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Finding the image: a 1/z transformation

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the image of the region x > 1, y > 0 under the transformation w = 1/z. The answer is given in the book: (u - 1/2)^2 + v^2 < (1/2)^2, v < 0


    2. Relevant equations
    A(x^2 + y^2) + Bx + Cy + D = 0
    D(u^2 + v^2) + Bu - Cv + A = 0.
    (u,v) = (x/(x^2 + y^2), v = -y/(x^2 + y^2)



    3. The attempt at a solution
    A = 0 because this is essentially just a collection of lines (that do not pass through the origin, hence D not equaling 0). I chose x = 1 and y = 0 (planning on introducing the inequality later) in order to solve for the constants. I plugged these values into (u,v), then those values into the second equation. Moving things around didn't help much, and I ended up with D = -B and B = B. I couldn't get a definite value for my constants. How should I go about solving for the constants?
     
  2. jcsd
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