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Complex numbers: Conformal mapping

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I have seen a conformal mapping of z = x+iy in MAPLE, and it consists of horizontal and vertical lines in the Argand diagram (i.e. the (x,y)-plane).

    On the Web I have read that a conformal map is a mapping, which preserves angles. My question is how this mapping is constructed? I.e., how is it that I find that the conformal mapping of z = x+iy is horizontal and vertical lines?

    Regards,
    Niles.
     
  2. jcsd
  3. Apr 20, 2009 #2

    marcusl

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    To understand conformal mapping, you need to have studied complex variables at least through the properties of an analytic function. Analytic functions are the only ones that possess conformal maps (conformal meaning that angles between two curves are the same in both original and mapped geometries). Since all books on complex variables also cover conformal mapping, I presume that you haven't had a chance to study this topic yet. I recommend starting with Churchill's "Complex Variables" which is written very clearly.
     
  4. Apr 20, 2009 #3

    Dick

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    What do the contours of the lines x=constant and y=constant in the x-y plane look like in the z=x+iy Argand plane? This is a pretty easy conformal mapping.
     
  5. Apr 21, 2009 #4
    I do know about analytic functions and the Cauchy-Riemann conditions.


    So I just have to plot Re(z) = constant and Im(z)=constant, and then I get the conformal mapping?
     
  6. Apr 21, 2009 #5

    Dick

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    Not quite. If z=x+iy and f(z)=u(x,y)+iv(x,y), then mapping that takes the point (x,y) in the plane to (u(x,y),v(x,y)) is the conformal mapping. It's the same as taking z->f(z) in the Argand plane.
     
  7. Apr 21, 2009 #6
    Ok, so if we take f(z) = z = z+iy, then we have (x,y) -> (u,v) = (x,y). How is it that this results in constant curves?
     
  8. Apr 21, 2009 #7

    Dick

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    If you plot, say u(x,y)=constant in the xy plane the resulting curve will map to a vertical line in the uv plane, right? v(x,y)=constant corresponds to a vertical line in uv. Try it with f(z)=z^2. f(z)=z is almost too easy to convey the idea clearly. You should get two sets of orthogonal hyperbolae.
     
  9. Apr 21, 2009 #8
    For f(z)=z^2 we have that u(x,y) = x^2-y^2 and v(x,y) = 2xy.

    I want to plot this in (u,v), so I insert u(x,y) in v(x,y) and get

    v(x,y)=2*sqrt[u(x,y)+y^2]*y.

    Is this correct?
     
  10. Apr 21, 2009 #9

    Dick

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    Nooo. A line of constant u, u=C corresponds to x^2-y^2=C in the xy plane right? That's a hyperbola, right?
     
  11. Apr 21, 2009 #10
    Ok, a few issues:

    1) When you say (x,y)-plane, do you mean the Argand-diagram or the usual (x,y)-coordinate system?

    2) Ok, so the reason why we equal u and v to constants is because we want the contours (atleast I think this is the reason).

    So u(x,y) gives us 1 hyperbola, and 2xy = constant is the other one?
     
  12. Apr 21, 2009 #11

    Dick

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    1) either one. It doesn't make much difference. They both look alike. 2) u and v are CONSTANT because those are horizontal and vertical lines in the uv plane. You want to find the corresponding curves in the xy plane. Yes, 2xy=constant is the other. Except they aren't just '1 hyperbola'. There is a different hyperbola for every different choice of the constant.
     
  13. Apr 21, 2009 #12
    What I meant was that in the Argand diagram x and y are part of z, but in the normal (x,y), then x is input and y is output. I was just wondering if I had to isolate y or something when doing this, but I guess it is apparent what the curves look like (like the hyperbola).
     
  14. Apr 21, 2009 #13

    Dick

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    No, don't think of y as a function of x. They are both independent variables. w is a function of z.
     
  15. Apr 21, 2009 #14
    Ok, great. Thanks for helping - I really appreciate it.
     
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