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Complex numbers

  1. Jun 1, 2014 #1
    Hello,
    please I need help. I have no idea how to start this. Can someone guide me? This is not homework, I'm just studying on my own and I really don't know how to begin this.
     

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  3. Jun 1, 2014 #2

    benorin

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    probably ought to start with [itex]\cos z = \frac{1}{2}\left(e^{iz}+e^{-iz}\right)[/itex], substituting into the given equation we have

    [tex] \frac{1}{2}\left(e^{iz}+e^{-iz}\right)=2[/tex]

    now multiply both sides by [itex]2e^{iz}[/itex] and collect all the terms on one side...
     
  4. Jun 2, 2014 #3
    Why must we multiply both sides by [itex]2e^{iz}[/itex] ?
     
  5. Jun 2, 2014 #4
    Do it, and see what you end up with. I think you'll agree that it becomes much easier find the set of solutions for z.
     
  6. Jun 2, 2014 #5
    Is this ok?
     

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  7. Jun 2, 2014 #6
    Very close! You're just missing one part to capture the periodic nature of cosine.
     
  8. Jun 2, 2014 #7
    Which part? I don't understand. I thought that would be enough?
     
  9. Jun 2, 2014 #8
    If cos(z) = 2, what is cos(z+2pi)?
     
  10. Jun 2, 2014 #9
    I'll be honest and say that I don't know how to produce the extra term from that particular solution, but that's only because I'm missing something that someone else will probably pick up. The term is there nonetheless, which you will see (I believe) if you solve it with Euler's identity instead of u-substitution.
     
  11. Jun 2, 2014 #10

    CAF123

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    From ##e^{iz} = 2\pm \sqrt{3}\Rightarrow iz = \ln (2 \pm \sqrt{3})##, you can expand the RHS, taking into consideration the fact the complex logarithm is multiple valued.
     
  12. Jun 2, 2014 #11
    So how should I solve this question? I guess my working is alright up to this part, right?
     
  13. Jun 2, 2014 #12

    lurflurf

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    in fact by convention the principle value is
    $$\mathrm{Cos}^{-1}(2)=\imath\, \mathrm{Cosh}^{-1}(2)=\imath\, \mathrm{Ln}(2+\sqrt{3})$$
    to find all solutions recall that
    $$\cos(-x)=\cos(x) \\
    \text{and for any integer k}\\
    \cos(x +2k\, \pi)=\cos(x)$$
     
  14. Jun 2, 2014 #13

    CAF123

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    It is, the method is fine. Given ##z \in \mathbb{C},## what is ##\log_{e} z##?
     
  15. Jun 2, 2014 #14
    I don't get that part. How do you calculate that?
     
  16. Jun 2, 2014 #15
    ##\log_{e} z## = loge(x+iy)

    Is that correct?
     
  17. Jun 2, 2014 #16

    CAF123

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    It is not what I meant for you to write down - have you studied the complex logarithm?
     
  18. Jun 2, 2014 #17

    lurflurf

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    Same way you did.

    $$\mathrm{Cos}^{-1}(2)=\imath\, \mathrm{Cosh}^{-1}(2)=\imath\, \mathrm{Ln}(2+\sqrt{3})\sim 1.316957896924816708625046347307968444026981971\imath
    $$
    is the principal value.
    That is the agreed reference value we give when we give only one value.
    To find all values you do not need to deal with complex logs (though you could).
    Just think back to real trigonometry.
    We found one value call it P
    then -P was another
    then we could add 2n ∏

    Once we find one solution, we then find them all
    what are all solutions to
    $$\cos(x)=2$$
    is done the same way as
    what are all solutions to
    $$\cos(x)=0.2$$
     
  19. Jun 2, 2014 #18
    Yes, I have. But I don't know how to use that here.
     
  20. Jun 2, 2014 #19
    Ok, so what do I do next? Am I supposed to expand that log part?
     
  21. Jun 2, 2014 #20

    lurflurf

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    how would you do the one with 0.2?
    It is the same.

    If t is any solutions and k is an integer

    $$2k\, \pi\pm t$$
    is all the solutions because
    $$\cos(2k\, \pi\pm t)=\cos(t)$$
    by properties of cos(x)
     
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