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