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Math Help - proof

  1. #1
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    Talking proof



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  2. #2
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    Re: proof

    To prove this you need to know the principal value of a complex exponential.
    z^w=\exp(w\cdot\text{Log}(z))

    Recall that \text{Log}(i)=\ln(1)+\frac{i\pi}{2}.
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    Re: proof

    Ok,but how can i start?
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    Re: proof

    Quote Originally Posted by Mhmh96 View Post
    Ok,but how can i start?
    What do you mean by that?

    What does i\text{Log}(i)=~?.

    Then what is \exp(i\text{Log}(i))~?
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  5. #5
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    Re: proof

    Yes, i have no background about how to prove it so i want to know the proof in details ,if that ok.
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    Re: proof

    Quote Originally Posted by Mhmh96 View Post
    i have no background about how to prove it so i want to know the proof in details ,if that ok.
    Well that is not going to happen here.

    Because if you have have no background then you would not begin to understand the proof.
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  7. #7
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    Re: proof

    I guess that is right,thanks anyway.
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  8. #8
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    Re: proof

    I think i found the proof ,but i still have one more problem

    when we raise i to the power i ,the result should be real number,right?
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  9. #9
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    Re: proof

    Hello, Mhmh96!

    \text{Prove: }\:i^i \:=\:\dfrac{1}{\sqrt{e^{\pi}}}

    \text{Let }\,z \:=\:i^i

    \text{Take logs: }\:\ln(z) \:=\:\ln(i^i) \quad\Rightarrow\quad \ln(z)\:=\:i \ln(i) \;\;{\bf[1]}


    \text{Given: }\:e^{i\theta} \:=\:\cos\theta + i\sin\theta

    \text{Let }\theta = \tfrac{\pi}{2}\!:\;\;e^{i\frac{\pi}{2}} \;=\;\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} \quad\Rightarrow\quad e^{i\frac{\pi}{2}} \:=\:i

    \text{Take logs: }\:\ln\left(e^{i\frac{\pi}{2}}\right) \:=\:\ln(i) \quad\Rightarrow\quad i\tfrac{\pi}{2}\ln(e) \:=\:\ln(i)

    . . \text{Hence: }\:\ln(i) \:=\:i\tfrac{\pi}{2}


    \text{Substitute into }{\bf[1]}\!:\;\;\ln(z) \:=\:i\left(i\tfrac{\pi}{2}\right) \quad\Rightarrow\quad \ln(z)\:=\: -\tfrac{\pi}{2}

    \text{Therefore: }\:z \;=\;e^{-\frac{\pi}{2}} \;=\;\dfrac{1}{e^{\frac{\pi}{2}}} \;=\; \dfrac{1}{(e^{\pi})^{\frac{1}{2}}} \;=\;\dfrac{1}{\sqrt{e^{\pi}}}

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