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Referring to precisely the comments on the original question posted here, one user says,

I suppose you are talking about exponentiation for complex numbers, rather than for reals. Also, isn't it the case that $a^b$ is not unique for any $b$ that is not an integer (assuming we're working with complex numbers)?

And it seems like the resolution was that this is true for complex $a$ and $b$, but I didn't think this was correct. I thought exponentiation defined for a base $a \in \mathbb{C}$ and exponent $b \in \mathbb{C}$, that $a^b$ must be unique. Am I misinterpreting the discussion or is there something that I'm missing?

rb612
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2 Answers2

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$a^b$ is defined as $\exp(b \log (a))$ for any branch of the logarithm. The complex logarithm is multi-valued: if $L$ is one value, then $L+2\pi i n$ is a value for any $n$. Unless $b$ is an integer, this makes $a^b$ multivalued as well.

To single out a particular value, one sometimes takes the principal branch where $-\pi < \text{Im}(\log(a)) \le \pi$ (and this is almost always done when $a$ is a positive real).

Robert Israel
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  • Thank you! Although I'm clear on how if $L$ is a value then $L+2\pi i n$ is a value, but I only see this as a consequence of complex exponentiation not being injective rather than it being multivalued. What exactly makes $a^b$ multivalued also? – rb612 May 28 '18 at 07:38
  • Just a clarification on notation. Isn't $\mathrm{Im}(\log (a))$ real? – ty. May 28 '18 at 08:05
  • @rb612 Stay in the positive reals for a moment. Calculate $a^b$ by taking the log of $a$, multiplying it by $b$ and taking the antilog. Nice and simple and you get a single answer. Now, move to the complex numbers and try. Here, you will get multiple values for the log. If $b$ is an integer then after the antilog, these will all yield the same result. If $b$ is rational, say $\frac{p}{q}$ in simplest terms, then you will find $q$ many different results. If $b$ is not rational then you find that all of the (countably) infinitely many logarithms give different values. – badjohn May 28 '18 at 08:34
  • @ty Thanks, edited. – Robert Israel May 28 '18 at 18:01
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Because $a^b$ is easy to write, it is tempting to assume that it must be easy to define in some sensible and useful way. However, there is no reason why this must be so. $\frac{1}{0}$ is also easy to write but not easy to define in a useful way. Division has a fairly simple restriction, the restrictions on exponentiation are more complicated. This link in one of the answers to the question that you quote describes the situation quire well. https://en.m.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers

badjohn
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