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Sinhz = 0 iff z=n(pi)i (n=0, +/- 1, +/- 2 ) ~ a question about this?

  1. Mar 21, 2007 #1
    1. The problem statement, all variables and given/known data
    I have this weird quesiton in my text book I'm not even sure which part is the question and what i need to do...

    It says

    "Give details showing that the zeros of sinhz and coshz are as in statements (14) and (15) "


    2. Relevant equations

    (14) sinhz = 0 iff z=n(pi)i (n=0, +/- 1, +/- 2...)

    (15) coshz = 0 iff z=(pi/2 + n(pi)) (n=0, +/- 1, +/- 2...)

    -sin(iz) = sinh(z)
    cos(iz) = cosh(z)
    sinh(z) = [e^z - e^(-z) ] /2
    cosh(z) = [e^z + e^(-z)]/2

    3. The attempt at a solution

    I haven't made a start on the quesiton because I don't understand the words and what it wants me to calc? :grumpy:
     
  2. jcsd
  3. Mar 21, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Substitute the given values into your formulas for sinh and cosh and show they are zero. The words look pretty clear to me. I think you've forgotten an i in your cosh zeros.
     
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