary parts). The reason for this is that there are infinitely many ways to represent the same complex number in exponential form, because of the periodicity of the trigonometric functions. That is, we have that for any complex $z=re^{i\phi},$ we also have $$re^{i\phi}=re^{i\phi+2πki},$$ where $k$ can be any integer.

To see why this causes a problem for your example, we must have $$e^x=(e^{2πi})^{x/2πi}=(e^{2πki})^{x/2πi}=e^{kx},$$ which is obviously false.

So, again, the problem is with assuming that $(a^m)^n=a^{mn}$ is valid here. Thus, what we have is that $$e^x\ne (e^{2πi})^{x/2πi}.$$

Answer 2

Yes this is a confusing thing. The property $(x^{a})^b=x^{ab}$ holds when $x$ is a positive real number and $a$ and $b$ are real numbers. More generally this property fails. This leads to many fake "proofs" like \begin{align} 1&=1^{1/2}\\ &=((-1)^2)^{1/2}\\ &=(-1)^{2/2}\\ &=\left(\left(-1\right)^{1/2}\right)^2\\ &=i^2\\ &=-1. \end{align}

With complex exponentiation, you have to be extra careful because expressions like $z^{1/n}$ are usually not well-defined. When you switch to complex numbers, for example, $16$ suddenly has $4$ fourth roots. So by convention, $\sqrt[4]{16}=2$, but the equation $z^4=16$ has the set of solutions $\{2,-2,2i,-2i\}$.

Also see the responses here.