Wikipedia states that an immediate corollary of the Gelfond Schneider theorem is that $i^i$ is a transcendental number. To me it not so obvious, because it first has to be shown that i is both algebraic and irrational. I have not seen this proved anywhere. Wikipedia only offers a token one line "proof" stating:
$i^i = \left( e^{(i\pi / 2)} \right)^i = e^{(-\pi / 2)} = 0.2078….$
I can see that $i = \left( e^{(i\pi / 2)} \right)$ is just a rearrangement of Euler's identity $e^{(i\pi)} = -1$, but I don't see how $ e^{(-\pi / 2)} $ proves $i^i$ is transcendental because it is basically stating that $i^i$ can be rearranged into the form $a^b$ where both a and b are transcendental, so $i^i$ must be transcendental because $a^b$ is transcendental, but I do not think that is necessarily the case.
Euler's identity actually provides a counter example to $a^b$ of the above form being transcendental, since $e^{(i\pi)}$ is a transcendental raised to the power of a transcendental and the result is algebraic (-1).
As an aside, $(i\pi)$ must be transcendental because if it were not, $e^{(i\pi)}$ would be transcendental (which is not the case) since e raised to the power of any algebraic results in a transcendental number.
The other issue is that algebraic numbers are normally considered to belong to the set of real numbers and I obviously is not in that set.
So is there any supporting evidence that i is algebraic and irrational, so $i^i$ must be transcendental?
Best Answer
A rational number is a (necessarily real) number that can be expressed as the quotient of two integers. $i$ is not real, so it is not rational.
An algebraic number is a complex number which is a root of a polynomial with rational coefficients. $i$ is algebraic because it is a root of the polynomial $x^2+1$, which has rational coefficients.
Your assertion that algebraic numbers "are normally considered to belong to the set of real numbers" is just not the case. I would request that you provide any evidence for this assertion; I am not aware of any standard definition that would restrict "algebraic number" to only real numbers. It just isn't true. While some folks may say that a real number is algebraic if it satisfies a polynomial with rational coefficients, that is not a definition of "algebraic number"; it is a definition of real algebraic number.
$i^i$ is therefore transcendental by the Gelfond-Schneider Theorem; that theorem states that if $a$ is an algebraic number different from $0$ and from $1$, and $b$ is an algebraic number that is not rational, then $a^b$ is transcendental. Since $i\neq 0$, $i\neq 1$, and $i$ is algebraic, and $i$ is not rational, it follows that $i^i$ is transcendental. No need for "supporting evidence": you have a theorem that tells you so.
$e^{i\pi}=-1$ has absolutely no bearing on the Gelfand-Schneider Theorem, as it does not satisfy the hypotheses: the base, $e$ is not algebraic; and the exponent, $i\pi$, is also not algebraic (both $e$ and $\pi$ are transcendental). The Gelfond-Schneider Theorem is silent about powers that do not satisfy the hypotheses of the theorem, just like every theorem is silent about situations that do not meet the hypotheses of that theorem.