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$ai^{bi}$ 1 immediately appears as an imaginary number: How could an imaginary to the power of yet another imaginary not result in an imaginary?

But, for example, $3i^{2i}≈0.12964$ (truncated).

How is it possible that an imaginary to the power of an imaginary is a real number?

Notes:
1: $a$ and $b$ have the domain $\{0∉ℝ\}$ 2
2: Did I format the domain correctly? I've never used set theory before.

CATboardBETA
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1 Answers1

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The number $i^{i}$ is real, which immediately implies the result you're interested in. "Proof" : the complex conjugate of a complex number is obtained by replacing every occurrence of $i$ with $-i$; so the complex conjugate of $i^{i}$ is $(-i)^{-i}=(-1/i)^{i}=i^{i}$; but the only way that complex conjugation can give back the same number is if the number is actually real.

Addendum: strictly speaking, raising imaginary numbers to imaginary powers is not well-defined. Consider the following. We have $i=e^{i\pi/2}=e^{5i\pi/2}.$ Hence $i^{i}=e^{-\pi/2}=e^{-5\pi/2},$ but these two real numbers are clearly different. This is a delicate issue, in that resolving this contradiction requires choosing a "branch" of the complex logarithm.

Will R
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  • Can you expand on this? I don't quite understand what half of what you just said means. – CATboardBETA Dec 06 '21 at 19:44
  • @CATboardBETA: Do you know what complex conjugation is? – Will R Dec 06 '21 at 19:45
  • No, looking at it right now... – CATboardBETA Dec 06 '21 at 19:46
  • The complex conjugate of a complex number $a+ib$ is simply $a-ib$. This turns out to be an incredibly useful and important tool. I suggest you begin by convincing yourself that the only way the complex conjugate can "have no effect" is that the complex number is actually a real number. – Will R Dec 06 '21 at 19:49
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    @WillR Exactly, so why should $(-i)^{(-i)}$ be the complex comjugate of $i^i$? – Filippo Dec 06 '21 at 19:50
  • As you already noticed, $i^i$ is not defined, so nothing can be said – Filippo Dec 06 '21 at 19:52
  • @Filippo: There is a reason I put the word proof in quotation marks. But I did actually justify that step. I learned this proof from a book written by Ian Stewart. – Will R Dec 06 '21 at 19:59
  • @WillR If it can be calculated, than how is it not defined? Some sort of Schrödinger's imaginary conjugate thing? – CATboardBETA Dec 06 '21 at 20:12
  • The "calculation" gives more than one sensible value. A proper definition should give only one value. – Will R Dec 06 '21 at 20:23
  • @Filippo: Although $i^{i}$ is not well-defined without making a choice beforehand, something can still be said. What can be said, and what this argument tries to say, is: however you choose to define $i^{i}$, it had better be a real number, otherwise you'll find contradictions. In other words, if $i^{i}$ exists, then it is real; whether it exists is another matter entirely. – Will R Dec 06 '21 at 23:40
  • @WillR Indeed, the real numbers are precisely the complex numbers that are equal to their conjugate. But I don't see any motivation for assuming that $(-i)^{-i}$ is the conjugate of $i^i$, which - as far as I understand - is what you do when you say "if $i^i$ exists, then it is real" – Filippo Dec 07 '21 at 07:30
  • @Filippo: I said it right there: the complex conjugate of a number is obtained by replacing all occurrences of $i$ with $-i.$ The reason the word "Proof" is in quotation marks is that this is brushing some analysis under the rug (conjugation is a reflection, so it's continuous under the Euclidean metric on $\mathbb{C}$, so this purely algebraic way of determining the complex conjugate of a polynomial extends to convergent power series, and hence to imaginary powers since $\exp$ is entire); but going into this much depth is clearly an inappropriate response for someone asking this question. – Will R Dec 07 '21 at 07:35
  • @WillR Thank you for the reply. Now I am satisfied. – Filippo Dec 07 '21 at 07:44
  • @CATboardBETA: Can I provide any more help? Does my answer make at least some sense to you? (This is one of those moments where a seemingly innocuous question is actually the entrance to a very deep rabbit hole, so it's perfectly alright if you don't understand my addendum, particularly my use of the word "branch".) – Will R Dec 07 '21 at 08:21
  • @WillR How does it work that there is more than one, aws you said, "branch" of the complex logarithm. What is the complex logarithm (I'm pretty much self-taught when it comes to things like imaginaries, apologies) – CATboardBETA Dec 07 '21 at 13:32
  • The function $\exp\colon\mathbb{R}\to\mathbb{R}{>0}$ is bijective, so it has an inverse function. That's the function $\log\colon\mathbb{R}{>0}\to\mathbb{R}.$ When we extend $\exp$ to complex numbers, the resulting complex function $\mathbb{C}\to{z\in\mathbb{C} : z\neq0}$ is no longer bijective. It's surjective, but it's not injective: $1=\exp(0),$ but $1=\exp(2\pi i)$ as well. So there is no inverse function, but we can try to fix this by restricting the range of $\exp$ to a portion of $\mathbb{C}$ where it is injective. Doing this in different ways, the inverses are "branches" of $\log$. – Will R Dec 07 '21 at 18:12
  • I will also say that this is not an issue that can really be cleared up in comments. I would advise that you have a good think, perhaps look for some explanations on YouTube or other online resources, and ask more questions on this site to clear-up any confusion you may have. – Will R Dec 07 '21 at 18:15
  • Sorry, I said "the range of $\exp$" when I meant to say "the domain of $\exp$". I hope this doesn't cause too much confusion. – Will R Dec 07 '21 at 18:52