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Math Help - Complex Directional Derivative.

  1. #1
    Senior Member vincisonfire's Avatar
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    Complex Directional Derivative.

    I have the following definition of the complex directional derivative :
     D_{w_1} f(z) = lim_{t \rightarrow 0} \frac{f(z+tw_1)-f(z)}{t} where  |w_1| = 1 .
     D_{w_1} f(z) = lim_{t \rightarrow 0} \frac{u(x+tx_1,y+ty_1)-u(x,y)}{t}+i\frac{v(x+tx_1,y+ty_1)-v(x,y)}{t}
    From the definition of real directional derivative
     D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d  x}+y_1\frac{dv}{dy})
    But now if  w_1 = i ,  x_1 = 0 ,  y_1 = 1 and  D_{w_1} f(z) = \frac{du}{dy}+i \frac{dv}{dy} which does not agree with Cauchy Riemann equations.
    Could someone tell me where I am mistaken please?
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  2. #2
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    The directional derivative does not need to "agree with" the Cauchy–Riemann equations. In fact, it has nothing to do with them. For the directional derivative to exist, the function does not even have to be analytic. For example, the complex conjugate function f(z) = \bar{z} is not analytic and so does not satisfy the C–R equations, but it has directional derivatives in every direction.
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  3. #3
    Senior Member vincisonfire's Avatar
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    Thanks I think I understand why.
    So is <br /> <br />
D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d  x}+y_1\frac{dv}{dy})<br />
correct, if I may ask?
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  4. #4
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    Quote Originally Posted by vincisonfire View Post
    Thanks I think I understand why.
    So is <br /> <br />
D_{w_1} f(z) = x_1\frac{du}{dx}+y_1\frac{du}{dy}+i(x_1\frac{dv}{d  x}+y_1\frac{dv}{dy})<br />
correct, if I may ask?
    Yes (the derivatives should be partials, of course).
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  5. #5
    Senior Member vincisonfire's Avatar
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    Yes I didn't take the time to put the partial symbols.
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