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?

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    $\begingroup$ $i$ is not a rational number because it is not equal to a quotient of two integers. $i$ is algebraic because it is a root of the polynomial $x^2+1$ which has rational coefficients. Gelfond-Schneider's Theorem tells you that $i^i$ must be transcendental, since the base is not $0$, not $1$, and is algebraic; and the exponent is algebraic and not rational. It's a theorem. There's no need for "supproting evidence". $\endgroup$ Commented Oct 19, 2023 at 17:48
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    $\begingroup$ Why is this labeled "solution verfication"? You aren't talking about any proof. $\endgroup$ Commented Oct 19, 2023 at 17:54

1 Answer 1


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.

  • $\begingroup$ I know that technically all real numbers are complex numbers but I would still say "an algebraic number is a real number"... $\endgroup$
    – Nate
    Commented Oct 20, 2023 at 3:28
  • $\begingroup$ @Nate And then you would be completely, absolutely, totally, 100% wrong. Algebraic numbers are not restricted to being real numbers, regardless of what you decide to say or not say. Give one reputable reference that says algebraic numbers must be real. But don't try to hold your breath while you search for it. $\endgroup$ Commented Oct 20, 2023 at 4:02
  • $\begingroup$ ^I guess "real or complex number" then. Since most of them are real numbers that gives people a better sense of how they're defined. $\endgroup$
    – Nate
    Commented Oct 20, 2023 at 4:46
  • $\begingroup$ @Nate Where do you get "most of them are real numbers" from? That's also completely, totally, wrong. There are just as many real as purely imaginary as complex algebraic numbers, namely $\aleph_0$ of them. For starters, if $a\neq 0$ is a real algebraic number, then $ai$ is a purely imaginary algebraic number, and $a+ai$ is a complex, non-real, not purely imaginary algebraic number. You just don't know what you are talking about, so perhaps learn before saying so many incorrect things. $\endgroup$ Commented Oct 20, 2023 at 5:16
  • $\begingroup$ @Nate As to "how they are defined", it's simple and I already gave the definition. They are the roots of polynomials with rational coefficients. That is how they are defined. $\endgroup$ Commented Oct 20, 2023 at 5:21

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