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Math Help - show sin(z-bar) is not differentiable everywhere

  1. #1
    Junior Member rubix's Avatar
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    show sin(z-bar) is not differentiable everywhere

    z is a complex number; z = x+iy
    z-bar is a complex conjugate; z-bar = x-iy

    need to show: sin(z-bar) is not differentiable everywhere

    formulas (if needed):

    1) sinz = (e^(iz) - e^(-iz))/2i

    2) Cauchy Riemann equations

    f = u(x,y) + iv(x,y)

    u_x = vy; u_y = -v_x

    ----

    problem is i dunno how to express sin(z) as real u(x,y) + iv(x,y). I'm not even sure if C-R is the best way to go about solving this problem.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by rubix View Post
    z is a complex number; z = x+iy
    z-bar is a complex conjugate; z-bar = x-iy

    need to show: sin(z-bar) is not differentiable everywhere

    formulas (if needed):

    1) sinz = (e^(iz) - e^(-iz))/2i

    2) Cauchy Riemann equations

    f = u(x,y) + iv(x,y)

    u_x = vy; u_y = -v_x

    ----

    problem is i dunno how to express sin(z) as real u(x,y) + iv(x,y). I'm not even sure if C-R is the best way to go about solving this problem.
     sin( \bar z ) = sin ( x - iy)= sin(x)cos(iy) - sin(iy)cos(x)

    Let  f(x) = sin(x)cos(iy) - sin(iy)cos(x)

    Let us use the fact that  sin(iy) = i sinh(y) and  cos(iy) = i cosh(y)

     f(x) = isin(x) cosh(y) -isinh(y)cos(x)

    To me this looks like we left with the form

     f(x) = 0 + i [ sin(x)cosh(y) - cos(x)sinh(y) ]

    This would plot a straight line up the imaginary axis and could be considered in the same light as a vertical asymtote. We cannot differentiate a vertical line.

    I havent delt with imaginary numbers all that much (besides the basic principles involved with calculus, so perhaps a more indepth assesment is required). However, I think the above should suffice as proof enough.
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