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Eulers idenity

  1. Jul 14, 2009 #1

    could someone tell me where the i^i=e^(pi/2) came from??
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Jul 14, 2009 #2
  4. Jul 15, 2009 #3
    Ha. They're wrong. [itex]i^i[/itex]=0.20787 and [itex]e^{\pi/2}[/itex]=4.81047. They're missing a negative sign. Remember that taking the square root of something is the same thing as raising that something to the 1/2 power. Then [itex]\sqrt{-1} = i = (-1)^{1/2}[/itex]. Also, 1/i=-i. Then
    [tex]
    \begin{align*}
    e^{i\pi}+1 &= 0 \\
    e^{i\pi} &= -1 \\
    \left( e^{i\pi} \right)^{1/2} &= (-1)^{1/2} \\
    e^{i\pi/2} &= i \\
    \left( e^{i\pi/2} \right)^{1/i} &= i^{1/i} \\
    e^{\pi/2} &= i^{-i}
    \end{align*}
    [/tex]
     
  5. Jul 15, 2009 #4
    I do not like the idea that [tex]i ^{\frac{1}{i}}[/tex] is [tex]i ^{-i}[/tex]. I would of thought it would be [tex]i ^{i^{-1}}[/tex] which are not the same thing as far as I know.
     
  6. Jul 15, 2009 #5

    Cyosis

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    Homework Helper

    It is true that most often they are not the same, however for i they are.

    Proof:
    [tex]
    \frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i
    [/tex]
     
  7. Jul 15, 2009 #6
    Oh cool, I didn't know that. That is pretty nice to know. And it makes sense when put like that.
     
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