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Complex numbers with trig functions

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data

    prove: arctanh(z) = 1/2 ln( (1+z)/(1-z) )


    2. Relevant equations

    cosh z = (ez + e-z)/2
    sinh z = (ez - e-z)/2
    ez = ex + iy = ex(cosy + siniy)

    3. The attempt at a solution

    cosh z / sin hz = ez+e-z/ez-e-z
     
  2. jcsd
  3. Apr 4, 2009 #2

    Dick

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    tanh(z)=sinh(z)/cosh(z). Not cosh(z)/sinh(z). This is just like the problem of finding the inverse of y=f(x). First you try and solve for x in terms of y then interchange x and y. Try and solve y=(e^x-e^(-x))/(e^x+e^(-x)) for x in terms of y and then interchange x and y. Hint: multiply numerator and denominator by e^x.
     
  4. Apr 4, 2009 #3
    sorry, the problem specifies artanh . . .
    solving for e^x wont give me the same properties that using z implies (since z = xi + y)
     
  5. Apr 4, 2009 #4

    Dick

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    The inverse function of y=tanh(z) is arctanh. You don't need to break z down to it's real and imaginary components. You need to solve that equation for z in terms of y and then replace y with z.
     
  6. Apr 7, 2009 #5
    thanks! I finally got it. i appreiate your help. cool problem.
     
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