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Complex Variables Algebra Solutions / Argument/Modulus

  1. Feb 26, 2012 #1

    gbu

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

    Solve for a, [itex] a \in \mathbb{C}[/itex]

    [itex]
    \frac{2\ln(a^2 - 1)}{\pi i} = 1
    [/itex]

    2. Relevant equations
    N/A.


    3. The attempt at a solution

    Reorganizing the equation.
    [itex]2\log(a^2 - 1) = \pi i[/itex]
     
    Last edited: Feb 26, 2012
  2. jcsd
  3. Feb 26, 2012 #2
    so a = x+yi. Then (x+yi)^2-1 = (x^2-y^2-1)+2xyi. What is the modulus of that?

    Put your i's together as well.
     
  4. Feb 26, 2012 #3

    gbu

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    It's

    [itex]
    \sqrt{(4x^2 y^2 + (x^2-y^2-1)^2)}
    [/itex]

    Which I suppose gets me to an answer, but its certainly not a pretty one. Wolfram Alpha gives a very simple answer to the question (a^2 = sqrt(1+i))
     
  5. Feb 26, 2012 #4
    There may be a better way to do this one. Let me think.
     
  6. Feb 26, 2012 #5
    Separate to log = pi i/2 and raise use the inverse function e.
     
  7. Feb 26, 2012 #6

    Dick

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    Science Advisor
    Homework Helper

    Look at what the parts of
    [itex] 2\ln(|a^2 - 1|) + 2 i \arg(a^2 - 1) = \pi i [/itex]
    are telling you. Equate real and imaginary parts, so [itex]\ln(|a^2-1|)=0 [/itex] and [itex]\arg(a^2 - 1)= \frac{\pi}{2}[/itex]. Without solving any equations and just thinking geometrically, what do those two things tell you about [itex]a^2-1[/itex]?
     
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