1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Very simple complex powers problem

  1. Sep 24, 2012 #1
    Write the following in x+iy form: (-1)^i

    my solution:
    (-1)i = (ei(π+2πk))i = e-(π+2πk)

    However, my book just states the answer as e-(π) .....but these are not the same....
  2. jcsd
  3. Sep 24, 2012 #2
    For integer [itex]k[/itex], sure they are. Any full turn around the unit circle doesn't change the value.
  4. Sep 24, 2012 #3
    but ei2πk and e2πk are different...the first one equals 1 but the second does not...

    i mean look just plug in k=1 and k=2...you get e^-3 and e^-5, which are not the same
  5. Sep 24, 2012 #4
    Oh, okay, I see what's going on. Yeah, in this case, you may be more correct than the book is. Someone with more recent experience than I in complex analysis may know something more specific to this, though.
  6. Sep 24, 2012 #5
    oh alright. idk my book is doing this same mistake for each and every related problem. so i was thinking that i must be doing something wrong. but maybe they are just giving the answer when k=0...although they never specify this....
  7. Sep 24, 2012 #6


    User Avatar
    Homework Helper

    Your answer is more complete than the book's. Exponentiation to a complex power can give a multivalued result. The book just gave the principal value, which, if you're asked for a single answer, is the usual value we take.
  8. Sep 24, 2012 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    (Those π's are hard to read !)

    Sure enough.

    Give WolframAlpha, [itex]\displaystyle \left(e^{\pi} \right)^{1/i}[/itex] or [itex]\displaystyle \left(e^{-3\pi} \right)^{1/i}\,,[/itex] and it gives -1.

    Of course, [itex]\displaystyle e^{-3\pi}\approx 0.0000806995175703\,,[/itex] [itex]\displaystyle e^{-\pi}\approx 0.04321391826377226\,,[/itex] [itex]\displaystyle e^{\pi}\approx 23.140692632779263\,,[/itex]

    So it looks like [itex]\displaystyle (-1)^i=e^{(\text{odd integer multiple of }\pi)}\ .[/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook