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Complex number problem

  1. May 11, 2009 #1
    Calculate (-1) ^ i

    I tried using the formula x^ni = cos (ln (x)^n) + i sin (ln (x)^n)

    but i cannot solve it. i used matlab to get this answer 0.0432139182637723 + 0i

    but i dunno how to solve it with steps.. can i get some assistance please.

    thank you.
     
  2. jcsd
  3. May 11, 2009 #2

    jbunniii

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    Do you know how to write -1 in polar form?
     
  4. May 11, 2009 #3
    yup the polar form will just be cos (theta) + i sin (theta) and the modulus here is 1.. correct?
     
  5. May 11, 2009 #4

    jbunniii

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    Well, that's a particular complex number, but it's actually expressed in rectangular form x + iy, where x = cos(theta) and y = sin(theta).

    Do you know how to write -1 in terms of "e", i.e., do you know what a complex exponential is? It would help to know what background can be assumed for this exercise.
     
  6. May 11, 2009 #5
    well what i did was... x = ln (-1)^i
    which is.. i ln (-1)
    then in terms of "e" i will get... e ^ i ln (-1)

    which gives me cos (ln (-1)) + i sin (ln (-1))
    but i cant go any further to get the answer
     
  7. May 11, 2009 #6

    jbunniii

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    What I was trying to get at is, have you been exposed to Euler's famous formula:

    [tex]e^{i\pi} = -1[/tex]

    If so, then you can easily use this to get the answer you want.
     
  8. May 11, 2009 #7
    yup i know that formula.. but how do i use it here?.. i tried to use that formula too but dint work.. maybe i did something wrong?
     
  9. May 11, 2009 #8

    jbunniii

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    Well, you're trying to find (-1)^i, right? So what is the natural thing do to both sides of Euler's formula?
     
  10. May 11, 2009 #9
    um not sure exactly.
     
  11. May 11, 2009 #10

    jbunniii

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    Oh, come on!

    What operation do you do to -1 to obtain (-1)^i? (This isn't a trick question!) Just do that operation to both sides of Euler!
     
  12. May 11, 2009 #11

    Matterwave

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    What jbunnii is trying to say is:

    [tex](-1) = (e^{i\pi})[/tex]
    [tex](-1)^i = ...[/tex]

    Use basic algebra here.
     
  13. May 11, 2009 #12
    we take logs of both sides
     
  14. May 11, 2009 #13
    ohhh yup i get what you asking
     
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