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Conformal maps

  1. Dec 18, 2007 #1
    Hello!, I was studing the conformal maps in complex analysis, I don't understand this definition:

    Definition: A map [tex]f:A\rightarrow\mathbb{C}[/tex] is called conformal at [tex]z_0[/tex] if there exist a [tex]\theta\in[0,2\pi][/tex] and [tex]r>0[/tex] such that for any curve [tex]\gamma(t)[/tex] which is differentiable at [tex]t=0[/tex], for which [tex]\gamma(t) \in A[/tex] and [tex]\gamma(0)=z_0[/tex], and which satisfisfies [tex]\gamma\prime(0)\neq0[/tex] the curve [tex]\sigma(t)=f(\gamma(t))[/tex] is differentiable at [tex]t=0[/tex] and, setting [tex]u=\sigma\prime(0)[/tex] and [tex]v=\gamma\prime(0)[/tex], we have [tex]\left|u\right|=r\left|v\right|[/tex] and [tex]\arg u =\arg v + \theta (\mod 2\theta)[/tex]

    I only know about the conformal maps, that the angle between the curves after the transform is equal to the before of the transformation. But I cannot find the relation, with the definition. I think that I don't undertand the definition.

  2. jcsd
  3. Dec 19, 2007 #2
    This says that the tangent vector of the curve is scaled by some number that does not depend on the direction of the tangent vector.

    This says that f rotates the tangent vector by some constant angle.

    Suppose you had two curves [itex]\gamma_1[/itex] and [itex]\gamma_2[/itex], so that [itex]\gamma_1(0)=\gamma_2(0)[/itex], their tangent vectors [itex]u_1:=\gamma'_1(0)[/itex] and [itex]u_2:=\gamma'_2(0)[/itex], and tangent vectors of the image paths [itex]v_1:=(f\circ\gamma_1)'(0)[/itex] and [itex]v_2:=(f\circ\gamma_2)'(0)[/itex].

    The result

    \arg u_2 - \arg u_1 = \arg v_2 - \arg v_1\quad\mod\;2\pi

    comes quite easily, and that is what the angle preserving means.
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