12
$\begingroup$

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 :)

$\endgroup$
2
  • $\begingroup$ While you may define $\log(z)$ in that way, other arithmetic operations might not want to work with $(\mathbb C/2\pi i \mathbb Z,+)$. What would you do with, say, $\log(z^2)/\log(z)$? $\endgroup$ Dec 25, 2014 at 1:14
  • $\begingroup$ Note that $\exp: \mathbb{C} \to \mathbb{C} - \{ 0 \}$ descends to a group isomorphism $\mathbb{C} / \ker \exp = \mathbb{C} / 2 \pi i \mathbb{Z} \to \mathbb{C} - \{ 0 \}$, and by construction $\log$ (as regarded above) is just the inverse of this; in particular, it too is an isomorphism. $\endgroup$ Dec 25, 2014 at 1:55

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\hat{z}^a = (\exp(aw), aw)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .