Conformal map

[T-O7]

Okay,
I'm having trouble understanding the following:
Given a region of a circular wedge with endpoints a and b, the mapping z_{1}=\frac{z-a}{z-b} transforms this wedge into an angular sector. Then, by an appropriate power \alpha, the map w = z_{1}^\alpha maps the angular sector onto a half plane. How exactly does this wedge turn into a nice angular sector just by z_1? :confused:

[ReyChiquito]

Im not sure what your region is, an image could help...

The frist transformation is making a go to 0 and b go to \infty. That means that all arcs passing trough b 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 a and b, there you have the angular sector.

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

The angle has been widen by \alpha.

Sorry for bad english

[T-O7]

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 \gamma_1 and \gamma_2) of the circular wedge that connect a and b 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 a and at b (i.e at 0 and \infty), but why straight lines? And how do you 'interpret' the angle preserving behaviour of \gamma_1 and \gamma_2 at the image of b(i.e at \infty)?

[T-O7]

Okay, so i think i may have somewhat understood what you were saying before, ReyChiquito. If the map z_1 is a Mobius transformation, then it must map circles to circles (considering lines as circles too). Since the edge curves \gamma_1 and \gamma_2 of the circular wedge is part of a circle, their images under z_1 must also be circles (or lines). But it can't be a real circle, otherwise it wouldn't map b to \infty, so it must be a straight line.

Is this half-logical thinking? :redface:

[ReyChiquito]

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?