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Exponential functions, and complex numbers

  1. Sep 10, 2009 #1
    hey im doing some questions outta a txt book, i sorta understand complex numbers, like multiplying and dividing, ..

    The question asks to rearrange for z,
    e^(iz) = i - 1

    im not sure what to do with the exponential function.

    thanks for the help
     
  2. jcsd
  3. Sep 10, 2009 #2
    Are you familiar with the log function for complex variables?
     
  4. Sep 10, 2009 #3
    hmmm no... that sounds like that could help
     
  5. Sep 10, 2009 #4
    .. does any 1 no how i could solve this?
     
  6. Sep 10, 2009 #5
    You want to take the log of both sides of the equation. Look in your text book for more information.

    Note: some books use ln instead of log.

    HTH

    Petek
     
  7. Sep 10, 2009 #6
    hmm ok so that gives...

    z = ln(i - 1)/i
    it cant be that easy?

    what about... since e^(iz) = cos(z) + isin(z)

    cos(z) + isin(z) = i - 1
    Then how would i solve for Z?
     
  8. Sep 10, 2009 #7

    lurflurf

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    Think about where the line segment between i - 1 and 1 intersects the unit circle. Use trigonometry since exp(z)=exp(x+y*i)=exp(x)*[cos(y)+i*sin(y)]
     
  9. Sep 10, 2009 #8

    lurflurf

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    z = ln(i - 1)/i
    is right, it depends on the form you want the answer in

    solving this will work but z will be complex
    cos(z) + isin(z) = i - 1

    write i-1=sqrt(2)(i*sqrt(2)/2-sqrt(2)/2)
    then solve
    exp(x)=sqrt(2)
    and
    cos(y) + isin(y)=(i*sqrt(2)/2-sqrt(2)/2)
    y real so
    cos(y)=-sqrt(2)/2
    sin(y)=sqrt(2)/2
    tan(y)=-1
     
  10. Sep 10, 2009 #9
    hmm ok .. just a question, when its iz = ln|i-1|

    wouldnt i go z = ln(i-1) * (0+i)
    (0 - i) * (0+i)
    because im just remembering when you divide complex numbers thats all.
     
  11. Sep 10, 2009 #10
    opps that didnt post right
    wouldnt i go z = ln(i-1)/(0 - i) *(0+i)/(0+i)
     
  12. Sep 10, 2009 #11

    lurflurf

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    exp(iz)=i-1
    take log
    log(exp(iz))=log(i-1)
    log cancels exp
    iz=log(i-1)
    divide by i
    z=log(i-1)/i
    that is the priciple value , there are others
     
  13. Sep 11, 2009 #12
    oh, ok thanks.
     
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