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Math Help - cauchy riemann equation

  1. #1
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    cauchy riemann equation

    Hi guys: just needed some verification on my working out for the following problem:

    Question:
    Using Cauchy Riemann prove that the function h(z)=sin(Imz) is not differentiable at any point of the strip \{ z:\frac{{ - \pi }}<br />
{2} < {Im} z < \frac{\pi }<br />
{2}\}

    ------------------------------------------------------------------
    My Solution: Please check to see whether it is correct or not.
    let z=x+iy
    h(z)=\sin (Im \,\, z) = \sin y
    u(x,y)= \sin y and v(x,y)=0


    \frac{\delta u}{\delta x}=0 ... \frac{\delta v}{\delta x}=0


    \frac{\delta u}{\delta y}=cos y, ... \frac{\delta v}{\delta y}=0

    for CR conditions to be fulfilled, u_x = v_y and v_x = -u_y

    0 = 0 but
     0 \neq -cos y

    both the CRE conditions aren't fulfilled so the function is not differentiable at any point of the strip
    ------------------------------------------------------------------

    is this correct?
    thanks
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  2. #2
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  3. #3
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    Well the C-R equations are satisfied for multiples of y = \frac  {\pi} {2}.
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  4. #4
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    hi perfecthacker, that was a continual of your last post on that other thread. couldn't find that thread so asked it here.
    i've attempted it here. any suggestions? thanks
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  5. #5
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    Quote Originally Posted by Plato View Post
    Well the C-R equations are satisfied for multiples of y = \frac  {\pi} {2}.
    thanks. yes i see what you mean. cos(pi/2) = 0 in which case the CR are satisfied.

    any suggestions then on how to go about with the problem?
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  6. #6
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    Quote Originally Posted by mathfied View Post
    thanks. yes i see what you mean. cos(pi/2) = 0 in which case the CR are satisfied.any suggestions then on how to go about with the problem?
    Well that is the answer.
    There is no solution for -\frac {\pi} {2} < y < \frac {\pi} {2}.
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  7. #7
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    im confused here. the question says show that the function h(z) is not differentiable at any point of the range.

    but we have just proved that the CR equations are satisfied in the range specified and therefore the function is infact differentiable.

    this contradicts what the question is asking me to prove???

    does the fact that cos y = 0 (for the range specified for y) mean that it is not differentiable?
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  8. #8
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    I do not think that you fully understand my reply!
    There are places where the S-R apply!
    But none of them is in  -\frac {\pi} {2}<y<\frac{\pi}{2}.
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  9. #9
    Super Member wingless's Avatar
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    A function is differentiable at a point if and only if the Cauchy-Riemann equations hold at that point.
    (For continuous u_x, u_y, v_x and v_y)

    Where do the Cauchy-Riemann Equations hold?
    Do they hold at any points in the given range?
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