How do I formally define complex exponentiation? I'm trying to make basic complex analysis tools more concrete. That is, I'm trying to eliminate the term "multi-valued function" in my language.
For example, $\log$ can be viewed as a group homomorphism $(\mathbb{C}^*,•) \rightarrow (\mathbb{C}/2\pi i \mathbb{Z},+)$, so that $\log$ can be viewed as actually a function. Similarly, argument can also be viewed as a group homomorphism.
The problem is, I don't know how to view $z^w$ as a function so that arithmetic including this can be done while viewing this as a single object, not a multivalued function.
I was trying to fix $z$ and view $z^w$ as a function where $w$ ranges over $\mathbb{C}$ and make $z^w$ as a group homomorphism so that $z^{w+a}=z^w • z^a$ can be done, but I don't know how to quotient the range of $z^w$.
More precisely, let $f(w)=z^w$.
I wanted to view $f$ as a homomorphism from $(\mathbb{C},+)$ to some quotient $(\mathbb{C}/H, •)$, to make $f(w+a)=f(w)f(a)$. My question is what would be a natural choice of $H$? Or is this approach completely wrong? If so, what would be a nice viewpoint of exponentiation?
If there is a complex-analysis text introducing the theory in this way, please recommend me one. Thank you in advance :)
 A: The only such quotient of $\mathbb{C}^*$ is the trivial, one-element group (NB, we don't look for a quotient of $\mathbb{C}$, essentially because of the multiplicative behavior of $\exp$).
For a fixed base $z \in \mathbb{C}$ and any choice of branch $\log$ of the logarithm function, we can define a branch of the exponential function with that base by
$$f(w) := \exp(w \log z).$$
As usual, any other branch $\widetilde{\log}$ of the logarithm function can be written as
$$\widetilde{\log}(\zeta) := \log \zeta + 2 \pi i g(k)$$ for some integer-valued (and not necessarily continuous) function $g$.
The corresponding branch of the exponential function with base $z$ is
\begin{align}
\widetilde{f}(w)
&:= \exp(w \widetilde{\log} z)\\
& = \exp(w (\log z + 2 \pi i g(w)))\\
& = \exp(w \log z) \exp(2 \pi i g(w) w)\\
& = f(w) \exp(2 \pi i g(w) w) \textrm{.}
\end{align}
We can now formalize the question: We are looking for a subgroup $\Gamma \subset (\mathbb{C}^*, •)$ such that the composition
$$\mathbb{C} \stackrel{f}{\to} \mathbb{C}^* \to \mathbb{C}^* / \Gamma$$
is independent of the choice of branch $f$ of the exponential function with the base $b$, or equivalently, of the choice of branch $\log$ of the logarithm. (The second map in the composition is just the canonical quotient map.)
Now, by the above computation, $\Gamma$ must contain every possible value of $f(w)^{-1} \widetilde{f}(w)$, that is, it must contain
$$\exp(2 \pi i k w)$$
for every value of $w$ and every integer $k$, but the set of such values is the image of $w \mapsto \exp(2 \pi i w)$, which is $\mathbb{C}^*$ itself. Thus, the only admissible map $\mathbb{C}^* \to \mathbb{C}^* / \Gamma$ is the trivial map $\mathbb{C}^* \to \{ 1 \}$ itself.
Of course, there is still a way to define complex exponentiation formally---this is precisely the definition of $f(w)$ in the first display equation above.
A: For variety, we could use a Riemann surface: the set of all points $(z,w)$ such that $\exp(w) = z$.
The points of this surface should be thought of as follows: the first coordinate expresses a point on the complex plane, and the second coordinate expresses additional information: a choice of the branch of the logarithm.
Let $\hat{z} = (z,w)$. The multi-valued function $\log z$ on the complex plane lifts to a well-defined, continuous, and differentiable function $\log \hat{z}$ on the Riemann surface above, given by $\log \hat{z} = w$, and this function satisfies $\exp(\log \hat{z}) = z$.
Furthermore, we can define $\hat{z}^a = \exp(a \log \hat{z})$ to get a well-defined exponentiation function everywhere defined on the Riemann surface.
We could actually make $\exp$ take values on your Riemann surface, so that $E(a) = (\exp(a), a)$. Then $\log$ and $E$ are inverse functions. The corresponding version of complex exponentiation is $\hat{z}^a = (\exp(aw), aw)$.
