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Conformal map

  1. Oct 24, 2004 #1
    I'm having trouble understanding the following:
    Given a region of a circular wedge with endpoints [tex]a[/tex] and [tex]b[/tex], the mapping [tex]z_{1}=\frac{z-a}{z-b}[/tex] transforms this wedge into an angular sector. Then, by an appropriate power [tex]\alpha[/tex], the map [tex]w = z_{1}^\alpha[/tex] maps the angular sector onto a half plane. How exactly does this wedge turn into a nice angular sector just by [tex]z_1[/tex]? :confused:
  2. jcsd
  3. Oct 25, 2004 #2
    Im not sure what your region is, an image could help...

    The frist transformation is making [itex]a[/itex] go to [itex]0[/itex] and [itex]b[/itex] go to [itex]\infty[/itex]. That means that all arcs passing trough [itex]b[/itex] will be straight lines. Given that it is conformal (is a Möbius transform), preserves angles. So the anlge in wich the two curves cut will preserve, and if the other curve passes trough [itex]a[/itex] and [itex]b[/itex], there you have the angular sector.

    The sedcond one is easy to see, remember that a complex variable can be written in the form [itex]z=Re^{i\theta}[/itex]. Consequently

    The angle has been widen by [itex]\alpha[/itex].

    Sorry for bad english
    Last edited: Oct 25, 2004
  4. Oct 26, 2004 #3
    Hmm....okay. I think I understand the second map now. For the first one though, I'm still a little unclear about why the two "boundary" curves (say [tex]\gamma_1[/tex] and [tex]\gamma_2[/tex]) of the circular wedge that connect [tex]a[/tex] and [tex]b[/tex] are mapped to straight lines. How do we know they don't get mapped onto non-straight lines? I understand that they must preserve the angle between each other at the image of [tex]a[/tex] and at [tex]b[/tex] (i.e at 0 and [tex]\infty[/tex]), but why straight lines? And how do you 'interpret' the angle preserving behaviour of [tex]\gamma_1[/tex] and [tex]\gamma_2[/tex] at the image of [tex]b[/tex](i.e at [tex]\infty[/tex])?
  5. Oct 26, 2004 #4
    Okay, so i think i may have somewhat understood what you were saying before, ReyChiquito. If the map [tex]z_1[/tex] is a Mobius transformation, then it must map circles to circles (considering lines as circles too). Since the edge curves [tex]\gamma_1[/tex] and [tex]\gamma_2[/tex] of the circular wedge is part of a circle, their images under [tex]z_1[/tex] must also be circles (or lines). But it can't be a real circle, otherwise it wouldn't map b to [tex]\infty[/tex], so it must be a straight line.

    Is this half-logical thinking? :redface:
  6. Oct 27, 2004 #5
    That would be correct. Think of the transformation in the Riemann Sphere. What does a circle that passes trough N looks like in the C-plane?
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