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Complex indices

  1. Sep 28, 2007 #1

    JPC

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    hey, was wondering what would happen if i do :

    a^b
    with a : real or complex number
    and b : a complex number

    like : 2^i
     
  2. jcsd
  3. Sep 28, 2007 #2
    The definition:

    [tex]a^b = \exp (b \ln a)[/tex]

    In general ln is multivalued (actually infinitely valued) so there is no unique answer.
     
  4. Sep 28, 2007 #3

    JPC

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    whats exp() ?
    i know log() , and ln() but never heard of exp()
     
  5. Sep 28, 2007 #4
    exponential
    e = 2.718281828
     
  6. Sep 28, 2007 #5
    Specifically, [tex]exp(x) = e^x[/tex].
     
  7. Sep 29, 2007 #6

    JPC

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    oh ok :

    so
    2^i = e^ ( i * ln(2))
    ?
     
  8. Sep 29, 2007 #7

    Gib Z

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    Homework Helper

    Yup that is perfectly correct. Or if you want it split into real and imaginary parts: [tex]\cos (\ln 2) + i \sin (\ln 2)[/tex]
     
  9. Oct 24, 2007 #8

    JPC

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    ok
    now that we started working on exp(x) in my class i know a bit more what you are talking about

    But how do u pass from e^(i * ln(2) ) to [tex]\cos (\ln 2) + i \sin (\ln 2)[/tex] ?
     
  10. Oct 24, 2007 #9
    remember that exp(x) and ln(x) are defined in a more general sense as complex functions.
    if a complex number is of the form z=a +ib then
    exp(z)=exp(a+ ib)=exp(a)exp(ib) by additivity of exponentials and then using euler's formula: exp(z)=exp(a)(cosb +isin(b))
    here the a is real so exp(a) is evaluating using the real definition.

    The complex definitions of the trancesendental functions extend that of it's real analogue.

    JPC: if you learning about the real exponential function for the first time you probably won't encounter euler's formula until later in the course - it's an application of power series.
     
  11. Oct 25, 2007 #10

    Gib Z

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    JPC, it seems my assumptions on your maths education are wrong, in Australia for some reason we do complex numbers after Exponential functions, forgive the pun but here they are seen as, well, more complex. Anyway, You obviously don't need to know what I say from here, so you can either choose to ignore my post completely or read more on what I say next: There is a famous relation that Euler derived through Taylor Series expansions that told him that [tex]e^{ix} = \cos x + i \sin x[/tex]. I Merely used that identity straight off.
     
  12. Sep 13, 2010 #11
    For [tex]b=x+iy[/tex] , [tex]a^b=a^{x+ i y}=a^x a^{i y} = a^x \exp(i y \ln (a)) = a^x (\cos (y \ln (a))+i \sin (y \ln (a))[/tex]
    by Euler's formula [tex]e^{i\theta}=\cos\theta+i\sin\theta[/tex].

    I wonder what happens to [tex]A^B[/tex] where both [tex]A[/tex] and [tex]B[/tex] are matrices
     
    Last edited: Sep 13, 2010
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