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Fermat's Library tweeted five hours ago that $e^\pi$ can be proved to be transcendental as follows. Let $A=-1$ and $B=-i$. Then $A$ and $B$ are algebraic numbers, $A\ne0,1$ and $B$ is not rational. Therefore, by Gelfond-Schneider theorem, $$A^B=e^\pi=(e^{i\pi})^{-i}=(-1)^{-i}\tag{1}$$ is a transcendental number.

The same equality $(1)$ is also mentioned in Wikipedia, but I am confused because an answer on this site stated that the rule $x^{yz}=(x^y)^z$ is valid only when $x>0$ and $y$ is real. Does the rule really apply here? Is the proof in the aforementioned tweet valid?

Ramen Nii-chan
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    The usual approach is to work with $i^i$ which is , if we use the proper branch , $e^{-\frac{\pi}{2}}$. Since this is transcendental, $e^\pi$ is transcedental as well. We can use Gelfond-Schneider here because we get a transcendental number for every possible branch. So, we can be sure that $e^{\pi}$ is actually transcendental. – Peter Dec 16 '19 at 20:06
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    Look at the explanation here Not everyone agrees with the position taken in the answer you cite. I well remember that an early exercise in Lars Ahlfors' complex analysis book is "Show that there are infinitely many values of $i^i$ and they are all real and positive." Ahlfors was a Fields medalist, specializing in complex analysis. – saulspatz Dec 16 '19 at 20:07
  • See Gelfond's Constant, mentioned by saulspatz. Wikipedia is (perhaps) a better reference than twitter. – Dietrich Burde Dec 16 '19 at 20:09
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    @saulspatz Thanks. "...then $\color{red}{\text{any}}$ value of $a^b$ is a transcendental number". This is much better than the over-simplified statement in the tweet! – Ramen Nii-chan Dec 16 '19 at 20:15
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    Better than "any value of" would be "every value of" or "all values of … are", in my opinion. – Daniel Fischer Dec 16 '19 at 20:17

1 Answers1

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Just because an identity does not hold generically, does not mean that there are no coincidences where it seems to work anyway.

Having chosen a branch of the logarithm, we may write the complex exponential $$ (-1)^{-\mathrm{i}} = \mathrm{e}^{-\mathrm{i} \ln(-1)} $$ and proceed from there.

If we do not wish to choose a branch of the logarithm, we instead write $$ (-1)^{-\mathrm{i}} = \mathrm{e}^{-\mathrm{i} (\ln(-1) + 2 \pi \mathrm{i} k)} $$ for some choice of integer $k$ (which choice is equivalent to a choice of branch). Then we can evaluate $\ln(-1)$ on any branch, for instance $\ln(-1) = \mathrm{i} \pi$. And finally, \begin{align*} -\mathrm{i} (\ln(-1) + 2 \pi \mathrm{i} k) &= -\mathrm{i}(\mathrm{i} \pi + 2\pi \mathrm{i}k) \\ &= \pi + 2 \pi k \text{.} \end{align*} So $$ (-1)^{-\mathrm{i}} = \mathrm{e}^{\pi + 2 \pi k} \text{,} $$ where the choices of $k$ correspond to different choices of branch of logarithm in defining the complex exponential. If we choose the branch with $k = 0$, we recover the claimed identity.

Eric Towers
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