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Is the following true for all complex $s$?

$$ 1^{-s} = 1 $$

My doubt arises from multivalued solutions to apparently simple complex equations such as $e^z=1$ where $z=2\pi i n$ for integer $n$.

I have tried the following:

$$ 1^{-s} = e^{-s \ln(1)} = e^0=1 $$

and can't see how multivalued solutions might enter into the logic.

Penelope
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1 Answers1

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Not necessarily. In general, the complex exponentiation $$z^w = e^{w\operatorname{Log}(z)}$$ requires a choice of branch for the complex logarithm.

For instance, if you choose a branch in which $\operatorname{Log}(1) = 2\pi i,$ then we'd have $$1^{-s} = e^{-2\pi i s}$$ which equals $1$ if and only if $s$ is a real integer.

Brian Moehring
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  • The original task was to find $\zeta(s)=\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\ldots$ as $\sigma \rightarrow +\infty$, where $s=\sigma+it$. The magnitude of each term $|n^{-s}|=n^{-\sigma}$ tends to 0, except for $n=1$, from which we have $1/1^s$ (using Tannery's theorem to justify swapping limit and sum operators). How does one choose a branch in this context, and what does it mean to choose a branch here? – Penelope Sep 24 '21 at 04:19
  • @Tariq I'll try to answer when I'm back at a keyboard. For future questions, though, try to give at least enough context to avoid the XY problem. – Brian Moehring Sep 24 '21 at 04:26
  • thanks Brian .. I find that long questions get ignored which is why I tried keeping this one short. Here's a long question https://math.stackexchange.com/questions/4252541/correct-application-of-tannerys-theorem-to-show-riemann-zeta-zetas-rightar – Penelope Sep 24 '21 at 04:43
  • @Tariq You don't need to make it a long question. All I needed to know is the first line of your comment, that you're trying to find the limit of $\zeta(s)=\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\ldots$ as $\Re{z} \to +\infty$. – Brian Moehring Sep 24 '21 at 05:16
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    @Tariq For the $\zeta$ function, I don't know if the branch is fixed by definition, but either way, here we'd want to choose a branch where the limit exists. This assumption forces $\operatorname{Log}(1) = 0$ (for instance, we can use the principal branch) and therefore $1^{-s} = 1$ for all complex $s$. – Brian Moehring Sep 24 '21 at 05:20
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    @Tariq For a little more discussion, see https://math.stackexchange.com/q/290808/694754 – Brian Moehring Sep 24 '21 at 05:21
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    Additionally, your claim that $|n^{-s}| = n^{-\sigma}$ in your first comment is false with some branches, so I have to assume the whole argument is done with the principal branch, where $n^{-s} = e^{-s\ln(n)}$ and $\ln$ is the natural logarithm you know from real analysis. – Brian Moehring Sep 24 '21 at 05:31
  • thanks I will read the suggested link. The comment that $|n^{-s}| = n^{-\sigma}$ is wrong unless the principal branch is chosen came as a surprise to me given so many textbooks and credibly authors repeat it without discussion branches. Is this because all complex functions by definition assume the principal branch unless otherwise stated? Apologies for the naive questions - I did not study maths to university and am self-teaching. – Penelope Sep 24 '21 at 12:38