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):

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

2) Cauchy Riemann equations

$\displaystyle f = u(x,y) + iv(x,y)$

$\displaystyle u_x = vy; u_y = -v_x$

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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.

Quote:

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Originally Posted by rubix

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):

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

2) Cauchy Riemann equations

$\displaystyle f = u(x,y) + iv(x,y)$

$\displaystyle u_x = vy; u_y = -v_x$

----

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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$\displaystyle sin( \bar z ) = sin ( x - iy)= sin(x)cos(iy) - sin(iy)cos(x) $

Let $\displaystyle f(x) = sin(x)cos(iy) - sin(iy)cos(x) $

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

$\displaystyle f(x) = isin(x) cosh(y) -isinh(y)cos(x) $

To me this looks like we left with the form

$\displaystyle 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.