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Math Help - Cauchy-Riemann equations in polar form

  1. #1
    MHF Contributor arbolis's Avatar
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    Cauchy-Riemann equations in polar form

    I'm totally stuck on this exercise. I'd like a hint rather than a full answer.

    Let r and \theta be the polar coordinates in the plane (complex I guess?), let f(z) be a function of the complex plane in itself. (I guess they mean f:\mathbb{C} \to \mathbb{C}.)
    Using the fact that f(re^{i\theta})=R(r,\theta)e^{i\Theta (r, \theta)} in which R(r,\theta) and \Theta (r,\theta) are real differentiable functions of r and \theta, show that the Cauchy-Riemann equations in polar coordinates are written as \frac{\partial R}{\partial r}=\frac{R \partial \Theta}{r \partial \theta} and \frac{\partial R}{r \partial \theta}=-R \frac{\partial \Theta}{\partial r}.

    I must also show another equation, but I'll ask help if I get stuck.


    I don't know how to start the problem. I know the Cauchy Riemann equations, but I don't think I should start by writing them down.
    Last edited by arbolis; August 20th 2009 at 03:55 PM.
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by arbolis View Post
    I'm totally stuck on this exercise. I'd like a hint rather than a full answer.

    Let r and \theta be the polar coordinates in the plane (complex I guess?), let f(z) be a function of the complex plane in itself. (I guess they mean f:\mathbb{C} \to \mathbb{C}.)
    Using the fact that f(re^{i\theta})=R(r,\theta)e^{i\Theta (r, \theta)} in which R(r,\theta) and \Theta (r,\theta) are real differentiable functions of r and \theta, show that the Cauchy-Riemann equations in polar coordinates are written as \frac{\partial R}{\partial r}=\frac{\partial R}{r \partial \theta}=-R \frac{\partial \Theta}{\partial r}.

    I must also show another equation, but I'll ask help if I get stuck.


    I don't know how to start the problem. I know the Cauchy Riemann equations, but I don't think I should start by writing them down.
    I would first find differential operators linking

    \frac{ \partial }{\partial x} and \frac{ \partial }{\partial y} to \frac{ \partial }{\partial r} and \frac{ \partial }{\partial \theta}

    then identify u and v from

    u + i v = f(re^{i\theta})=R(r,\theta)e^{i\Theta (r, \theta)}

    and then substitute everything into the CR equations and your new equations in polar form should merge.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Danny View Post
    I would first find differential operators linking

    \frac{ \partial }{\partial x} and \frac{ \partial }{\partial y} to \frac{ \partial }{\partial r} and \frac{ \partial }{\partial \theta}

    then identify u and v from

    u + i v = f(re^{i\theta})=R(r,\theta)e^{i\Theta (r, \theta)}

    and then substitute everything into the CR equations and your new equations in polar form should merge.
    Ok thanks. By the way I made a typo in the equations, I'll edit it right now.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Danny View Post
    I would first find differential operators linking

    \frac{ \partial }{\partial x} and \frac{ \partial }{\partial y} to \frac{ \partial }{\partial r} and \frac{ \partial }{\partial \theta}

    then identify u and v from

    u + i v = f(re^{i\theta})=R(r,\theta)e^{i\Theta (r, \theta)}

    and then substitute everything into the CR equations and your new equations in polar form should merge.
    I did something like this for homework in my complex analysis class. The question was asked differently (but the final result was to give the polar form of the C-R equations).

    @ arbolis: Look at the attachment at your own discretion. I have the whole thing worked out...but it may be different from what you're supposed to do.
    Attached Files Attached Files
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