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What is $0^{i+x}$, where $x$ is negative and $i$ is the imaginary unit?

I saw that Wolfram Alpha says its “complex infinity”, something it often gives when the answer is undefined. Is it undefined and if so, why? (Notably it also says that if $x$ is positive then $0^{i+x}=0$)

Gabriel Tellez
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  • Well $z$ approaches $\infty$ if and only if $1/z$ approaches $0$ so if they're claiming it's undefined and complex infinity I'd assume you'll looking at a division by zero situation. – CyclotomicField Nov 13 '23 at 13:28
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    "What is...?" questions always beg the question of, how is "..." defined? Maybe we're aliens. Explain to us what the definition of $0^{i+x}$ is, for you, before expecting an answer. One very valid definition of exponentiation simply omits zero from the domain, so... – FShrike Nov 13 '23 at 13:28
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    Do you know about the Reimann Sphere? Presumably the question is interpreted as $\lim_{z\to 0} z^{x+i}$, where the limit is taken on the sphere – Guy Nov 13 '23 at 13:29
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    $0^x=0$ for positive real $x$. $0^0=1$ by convention as it makes sense in combinatorics and generating functions and probability. $0^x$ is usually not defined for negative real $x$ as there is no plausible finite result. $0^i$ is usually not defined and similarly $0^z$ for complex $z$ with non-zero imaginary part. – Henry Nov 13 '23 at 13:33
  • @Henry It is worth pointing out, however, that there are circumstances in which the definition $0^0 = 0$ makes sense. It is question about whether you are interested in $\lim_{x\to 0} 0^x$ or $\lim_{x\to 0} x^0$. There are contexts in which the former is the more relevant limit (e.g. my advisor's work on fractal zeta functions, and my own phd thesis in the same area). – Xander Henderson Nov 13 '23 at 14:11
  • It depends on your definition, and the usual definition leaves this undefined. In particular, for complex powers the definition is usually $z^w = \exp(w \text{Log}(z))$ where $\text{Log}$ is the principal branch of the complex logarith, which is usually not defined at 0. – preferred_anon Nov 13 '23 at 14:17
  • @XanderHenderson I hope you had $\lim_{x\to 0^+} 0^x$ as $\lim_{x\to 0^-} 0^x$ would hit the same issue. On the other hand $\lim_{x\to 0^+} x^0 =\lim_{x\to 0^-} x^0$ happily, another argument for the convention $0^0$ unless defined otherwise. – Henry Nov 13 '23 at 14:18
  • @Henry Indeed. In both cases, $x > 0$. I'm was sloppy because I didn't want to type superscripts in my subscripts. Though, honestly, it was more like $\lim_{\Re(s) \searrow 0} 0^s$, where $s\in \mathbb{C}$. – Xander Henderson Nov 13 '23 at 14:21
  • @FShrike If I knew the definition then I would not have asked the question lol – Gabriel Tellez Nov 13 '23 at 18:25
  • @GabrielTellez My point was that the question isn't necessarily well posed. And that is not great, because it means that maybe there is no right answer. We can define it to be complex infinity, and give justification for that. Or, we can say: this is not defined. Which one is right? Well, .. both of them are? Questions about the value of some expression often fall into the trap of not really defining the expression itself, which makes questions about the value irrelevant. – FShrike Nov 13 '23 at 18:32
  • Well no one has given an answer – Gabriel Tellez Nov 13 '23 at 18:33
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    Guy's, Henry's and preferred_anon's comments all give great answers. If you want to understand the "complex infinity" answer, then Guy's comment should be helpful for you. It is formally correct that $\lim_{z\to 0}z^{x+i}$ is the point at infinity, if the limit is considered in the sphere $S^2=\Bbb CP^1$, for $x<0$. Regardless of the branch of the logarithm. So, that makes complex infinity a reasonable answer – FShrike Nov 13 '23 at 18:35

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