1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Bacle2

    User Avatar
    Science Advisor

    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

    Bacle2

    User Avatar
    Science Advisor

    Yes, you're right, my bad.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cauchy-Riemann Conditions in Polar Coordinates
Loading...