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Cauchy-Riemann Equations

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that u(x,y) and v(x,y) satisfy the Cauchy-Riemann Equations.


    x = r cos(Θ) y= r sin(Θ)



    2. Relevant equations

    Cauch-Riemann sum
    ∂u/∂x = ∂v/∂y

    3. The attempt at a solution


    My questions are:
    1. Do I convert the x and y values to rectangular co-ordinates first ?


    2. is u(x,y) equivalent to u(r cos(Θ)+r sin(Θ)) u(x^2+y^2) or u=R^2 ?

    regards
    Brendan
     
  2. jcsd
  3. Mar 31, 2009 #2

    CompuChip

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    I don't get the question at all. What are u and v? Why are you using polar coordinates?
     
  4. Mar 31, 2009 #3
    I have to take the partial derivatives of the functions u(x,y) and v(x,y).
    I have been told that x = r cos(Θ) y= r sin(Θ)
    So I suppose I have to find the partial derivatives of: u(x,y) and v(x,y).

    And they should equal each other if they are Cauchy-Riemann Equations.

    ∂u/∂x = ∂v/∂y

    My problem is I don't know how to represent both the functions u(x,y) and v(x,y) using

    x = r cos(Θ) y= r sin(Θ)

    To start and find the partial derivatives
     
  5. Mar 31, 2009 #4

    HallsofIvy

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    You can't do that if you don't know what u and v are! There is nothing in this that tells us what functions u and v are!


    This is just x and y in polar coordinates. Nothing is said about u and v.



    x and y are rectangular coordinates!


    2. is u(x,y) equivalent to u(r cos(Θ)+r sin(Θ)) u(x^2+y^2) or u=R^2 ?[/quote]
    u(x,y)= u(rcos(Θ), rsin(Θ)) of course. But even if you had "u(x^2+y^2)" that would not say that u was equal to x^2+ y^2.

    Surely somewhere in that problem, or perhaps in an earlier problem, u and v are defined!
     
  6. Mar 31, 2009 #5
    Sorry Guys,
    My mistake The functions for Cauchy-Riemann Equations are:


    u(x,y)= excos(y)
    v(x,y) = exsin(y)

    Your were right they were in a previous question.
     
  7. Apr 1, 2009 #6

    CompuChip

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    Good. Now you can simply calculate
    ∂u/∂x = ∂v/∂y
    right? I don't see why you would want to go to polar coordinates for this?
    u directly depends on x, and v directly depends on y, so you can just take the partial derivatives... :confused:

    OK, so I suppose the point is that once you show that u(x, y) and v(x, y) satisfy the Cauchy-Riemann equations, you have proven that
    [tex]f(x, y) = e^x \left( \cos(y) + \mathrm i \sin(y) \right) = e^{x + i y}[/tex]
    is complex differentiable...
     
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