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Math Help - Complex analysis: Evaluating an i^i

  1. #1
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    Complex analysis: Evaluating an i^i

    "Evaluate i^{1+i}. Put the result in the form u+iv when u,v are real."

    i = e^{i(\frac{\pi}{2}+2n\pi)}
    i^{1+i} = [e^{i(\frac{\pi}{2}+2n\pi)}]^{1+i} = e^{-(2n+\frac{1}{2})\pi}e^{i(2n+\frac{1}{2})\pi}
    = e^{-(2n+\frac{1}{2})\pi}[\cos(2n+\frac{1}{2})\pi + i\sin(2n+\frac{1}{2})\pi] = e^{-(2n+\frac{1}{2})\pi} for n = 0, \pm1, \pm2,...

    Looks good or did I not need to start with that 2n\pi?
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  2. #2
    MHF Contributor

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    Looks good to me. If you want the most general solution it doesn't hurt to have that 2n\pi from the start.
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  3. #3
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    An adjustment. Pick up from ... e^{-\left( 2n+\frac{1}{2}\right) \pi} e^{i \left( 2n+\frac{1}{2}\right) \pi} =  e^{-\left( 2n+\frac{1}{2}\right) \pi}  e^{i 2n \pi} e^{i \frac{1}{2} \pi} =  e^{-\left( 2n+\frac{\pi}{2}\right) } e^{i \frac{\pi}{2} } = i e^{-\left( 2n+\frac{\pi}{2}\right) } .
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