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Math Help - Holomorphic function in poloar coordinates?

  1. #1
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    Holomorphic function in poloar coordinates?



    Okay, so if f(z) = u(x,y) + iv(x,y), then in polar coordinates, since z = \rho e^i\theta , then...

    f(z) = f(\rho e^i\theta ) = u(\rho ,\theta ) + iv(\rho ,\theta )?
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  2. #2
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    Quote Originally Posted by davismj View Post


    Okay, so if f(z) = u(x,y) + iv(x,y), then in polar coordinates, since z = \rho e^i\theta , then...

    f(z) = f(\rho e^i\theta ) = u(\rho ,\theta ) + iv(\rho ,\theta )?
    After thinking about it, I realized that while the functions u and v would be dependent on \rho and \theta, the real and imaginary parts would both be dependent on \rho and \theta, and therefore it's probably more accurate to represent them thus:

    f(z) = f(e^i\theta ) = u(\rho cos \theta ,\rho sin \theta) + iv(\rho cos \theta ,\rho sin \theta)

    However, taking the limit of [f(x+h,y) - f(x,y)]/h in the purely real or imaginary sense still isn't getting me anywhere, or is it?
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  3. #3
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    Quote Originally Posted by davismj View Post
    After thinking about it, I realized that while the functions u and v would be dependent on \rho and \theta, the real and imaginary parts would both be dependent on \rho and \theta, and therefore it's probably more accurate to represent them thus:

    f(z) = f(e^i\theta ) = u(\rho cos \theta ,\rho sin \theta) + iv(\rho cos \theta ,\rho sin \theta)

    However, taking the limit of [f(x+h,y) - f(x,y)]/h in the purely real or imaginary sense still isn't getting me anywhere, or is it?
    \frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} =\frac{u_{r}}{\sin \theta} + i \frac{v_{r}}{\sin \theta}

    i \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} = - i \frac{u_{\theta}}{r \sin \theta} + \frac{v_{\theta}}{r \sin \theta}

    hence,

    r u_r = v_\theta

    r v_r = -u_\theta
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  4. #4
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    Quote Originally Posted by dedust View Post
    \frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} =\frac{u_{r}}{\sin \theta} + i \frac{v_{r}}{\sin \theta}

    i \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} = - i \frac{u_{\theta}}{r \sin \theta} + \frac{v_{\theta}}{r \sin \theta}

    hence,

    r u_r = v_\theta

    r v_r = -u_\theta
    Thanks so much!
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