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Homework Help: Cauchy-Riemann Conditions in Polar Coordinates

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Using f(z) = f(re^iθ) = R(r,θ)e^iΩ(r,θ), show that the Cauchy-Riemann conditions in polar coordinates become

    ∂R/∂r = (R/r)∂Ω/∂θ

    2. Relevant equations

    Cauchy-Riemann in polar coordinates
    Hint: Set up the derivative first with dz radial and then with dz tangential

    3. The attempt at a solution

    df/dz = (∂R/∂r)(∂r/∂z)e^iΩ + R(∂Ω/∂θ)(∂θ/∂z)e^iΩ

    Now, I have no idea what dz tangential should be. I'm guessing that I should set the radial df/dz equal to the tangential df/dz, but I have no idea about the tangential or if my radial is right. Functionals are confusing to me, and complex functionals even more so.
  2. jcsd
  3. Jan 25, 2012 #2
    I was originally correct, but my logic was flawed. One should take the derivative of the functional with respect to z. Then, by the product rule, you have two terms. When R is constant -- the tangential derivative -- one term is eliminated and vice versa. Set the two derivatives equal to each other, since it is an analytic functional. Then, your dz should have two terms as well, a real and imaginary component; You can eliminate some terms from your equality using this result and the chain rule.
  4. Sep 1, 2012 #3
    where i can find the solution of

    Using f(z) = f(re^iθ) = R(r,θ)e^iΩ(r,θ), show that the Cauchy-Riemann conditions in polar coordinates become ∂R/∂r = (R/r)∂Ω/∂θ
  5. Sep 1, 2012 #4


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    If you know the coordinate change between standard reals and polars,it then comes

    down to using the chain rule from f(x,y) to f^(r,θ)
  6. Sep 2, 2012 #5
    Yes, you use the chain rule for both derivatives, but you take the derivative once with R constant, and again with Ω constant. You should now be left with two expressions: one is imaginary, and one is real. Drop the i, and set the two equal together and manipulate to get the answer. Sorry if I wasn't clear seven months ago.

    Looking at my original reply, I think it was pretty clear, actually.
    Last edited: Sep 2, 2012
  7. Sep 2, 2012 #6


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    Science Advisor

    Yes, you're right, my bad.
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