Are numbers : $(-1)^{i} , 1^{-i} , 1^{i} $ transcendental numbers?

Question

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What is the value of 1^i?

According to Euler's formula : $e^{ix}=\cos x + i\cdot \sin x$ we may write :

$$e^{i\cdot \frac{\pi}{2}}=i \Rightarrow \left(e^{i\cdot \frac{\pi}{2}}\right)^{i}=i^{i}\Rightarrow e^{\frac{-\pi}{2}}=i^{i}$$

So,

$$e^{\pi}=\left(e^{\frac{\pi}{2}}\right)^2=\left(\frac{1}{e^{\frac{-\pi}{2}}}\right)^2=\left(\frac{1}{i^{i}}\right)^2=\left(i^{-i}\right)^2=(-1)^{-i}$$

Since $e^{\pi}$ is proven by Gelfond–Schneider theorem to be transcendental number it follows that $(-1)^{-i}$ is a transcendental number. So,my question is :

Are numbers : $(-1)^{i} , 1^{-i} , 1^{i} $ transcendental numbers ?

The "number" $i^i$ does not have a universal definition because exponentiation is not necessarily injective. It could be $e^{-\frac{\pi}{2}}$ or even $$e^{-\frac{1001\pi}{2}}.$$ It all depends on what branch of the logarithm you are using. Specifically you cannot say that $i^i = e^{-\frac{\pi}{2}}$. — Eric Naslund, Nov 24 '11 at 08:52
Maybe I'm missing something... $1^i=(e^{\ln 1})^i=e^{i\ln 1}=\cos(\ln 1)+i\sin(\ln 1)=1$. — David Mitra, Nov 24 '11 at 09:10
@pedja: You should read either my answer or this one: http://math.stackexchange.com/q/3674/6075 you are missing a very important point... — Eric Naslund, Nov 24 '11 at 09:12
@DavidMitra: The problem is $\log(1)$ is not well defined for complex numbers. It could be $2\pi i$ or $0$, or $2\pi k i$ since $e^{2\pi k i}=1$ for any integer. — Eric Naslund, Nov 24 '11 at 09:13
@David: As Eric notes, the complex logarithm is multibranched... — J. M. ain't a mathematician, Nov 24 '11 at 09:13

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

This is nitpicking, but it is important to understand. You have to be careful when dealing with the complex numbers as exponentiation is not always injective, and it does not mean the same thing as it does for real numbers.

Specifically, what does $z^w$ even mean for complex $z$, $w$? We have to start somewhere, what is its definition? We define $$z^w:= e^{w\log z}.$$ Everything here makes sense, as $e^z$ can be given by its power series, and we understand multiplication. But we have to remember that in the complex plane, $\log z$ is a multivalued function. We want it to be the inverse of the exponential, but the exponential function is not surjective. For example what is $\log(1)$? It could be $0$, as $e^0 =1$, but also $2\pi i$ as $e^{2\pi i}=1$. So we see that $\log (1)=2\pi ki $ for any integer $k$. This means that there are many different possibilities for $1^i$, specifically $$1^i=e^{i\log (1)}=e^{-2\pi k}\ \text{for any integer } k. $$

In particular, $1^i$ is transcendental if you take any branch, except the principal branch, in which case it is an integer.

Also see the following Math Stack Exchange posts:

Non-integer powers of negative numbers

How to combine complex powers?

What is the value of $1^i$?

On the other hand, $i^i$ is transcendental no matter which branch of the logarithm you use. Perhaps OP would be happy to know whether the expressions asked about are of the $1^i$ type or of the $i^i$ type. — Gerry Myerson, Nov 24 '11 at 11:51

Are numbers : $(-1)^{i} , 1^{-i} , 1^{i} $ transcendental numbers?

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