show sin(z-bar) is not differentiable everywhere

• Jul 6th 2010, 05:08 PM
rubix
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.
• Jul 7th 2010, 09:21 AM
AllanCuz
Quote:

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):

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

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.