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Zeros of sinh()

  1. May 27, 2008 #1

    How are the sinh() zeros defined?

    Unfortunately I don't understand the definitions found, google and wikipedia was searched,
    any help would be appreciated

    Thank you and best Regards
  2. jcsd
  3. May 27, 2008 #2
    you could use the fact that [tex] sin(-ix)(i)=sinh(x) [/tex] so the zeros of hyperbolic sine are [tex] in\pi [/tex] for every integer 'n' (complex zeros)
  4. May 27, 2008 #3


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    It is not clear to me what you mean by "defining" zeros.

    sinh(x), for real numbers, is defined as
    [tex]\frac{e^x- e^{-x}}{2}[/tex]
    The zeros are 'defined', of course, by
    [tex]sinh(x)= \frac{e^x- e^{-x}}{2}= 0[/tex]
    which is the same as saying ex= e-x or e2x[/itex]= 1. That leads immediately to x= 0 as the only real zero of sinh(x).

    If you expand to complex numbers, it is not to easy to see, from the fact that sinh(x) is defined as
    [tex]sinh(z)= \frac{e^z- e^{-z}}{2}[/tex]
    and the Cauchy formula
    [tex]sin(x)= \frac{e^{iz}- e^{-iz}}{2i}[/tex]
    that sinh(z)= (1/i) sin(iz). Since sin(z) is 0 if and only if z is an integer multiple of [itex]\pi[/itex], sinh(z) is 0 if and only if z is an integer multiple of [itex]2\pi[/itex], as mhill said.
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