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Homework Help: Complex Analysis - Transcendental Solutions Help

  1. Feb 23, 2012 #1

    gbu

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    This isn't really homework help. I'm working through a complex analysis textbook myself, and am stumped on the complex transcendentals, but I figured this was the best place for it. I would greatly appreciate any guidance here, I'm getting very frustrated!

    1. The problem statement, all variables and given/known data

    The problem is to find all solutions of [itex]e^{2iz} = 1[/itex] where [itex] z \in \mathbb{C}[/itex].

    The correct answer is, I believe [itex]z = n \pi[/itex] for any integer n.


    2. Relevant equations

    Euler's equation: [itex]-1 = e^{i \pi}[/itex]



    3. The attempt at a solution

    I tried turning the right hand side in to [itex]-e^{i \pi}[/itex] via Euler's equation, then taking a logarithm of both sides... gives [itex] 2iz = i \pi[/itex]... but Wolfram Alpha says the answer is [itex]n \pi[/itex] where [itex] n \in Z[/itex]. Clearly not where I got to.
     
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  3. Feb 23, 2012 #2

    Office_Shredder

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    If [itex]-1=e^{i\pi}[/itex], what is [itex] e^{2i\pi}[/itex]? (Then consider the rest of the solutions and how they fit in)

    You can find them more directly by just using Euler's formula [itex]e^{2iz} = e^{2i(a+bi)} = e^{-2b}e^{2ia} = e^{-2b}(\cos(2a)+i\sin(2a))[/itex]

    How can you solve for a and b here?
     
  4. Feb 23, 2012 #3

    gbu

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    I can see intuitively the correct answer from the first line (thanks a lot for that!), it makes it obvious that [itex]n \pi[/itex] is an answer... I'm still not sure how to solve a and b in that equation though.


    [itex]e^{-2b}(\cos(2a)+i\sin(2a)) = 1[/itex]
    [itex]\cos(2a)+i\sin(2a) = e^{2b}[/itex]

    I don't know where to go from here. Perhaps this is a simpler algebra problem than I'm making it, but it seems like there's two unknowns and only one equation?

    [strike]Edit: wait, [itex]\cos(2a)+i\sin(2a)[/itex] is a complex number itself... I'm still not sure where to go with that. Let [itex]w = \cos(2a) + i\sin(2a)[/itex] and we have

    [itex]e^{-2b} w = 1[/itex]

    But, still not sure how to get to n \pi.[/strike] I'm clearly very confused. I'll go back to the textbook I suppose.
     
    Last edited: Feb 23, 2012
  5. Feb 23, 2012 #4

    Office_Shredder

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    What is the absolute value of [tex] e^{-2b}(\cos(2a)+i\sin(2a))[/tex]? (Hint: it's [itex] e^{-2b}[/itex]
     
  6. Feb 23, 2012 #5

    gbu

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    Right. Which leaves me with

    [itex]
    e^{-2b} = 1
    [/itex]

    Which we can log both sides...

    [itex]
    -2b = \ln 1 = 0
    [/itex]

    Telling me b = 0.

    As for a, we have [itex]e^{-2b}(\cos(2a) + i\sin(2a)) = 1[/itex], but we know b is 0.

    [strike]So we have [itex ]\cos(2a) + i\sin(2a) = 1[/itex]

    [itex ]\cos(2a) + i\sin(2a) = 1[/itex] or [itex ]e^{2ia} = 1[/itex][/itex]

    Log both sides... nope?

    [itex ]2ia = 0[/itex][/strike]

    Still no idea what to do with a. :-( I greatly appreciate your help so far... do you have any recommended reading for this basic stuff? The textbook I'm using is a bit too "succinct" for my learning style. It was going great until we got in to something I didn't "get", and now I feel like I have no where to go with it.

    I get the feeling you can't go log(e^Z) = Z where Z is a complex variable. Is this correct?

    Edit:

    I think when we get to

    [itex]\cos(2a) + i\sin(2a) = 1[/itex]

    We can just solve by recognizing that cos and sin are periodic on [itex]2\pi[/itex] and thus 2a can be rewritten as [itex]2a + n\pi[/itex] giving

    [itex]\cos(2a + n\pi) + i\sin(2a + n\pi) = 1[/itex]

    I still don't know how to get rid of all that stuff though. :(
     
    Last edited: Feb 23, 2012
  7. Feb 24, 2012 #6
    What's wrong with just taking log of both sides?

    [tex]z=\frac{1}{2i}\left(\ln|1|+i(0+2k\pi)\right)[/tex]

    and that's just:

    [tex]z=k\pi,\quad k\in\mathbb{Z}[/tex]

    I think though maybe you're not recognizing that the complex log is infinitely-valued:

    [tex]\log(z)=\ln|z|+i(\text{Arg}(z)+2k\pi)[/tex]

    and that's what I'm doing implicitly in the first equation above.
     
    Last edited: Feb 24, 2012
  8. Feb 24, 2012 #7

    gbu

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    Brilliant, thank you jackmell! That's exactly what I was missing. It's funny, the textbook clearly says that, but I just didn't really "see" it.
     
  9. Feb 24, 2012 #8

    Office_Shredder

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    Two complex numbers are equal to each other, so the real and complex parts are:
    cos(2a)=1
    sin(2a)=0

    hopefully at that point it's clear that a is any integer
     
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