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Trig functions on complex numbers?

  1. Aug 2, 2007 #1
    Out of curiosity, what happens when you try to perform a trig function on a complex number? So, say, sin(4i+3)? Is this undefined since angles are only capable of being real numbers, or is there an agreed behavior for complex numbers?

    DaveE
     
  2. jcsd
  3. Aug 2, 2007 #2

    HallsofIvy

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    Even in terms of real numbers, the arguments for the functions sin(x) and cos(x) are NOT assumed to be angles!

    Do you know that eix= cos(x)+ i sin(x)?

    Then you also know that ex+ iy= ex(cos(y)+ i sin(y))
    Of course, then ex- iy= ex(cos(y)- i sin(y))

    We can, from those same formulas, derive
    cos(x)= (eix+ e-ix)/2 and
    sin(x)= (eix- e-ix)/(2i)
    While those are derived, originally, with x real, we can easily extend them as definitions for functions of complex x.

    In particular, if x= 4i+ 3, then
    cos(4i+3)= (e-4+ 3i+ e4- 3i)/2
    = (e-4(cos(3)+ i sin(3))+ e4(cos(3)- i sin(3))
    = (e-4+ e4)cos(3)/2 + i(e
    and
    sin(4i+ 3)= (e-4i+3- e-4+ 3i)/(2i)
    = (e-4(cos(3)+ i sin(3))/2i+ i(e4cos(3)+ i sin(3))
     
    Last edited by a moderator: Aug 3, 2007
  4. Aug 2, 2007 #3
    You might also want to look up the definitions of the hyperbolic trig functions, cosh and sinh. Their definitions look a lot like the ones HallsofIvy gave for cos and sin, but without the factors of i. Thus, putting an i in the argument of a sin or cos gives you a cosh or sinh, and vice versa.
     
  5. Aug 3, 2007 #4
    [tex]\sin (x+iy) = \sin x \cos iy + \cos x \sin iy[/tex]
    [tex]=\sin x \cosh y + i\cos x\sinh y[/tex]
     
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