If I understand correctly, $z^w$ for $z,w \in \mathbb{C}$ is a shorthand for $\exp(w\log(z))$. The problem is that every non-zero complex number has infinitely many logarithms. Say $\exp(y)=z$. Then, so is $\exp(y\pm2i\pi)$, $\exp(y\pm4i\pi)$, etc. Therefore, $z^w$ is a mutli-valued function. This is usually fixed by taking a branch cut, so that we require $\Im\left({\log(z)}\right) \in (-\pi,\pi]$, for instance. However, say we don't choose a branch of the logarithm. Is $z^w$ well-defined? Can I, for instance, write that $$ i^i=\{x \mid x=e^{-2 \pi n - \pi/2},n\in\mathbb{Z}\} \, ? $$ Another concern I have is about the notation $e^z$ being used in place of $\exp(z)$. I find this notation confusing, since $\exp$ denotes a single-valued function, whereas $e^z$ does not. If $z^w=\exp(w\log(z))$, then doesn't $$ e^z=\exp(w\log(e)) \, , $$ meaning that $e^z$ has infinitely many values?
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$\begingroup$ Mathematicians normally assume $e^z$ is short for $\exp z$, thereby taking the $\log e=1$ branch. Others are disinteresting, or we wouldn't be using the base $e$. $\endgroup$– J.G.Commented Jan 1, 2021 at 14:20
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$\begingroup$ @J.G. Thank you, this does seem to clear up my confusion with $e^z$ versus $\exp(z)$. Does $z^w$ even make sense if we don't specify what branch of the logarithm we are using? $\endgroup$– JoeCommented Jan 1, 2021 at 14:23
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1$\begingroup$ It does need to be specified, at least by context (a lot of complex analysis boils down to carefully considering whether a contour jumps branches) or convention. $\endgroup$– J.G.Commented Jan 1, 2021 at 15:04
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$\begingroup$ @J.G. In your experience, do people ever define $z^w$ as a multifunction? For example, do you ever see $$z^w=\{x : x = \exp(wr), \text{ where $r$ is a logarithm of $z$}\} \, ?$$And would this definition be useful? $\endgroup$– JoeCommented Jan 2, 2021 at 16:47
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$\begingroup$ @J.G. If you have the time to respond to my query, then I would be very grateful. But thanks anyway. $\endgroup$– JoeCommented Jan 5, 2021 at 15:40
1 Answer
You're right that every non-zero complex number has infinitely many logarithms, but it is possible that exponentiation will make all these values equal. For instance, let $ z= r e^{i \theta},$ then the possible values of $\log(z)$ are $ \log(r) + i \theta + 2 \pi i k$ for $k \in \mathbb{Z}.$ Now you would like $\exp(w\log z) $ to be a well-defined function on $\mathbb{C}$ i.e. you want all the possible values of the logarithm of $z$ to give the same value of $w^z.$ Writing this out, you get:
$$ \exp(w(\log r + i\theta + 2\pi i k )) = \exp ( w \log r + i\theta) \text{ for all } k \iff $$ $$ \exp( w \cdot 2\pi i k) = 1 \text{ for all } k \iff w \cdot 2\pi i k \in 2 \pi i \mathbb{Z} \text{ for all }k \iff w \in \mathbb{Z}.$$ Thus only exponentiating by integers is well defined without taking a branch cut, for example: $z^{1} $ is a well-defined analytic function.
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$\begingroup$ Thank you for this answer. I didn't know that only exponentiating by integers is well-defined (unless you take a branch cut). However, I'm still unsure about this: is it correct to write $$i^i=\{x \mid x=e^{-2 \pi n - \pi/2},n\in\mathbb{Z}\} \, ?$$ $\endgroup$– JoeCommented Jan 1, 2021 at 16:22
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$\begingroup$ I would take that to mean the set of all the possible values of $i^i$ defined by choosing a branch of the logarithm. In practice, however, you define a branch of $z^i$, then look at the value of this branch at $i,$ which is one of the values in your set. You can, however, define a "multi-valued" map: $\mathbb{C} \to \mathcal{P}(\mathbb{C}), z \mapsto$ $\{$ set of possible values...$\},$ but I have absolutely no idea how to deal with such object: What does mean for it to be continuous, analytic, etc? But we can answer these questions if we choose a branch and work with normal maps. $\endgroup$ Commented Jan 1, 2021 at 17:05