# Conformal map

1. Oct 24, 2004

### 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 $$z_1$$?

2. Oct 25, 2004

### 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 $\alpha$.

Last edited: Oct 25, 2004
3. Oct 26, 2004

### 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 $$\infty$$)?

4. Oct 26, 2004

### 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?

5. Oct 27, 2004

### 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?