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I just learned that $i^i \approx 0.20788$ is transcendental. Can anyone detail a proof?

Mini_me
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Joseph O'Rourke
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    By Gelfond-Schneider theorem... – Wojowu Aug 30 '17 at 11:58
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    The definition of $i^i$ however is not unique. It depends on the branch we use for the logarithm. But the gelfond-schneider-theorem guarantees that all those numbers are transcendental. The number chosen in the question can also be expressed by $\large e^{-\pi/2}$ – Peter Aug 30 '17 at 12:04

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As $i$ is algebraic ($\neq0,1$) and $i$ is irrational, a deepish theorem due to Gelfond and Schneider implies that $i^i$ is transcendental.

Jyrki Lahtonen
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  • Whether $i$ is irrational depends on your definition. See, e.g., this question (link). – user236182 Aug 30 '17 at 12:12
  • Sorry this was unnecessarily complicated at first. The reason was that I initially started tracing it back to the transcendence of Gelfond's constant $e^\pi$. For some reason I posted a version where G-S was applied to $(-1)^{i/2}$. – Jyrki Lahtonen Aug 30 '17 at 12:13
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    The definition used for Gelfond-Schneider: irrational means, not a quotient $a/b$ where $a,b \in \mathbb Z$, $b \ne 0$. So in particular $i/2$ is irrational. Makes sense for complex numbers, and is the standard definition. Makes sense for $p$-adic numbers, too. For example, if $\sqrt{-1}$ exists in a $p$-adic field, then its $p$-adic expansion is not eventually periodic. – GEdgar Aug 30 '17 at 12:13
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    @user236182 Whether Gelfond-Schneider theorem applies to the exponent $i$ most emphatically does not depend on your definition of an irrational number. – Jyrki Lahtonen Aug 30 '17 at 12:19
  • Though I agree that in this case for Gelfond-Schneider theorem $i$ is indeed irrational. – user236182 Aug 30 '17 at 12:20
  • Just to make this self-contained, the Gelfond-Schneider theorem is: If $a$ and $b$ are algebraic with $a \neq 0,1$ and $b$ irrational, then $a^b$ is a transcendental. – Joseph O'Rourke Aug 30 '17 at 13:18