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

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

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

    Quote Originally Posted by princeps View Post
    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
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    Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

    Quote Originally Posted by CaptainBlack View Post
    Yes, No, No.

    CB
    Excuse me my ignorance but how can you conclude that just looking at numbers ?
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    Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

    Quote Originally Posted by princeps View Post
    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
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    Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

    Quote Originally Posted by CaptainBlack View Post
    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 ?
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    Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

    Quote Originally Posted by princeps View Post
    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
    Last edited by CaptainBlack; December 1st 2011 at 12:56 AM.
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    Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

    Quote Originally Posted by CaptainBlack View Post
    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 ?
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    Re: Are numbers : (-1)^(i) , 1^(-i) , 1^(i) transcendental numbers?

    Quote Originally Posted by princeps View Post
    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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