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

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

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

So,

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

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

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

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

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

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

So,

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

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

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

Yes, No, No.

CB

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

**CaptainBlack** Yes, No, No.

CB

Excuse me my ignorance but how can you conclude that just looking at numbers ?

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

**princeps** Excuse me my ignorance but how can you conclude that just looking at numbers ?

I don't conclude that from "just" looking at them.

You have already demonstrated how you know the first is transcendental, the same form of argument can be applied to the others.

CB

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

**CaptainBlack** I don't conclude that from "just" looking at them.

You have already demonstrated how you know the first is transcendental, the same form of argument can be applied to the others.

CB

But complex logarithm is multibranched .This means that there are many different possibilities for$\displaystyle 1^i$, specifically

$\displaystyle 1^i=e^{i\log (1)}=e^{-2\pi k}$ for any integer $\displaystyle k$.

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

Am I correct ?

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

**princeps** But complex logarithm is multibranched .This means that there are many different possibilities for$\displaystyle 1^i$, specifically

$\displaystyle 1^i=e^{i\log (1)}=e^{-2\pi k}$ for any integer $\displaystyle k$.

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

Am I correct ?

No logarithms (explicitly) involved:

$\displaystyle 1=e^{2n\pi i},\ \ n= ..-1,0,1, ..$

$\displaystyle 1^i=e^{- 2n \pi}=1$ for $\displaystyle n=0$

Transcendental otherwise.

CB

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

**CaptainBlack** No logarithms (explicitly) involved:

$\displaystyle 1=e^{2n\pi i},\ \ n= ..-1,0,1, ..$

$\displaystyle 1^i=e^{- 2n \pi}=1$

CB

But :

If $\displaystyle n=1$ then $\displaystyle 1^{i}=e^{-2\pi}=(e^{\pi})^{-2}$ and $\displaystyle e^{\pi}$ is proven transcendental number

so number $\displaystyle (e^{\pi})^{-2}$ should be transcendental also .

Am I missing something ?

Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

Quote:

Originally Posted by

**princeps** But :

If $\displaystyle n=1$ then $\displaystyle 1^{i}=e^{-2\pi}=(e^{\pi})^{-2}$ and $\displaystyle e^{\pi}$ is proven transcendental number

so number $\displaystyle (e^{\pi})^{-2}$ should be transcendental also .

Am I missing something ?

Opp.. principle values are not transcendental ... Now I think about it since they are all real I'm not sure how one would characterise the principle value in this case.

CB