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

**princeps** · November 30th 2011, 05:24 AM

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 ?

**CaptainBlack** · November 30th 2011, 06:28 AM

Originally Posted by princeps

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 ?

Yes, No, No.

CB

**princeps** · November 30th 2011, 06:40 AM

Originally Posted by CaptainBlack

Yes, No, No.

CB

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

**CaptainBlack** · November 30th 2011, 12:49 PM

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

**princeps** · November 30th 2011, 09:30 PM

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 $1^i$ , specifically

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

Am I correct ?

**CaptainBlack** · November 30th 2011, 11:23 PM

Originally Posted by princeps

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

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

Am I correct ?

No logarithms (explicitly) involved:

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

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

Transcendental otherwise.

CB

**princeps** · November 30th 2011, 11:38 PM

Originally Posted by CaptainBlack

No logarithms (explicitly) involved:

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

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

CB

But :

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

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

Am I missing something ?

**CaptainBlack** · November 30th 2011, 11:53 PM

Originally Posted by princeps

But :

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

so number $(e^{\pi})^{-2}$ 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

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