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Simple complex number question

  1. Jan 6, 2010 #1
    Sorry i can tell im being stupid and missing something here.

    1. The problem statement, all variables and given/known data
    if 7^(3+2i)=re^(i[theta]) find values of the real numbers r and [theta]


    2. Relevant equations

    er^(i[theta])=r(cos[theta]+isin[theta])


    3. The attempt at a solution

    Ok you know that 7^(3+2i)=(7^3)(7^2i) but i dont know what to do from here. I think there must be a simple formula or concept i am missing :s. There are similar questions aswel such as (1+i)^i which i also dont know how to solve. I know it must involve getting the i off the top but am unsure how to go about this. Advice would be appreciated, Thanks!
     
  2. jcsd
  3. Jan 6, 2010 #2

    tiny-tim

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    Hi PhyStan7! :smile:

    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)

    Hint: 7i = (eln7)i :wink:
     
  4. Jan 6, 2010 #3
    Cheers tiny-tim!

    So is this right...

    343e(2i)ln7=re

    and as re=r(cos(θ)+isin(θ))

    θ=ln49 and r=343?

    Or have i gone wrong somewhere?

    Cheers
     
  5. Jan 6, 2010 #4

    tiny-tim

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    Yup, that's it! :smile:
     
  6. Jan 6, 2010 #5
    Hi, just thought I'd jump in and ask if there was a proof for 7i=(eln(7))i

    Cheers!

    edit: the better question is probably, should it proven like:

    7i = e
    i ln(7) = iθ
    ln(7) = θ

    ?
     
  7. Jan 7, 2010 #6

    tiny-tim

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    Hi tomeatworld! :wink:

    A more fundamental question is, when does ai = bi ?

    That's the same as (a/b)i = 1, so put a/b in polar form, and what do you get? :smile:
     
  8. Jan 7, 2010 #7
    Surely if ai = bi then a = b. Converting a/b to polar form, surely just leaves a/b. Rather confused...
     
  9. Jan 7, 2010 #8

    Hurkyl

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    Actually, in a typical formulation,
    7i=exp(ln(7) i) ​
    is the definition of what complex exponentiation means. (with exp defined by its power series)
     
  10. Jan 7, 2010 #9

    tiny-tim

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    That's if a and b are real. ok then … if a/b is real, when is (a/b)i = 1 ? :smile:
     
  11. Jan 7, 2010 #10
    Well, quick logging and algebra gives me it will equal 1 when: b=ei ln(a). Meaning that when a>0 it works. Am I missing anything or got it completely wrong?
     
  12. Jan 7, 2010 #11

    tiny-tim

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    uhh? That's b = ai. :confused:

    Try this … what are all the real solutions for xi = 1 ?
     
  13. Jan 7, 2010 #12
    So i ln(x) = 0. then if the imaginary part is 0, i spose you should say x=[tex]e^{2n \pi}[/tex]. Sound right?
     
  14. Jan 7, 2010 #13

    tiny-tim

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    Right! :smile:

    ok, so, for general complex numbers a and b, ai = bi when … ? :wink:
     
  15. Jan 7, 2010 #14
    When you generalise a complex number into a letter a, I can't understand how you deal with it :/ Unless it's just something like:

    a=b=[tex]e^{2n \pi}[/tex]
     
  16. Jan 7, 2010 #15

    tiny-tim

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    Sorry, but you'll have to get used to it.

    A general complex number usually is represented by a single letter (usually "z").

    (Just like a vector being represented by a single letter a)
     
  17. Jan 7, 2010 #16
    we usually use z = a + bi and work from there. but with two it's killing me...
     
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