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Math Help - Cauchy Riemann Equations

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    Cauchy Riemann Equations

    In deriving the necessary conditions for a function  f(x) = u(x,y) + iv(x,y) to be differentiable at a point  z_0 , why do we get  f'(z_0) = v_{y}(x_{0}, y_{0}) - iu_{y}(x_{0}, y_{0}) ? Would not it be  f'(z_0) = u_{y}(x_{0}, y_{0}) + iv_{y}(x_0, y_0) ?
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    Because the Cauchy Riemann equations are:

    \color{red}\frac{\partial{u}}{\partial{x}} = \frac{\partial{v}}{\partial{y}}

    \color{red}\frac{\partial{v}}{\partial{x}} = -\frac{\partial{u}}{\partial{y}}

    i\frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{y}}

    \frac{\partial{f}}{\partial{\bar{z}}} = 0 where \bar{z} is the complex conjugate of z = x + iy.

    Where u and v are the real and imaginary parts of:

    f(x + iy) = u(x,y) + iv(x,y)
    Last edited by Aryth; January 3rd 2009 at 06:34 AM.
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    Hello,
    Quote Originally Posted by Aryth View Post
    Because the Cauchy Riemann equations are:

    \color{red}\frac{\partial{u}}{\partial{x}} = \frac{\partial{v}}{\partial{y}}

    \color{red}\frac{\partial{v}}{\partial{x}} = \frac{\partial{u}}{\partial{y}}

    i\frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{y}}

    \frac{\partial{f}}{\partial{\bar{z}}} = 0 where \bar{z} is the complex conjugate of z = x + iy.

    Where u and v are the real and imaginary parts of:

    f(x + iy) = u(x,y) + iv(x,y)
    No, the second one is \frac{\partial v}{\partial x}={\color{red}-} \frac{\partial u}{\partial y}
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    Super Member Aryth's Avatar
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    You're right... I edited it.
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    Quote Originally Posted by manjohn12 View Post
    In deriving the necessary conditions for a function  f(x) = u(x,y) + iv(x,y) to be differentiable at a point  z_0 , why do we get  f'(z_0) = v_{y}(x_{0}, y_{0}) - iu_{y}(x_{0}, y_{0}) ? Would not it be  f'(z_0) = u_{y}(x_{0}, y_{0}) + iv_{y}(x_0, y_0) ?
    Fundamentally, the answer is the "chain rule". Since z= z+ iy, \frac{dv}{dz}= \frac{\partial v}{\partial y}\frac{\partial y}{\partial z} and \frac{\partial y}{\partial z}= \frac{1}{\frac{\partial z}{\partial y}}= \frac{1}{i}= -i

    Crucial point: 1/i= -i.

    Another way of looking at the Cauchy-Riemann equations is this: With f(z)= f(x+ iy)= u(x,y)+ iv(x,y), the derivative, at z= z_0, is the given by the limit \lim_{h\rightarrow 0} \frac{f(z_0+h)- f(z_0)}{h} just as for real numbers. And, just like limits for real numbers, in order that the limit exist, we must get the same result approaching z_0 along any path. In particular, if we approach z_0 along a line parallel to the real axis, h is real so we have
    \lim_{h\rightarrow 0}\frac{u(x_0+h,y_0)+ iv(x_0+ h,y_0)- u(x_0,y_0)- iv(x_0,v_0)}{h}
    \lim_{h\rightarrow 0}\frac{u(x_0+h,y_0)- u(x_0,y_0)}{h}+ \lim_{h\rightarrow 0}\frac{iv(x_0+h,y_0)- iv(x_0,y_0)}{h}
    [tex]= \frac{\partial u}{\partial x}+ i\frac{\partial v}{\partial u}

    Approaching instead along a line parallel to the imaginary axis, h is imaginary and we can use "ih", with i real, instead. Now we have
    \lim_{h\rightarrow 0}\frac{u(x_0,y_0+ h)+ iv(x_0,y_0+h)- u(x_0,y_0)- iv(x_0,y_0)}{ih}
    \lim_{h\rightarrow 0}\frac{u(x_0, y_0+ h)- u(x_0, y_0)}{ih}+ \lim_{h\rightarrow 0}\frac{v(x_0,y_0+ h)- v(x_0,y_0)}{ih}
    and the important difference is that "i" in the denominator. It will cancel "i" in the v part and remember that 1/i= -i so
    = -i\lim_{h\rightarrow 0}\frac{u(u_0,y_0+h)- u(x_0,y_0)}{h}+ i\lim_{h\rightarrow 0}\frac{v(x_0,y_0+h)- v(x_0,y_0)}{h}
    = -i\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial y}

    In order that the limit exist, those two must be the same. Equating real and imaginary parts,
    \frac{\partial u}{\partial x}= \frac{\partial v}{\partial y} and
    \frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}
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    Quote Originally Posted by HallsofIvy View Post
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    It's abuse of this wonderful site! Admins - you listening?
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