Gelfond’s Constant

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USSR mathematician Aleksandr Gelfond lived from 1906 to 1968.
Gelfond’s constant, e^\pi, is named in his honor, because he proved that the number is transcendental.

A transcendental number is a number that can neither be the root of a polynomial algebraic equation, nor found by a geometric construction. Every transcedental number is also an irrational number, the reverse is not correct.

Transcedental numbers showed mathematicians for the first time that problems like squaring the circle or trisection of an angle isn’t possible with a geometric approach, because numbers like \pi (which is transcendental) and sin(x) and other trigonometric functions (giving transcedental answers) were involved.

Not only \pi, but e is a transcedental number as well. But what about e^\pi? Gelfond took it upon him to prove this.
To be precise, Gelfond proved the 7th problem of Hilbert. On the 1900 Internatinal Mathematical Congress David Hilbert formulated 23 mathematical problems he thought should be solved the coming century.
The 7th is:

if \alpha is an algebraic nummer, not equal to 0 or 1, and \beta is both an algebraic and not a rational number, will \alpha^\beta be a transcendental number?

Gelfond proved this in 1934, and as often happens, Theodor Schneider, a German mathematician, proved it in 1935. So the proof is officially named after both of them, the Gelfon-Schneider theorem.

Gelfon showed that e^\pi can also be written as (-1)^{-i}. You can easily see that

    \[e^\pi=e^{1\cdot\pi}=e^{-i^2\pi}=(e^{i\pi})^{-i}\]

And as Euler’s Identity states that e^{i\pi}+1=0 or e^{i\pi}=-1 it’s clear that

    \[e^\pi=(e^{i\pi})^{-i}=(-1)^{-i}\]

which fulfills the conditions, as -1 is not equal to either 0 or 1, and -i is not a rational number.

Another, complicated, way of coming around, starting from Euler’s Identity, is

    \[e^{i\pi}=-1=i^2\]

Now

    \[\ln e^{i\pi}=\ln(i^2)\]

so

    \[i\pi=\ln(i^2)\]

    \[\pi=\tfrac{1}{i}\ln(i^2)=\ln(i^{2\tfrac{1}{i}})\]

which means that

    \[e^\pi=(i^2)^{\tfrac{1}{i}}\]

As \tfrac{1}{i}=\tfrac{-i^2}{i}=-i it follows that

    \[e^\pi=(-1)^{-i}\]

As a bonus, you can pick up that \ln(i^2)=2\ln(i)=i\pi so \ln(i)=\tfrac{1}{2}\pi i.

But as we also know that e^{i\theta}=cos(\theta)+isin(\theta) we have to realize that strictly speaking

    \[\ln(i)=\tfrac{1}{2}\pi i\pmod {2\pi i}\]

but that won’t deter us.