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Solve sin(z)=2

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve sin(z)=2.

    2. Relevant equations

    sin z=(e^i*z - e^-i*z)/2*i
    sin z=sin x * cosh y + i*cos x * sinh y (haven't tried this way yet)

    3. The attempt at a solution

    Starting with the first relevant equation, I got z=pi*(1/2 + 2*n) + i*ln(2+sqrt(3)).
    The book says that another solution is z=pi*(1/2 + 2*n) - i*ln(2+sqrt(3)).
    How do you get that ?
     
  2. jcsd
  3. Jul 1, 2011 #2
    Re: sin(z)=2

    because cosh is an even function.
     
  4. Jul 1, 2011 #3

    I like Serena

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    Homework Helper

    Re: sin(z)=2

    Assuming you solved your equation with the quadratic formula, you should have 2 solutions.

    If you work it out, you'll see that they match the 2 solutions that you gave.
     
  5. Jul 1, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: sin(z)=2

    The solution to the quadratic equation [itex]y^2- 4iy- 1= 0[/itex]
    is
    [tex]y= \frac{4i\pm\sqrt{-16+ 4}}{2}= 2i\pm i\sqrt{3}[/tex]
    Did you use both "+" and "-"?
     
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