Is $(-1)^x=(e^{\pi i})^x$

Question

In one of my questions here on mathematics about Graph of a negative exponential function One of the comments was that I could indeed have $(-1)^x$ since $(-1)^x = (e^{i\pi})^x = \cos(\pi x) + i\sin(\pi x)$. What I want to know is, is this identity $(-1)^x=(e^{\pi i})^x$ true? If yes, is there a proof for that?

Answer 1

It's true. First of all $(A)^x=(B)^x$ implies that the equality is true regardless the value of $x$ if $A=B$ .

You had just to prove that $e^{\pi i} = -1$ and it's the Euler's identity. She is bas on the definition of the exponential :

$e^{i x } = cos(x) + i sin(x)$

$cos(\pi) = -1$

and

$sin(\pi) = 0$

so $-1+i*0= -1 = e^{\pi i}$

Answer 2

There’s a problem, because you haven’t said what “$(-1)^x$” means.

Any definition of $a^b$ when $b$ is not an integer and $a$ is not positive real must pass through evaluation of $\log a$, and this is not a single-valued function unless $a$ is positive.

Even the “formula” $\log(e^x)$ must be considered to be ambiguous when $x$ is not real, even though it must be admitted that one particular evaluation of this expression is obvious.

EDIT:

To give an idea of how wrong your intended equation is, look at \begin{align} e^{-i\pi}&=-1=e^{i\pi}\\ \left(e^{-i\pi}\right)^{1/2}&=\left(e^{i\pi}\right)^{1/2}\quad\text{(ill-defined exponentiation)}\\ e^{-i\pi/2}&=e^{i\pi/2}\quad\text{(by your equation)}\\ -i&=i \end{align} It goes wrong at two places: in line 2, because the exponent is not an integer and the base is not positive; and in line 3, because your equation is not valid.