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Complex fraction as a ratio of sines?

  1. Feb 12, 2008 #1
    Is it possible for me to write this complex fraction as a ratio of two sines? Thanks.

  2. jcsd
  3. Feb 12, 2008 #2


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    It is possible, but the arguments of the sines will be complex numbers.
  4. Feb 12, 2008 #3
    I think you could use hyperbolic sine but I don't know much about hyperbolic trig f(x)s yet
  5. Feb 12, 2008 #4
    That's fine, Mathman. If we allow sin to be evaluated at complex numbers, using the standard definition of sin(z) in terms of the exponential, how do we we rewrite it? Thanks
  6. Feb 13, 2008 #5


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    Similar expression for 1-e101z. Then take ratio.
  7. Feb 13, 2008 #6
    I don't think that works though because sin(z) puts the exponent of e as iz and -iz, not z. For instance, if I plug that into a calculator, it doesn't come out equal. I am trying to factor out the real part, but the ratio will not cancel to the point that I have just sines.
  8. Feb 13, 2008 #7
    Nvm, I just realized you were talking about hyperbolic sines. Thanks for the tip!

    EDIT: However, now I get e[tex]^{50z}[/tex][tex]\frac{sinh101z/2}{sinhz/2}[/tex]

    Any way to simplify the exponent out front?
    Last edited: Feb 13, 2008
  9. Feb 14, 2008 #8


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    Sorry, my bad. I forgot to put the i factor in for the sin. Using sinh is better.
  10. Jul 13, 2008 #9
    A related question I have is can we express the complex fraction as a linear equation of Re()+Im()?
  11. Jul 13, 2008 #10


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    That's pretty standard isn't it? You "rationalize" the denominator by mutltiplying both numerator and denominator by the complex conjugate of the denominator. Exactly what the computations are depends on the particular complex fraction. Do you mean "complex fraction" in the sens "fraction with complex numbers" or "fraction with fractions in the numerator and denominator"?

    Can you give an example of what you had in mind?
  12. Jul 13, 2008 #11
    I am not sure what the complex conjugate would be for the denominator...



    exp(iaz) - exp(-iaz)
    exp(ibz) - exp(-ibz)


    Re() + Im()
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