Do any issues arise if we try to raise an element of $\mathbb{R}^+$ to an element of $\mathbb{C}$?

Question

If $a$ and $b$ are non-zero natural numbers, the definition of $a^b$ is clear. Now it seems to me that there are (at least) two distinct ways of generalizing to larger number systems.

Firstly, given $n \in \mathbb{N}$ and $z \in \mathbb{C},$ we can define $$z^n = \prod_{k=1}^n z$$

where the empty product is interpreted as $1 \in \mathbb{C}$. Furthermore, if $z \neq 0,$ then we can define $z^n$ for all integral $n$ via either of the following definitions.

1. For all integers $n>0$ we define $z^{-n}=(1/z)^n$.
2. For all integers $n>0$ we define $z^{-n}=1/(z^n)$.

Both definitions give the same result, and all our favorite identities hold under these definitions. Thus: so far, so good!

Now suppose we want more general exponents. E.g. Suppose we want to define $x^y$ for $y$ an arbitrary element of $\mathbb{R}$. To generalize in this new direction, we need to specialize in another. In particular, we require that $x$ be an element of $\mathbb{R}^+$. Then we may define $$x^y = \exp [y \log(x)]$$

My question is, are there any nasty surprises if we allow $y$ to be an arbitrary complex number in the above definition? That is, do any issues arise if we try to raise an element of $\mathbb{R}^+$ to an element of $\mathbb{C}$?

EDIT. I've just been toying with the numbers a bit, and it seems that the formula $x^{wz}=(x^w)^z$ fails for certain combinations of $w,z \in \mathbb{C}$. As a simple example, take $x=e$ and $w=i\pi,$ in which case $x^w = -1 \notin \mathbb{R}^+,$ and so $(x^w)^z$ is undefined (unless $z$ is integral).

For a more interesting example, take $x=e$, $w=2\pi i$ and $z=1/2$. Then $x^w=1,$ so $(x^w)^z=1$. But $wz=\pi i$, and so $e^{wz}=-1$. So $x^{wz}\neq (x^w)^z$, although both expressions are well-defined.

This begs the question, when is the identity $x^{wz}=(x^w)^z$ valid? Its easy to show that if $w$ is real, then the identity holds, since under these circumstances, $wz\log(x)=z\log(w^x).$ When else does it hold? I think that if $z$ is integral it will also work out, but I don't know how to show this.

Answer 1

Your definition of $x^z$ for real, positive $x$ and complex $z$ is consistent and natural. It works: we get $(xy)^z = x^zy^z$, $(1/x)^z = x^{-z}$ etc.

But when you try to iterate this function, as in $(x^w)^z$, you are stepping outside the domain of definition of your function, because $(x^w)$ is no longer a real, positive number. So you can't expect it to work so nicely.

Answer 2

In short, no. You're probably familiar with the famous $e^{i\pi}=-1$ or even $e^{i\theta}=\cos \theta + i \sin \theta$. Now we have $e^{a+bi}=e^a e^{bi}$ so finally we have:

$$x^{a+bi}=(e^{\ln x})^{a+bi}=e^{a\ln x}e^{i b \ln x}=e^{a \ln x}(\cos(b\ln x)+i\sin(b\ln x))$$