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Roots of sin h(z) = 1/2

  1. Dec 11, 2006 #1
    1. The problem statement, all variables and given/known data

    Find all the roots of sin h(z) = 1/2

    2. The attempt at a solution

    sin h(z) = [1/2](e^z - e^-z) = 1/2
    => e^z -e^-z = 1
    => e^2z - e^z - 1 = 0 {multiplied e^z bothsides}
    this is a quadratic equation in e^z using quadratic formula,

    e^z = [1+- sqrt(5)]/2

    taking 'ln' on bothsides

    z = ln[{1+- sqrt(5)}/2]

    And after this step I am confused, what to do next and how to find all the roots?

    can someone help me out in this?
     
  2. jcsd
  3. Dec 11, 2006 #2

    dextercioby

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    Just remember that

    [tex] e^{2k\pi i} =1 \ , \forall k\in\mathbb{Z} [/tex]

    Daniel.
     
  4. Dec 11, 2006 #3
    i can think of
    z = ln(1/2 +-(sqrt(5)/2)
    z = ln(1.61) and z = ln(-0.61)

    and ln(-ive) is not defined.

    :(
     
    Last edited: Dec 11, 2006
  5. Dec 11, 2006 #4

    dextercioby

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    No way is it true that ln(a+b)=ln a +ln b.

    Daniel.
     
  6. Dec 11, 2006 #5
    yupp u r right!
     
    Last edited: Dec 11, 2006
  7. Dec 11, 2006 #6

    cristo

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    Nowhere on that page does it state that ln(a+b)=lna+lnb.

    The correct result, which is stated, is ln(ab)=lna+lnb
     
  8. Dec 11, 2006 #7
    omg... cristo... I got ur point.I m making a blunder actually ...ok
    i come up wid sumthing new and edited my aforementioned post!
     
  9. Dec 11, 2006 #8
    z = ln(1/2 +-(sqrt(5)/2)

    z = ln(1+sqrt(5)/2) and z = ln(1-sqrt(5)/2)
    z = ln(3.24)-ln(2)+2k*pi*i and z = ln(1.24)-ln(2)+2k*pi*i

    0.48+2k*pi*i and -0.48+2k*pi*i

    we can say +-1/2 + 2k*pi*i

    is this solution correct?
     
  10. Dec 12, 2006 #9

    dextercioby

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    You can't approximate to 1/2. It's actually better if you leave it in the original form, the one still containing natural logs.

    Daniel.
     
  11. Dec 12, 2006 #10

    HallsofIvy

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    0.48 is not 1/2! Don't approximate.

    By the way, it is sinh(x) not sin h(x). I first thought you meant sine of some function h(x)!
     
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