Results 1 to 3 of 3

Thread: Complex analysis: Evaluating an i^i

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    14

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,434
    Thanks
    2528
    Looks good to me. If you want the most general solution it doesn't hurt to have that 2n\pi from the start.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    38
    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) } .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 4th 2011, 06:30 AM
  2. Replies: 6
    Last Post: Sep 13th 2011, 08:16 AM
  3. Replies: 1
    Last Post: Oct 2nd 2010, 02:54 PM
  4. Replies: 12
    Last Post: Jun 2nd 2010, 03:30 PM
  5. Replies: 1
    Last Post: Mar 3rd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum