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Cosh z=0 where z= x + iy

  1. Mar 13, 2006 #1
    cosh z=0 where z= x + iy

    we can solve this equation since its imposible coshz=0, is it right?

    how about sinh=0?

  2. jcsd
  3. Mar 14, 2006 #2
    sinh(z) = 0 is somewhat trivial z = 1 is an obvious solution.

    And if z = x + iy how is cosh(z) = 0 impossible? Do you know the formula for the inverse hyperbolic cosine? Have you tried using it?
  4. Mar 14, 2006 #3


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    How about using

    [tex] \cosh\left(a+b\right) =...?[/tex]

    That is addition theorems in hyperbolic trigonometry.

  5. Mar 14, 2006 #4
    I'm fairly sure that you mean z=0 ;)

    To the OP, you should use dexter's hint, and remember that a complex number is zero only when its real and imaginary parts are both zero.
  6. Mar 14, 2006 #5


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    Remember also that cosh(iy) = cos y and sinh(iy) = i sin y
  7. Mar 14, 2006 #6

    matt grime

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    So something is 'sovable' but 'impossible'? that makes no sense.

    In any case the notion that cosh(z) is always positive (greater than 1) is only true if we restrict to z a real number, which is expressly not what is happening here.
  8. Mar 19, 2006 #7
    that was a typo!:frown:
    it never occured to e that some people might think of as a stupidperson who'd say such a illogical thing.:eek:
  9. Mar 20, 2006 #8

    matt grime

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    But which of the parts is the typo? that is waht you're supposed to understand by my comment.
  10. Mar 9, 2007 #9
    cosh (x+iy)=0
    cosh x cosh iy + sinh x sinh iy = cosh x cos y + i sinh x siny =0

    cos y = 0 : y =(2n+1) [tex]\pi/2[/tex]
    sinh x =0 : x=1
    sin y =0 : y= 2k [tex]\pi/2[/tex]

    is it right?
  11. Mar 9, 2007 #10


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    Try to solve the "sinh x=0" again. You've got a wrong answer.
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