Mobius Transformations and Geometric Interpretations

moo5003
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Homework Statement


a) Show that you can split any transformation into a Translation, Dilation-Rotation, Inversion, and Translation.

b) Show using part a) that any straight line or circle is send to a straight line or circle when applying the mobius transformation.



Homework Equations



Mobius transformation is of the form:
(az+b)/(cz+d)
WHERE:
ad-bc != 0



The Attempt at a Solution


I believe I finished part a (mostly) though I'm a little unsure how to do the simple case.

First if c!=0 then we can split up the equation (az+b)/(cz+d)

into:

(a/c) - (ad-bc)/c^2 * 1/(z+d/c)
Thus
T_4(z) = z + (a/c)
T_3(z) = z * -(ad-bc)/c^2
T_2(z) = 1/z
T_1(z) = z + (d/c)

Where (az+b)/(cz+d) = T_4(T_3(T_2(T_1(z))))

If c = 0 then d!=0 and the mobius transformation takes the form:
(az+b)/d = (a/d)z + (b/d) which is just a rotation-dilation and a translation. Though I have no clue how to show this as 4 different functions (one of which being an inversion). I'm thinking that's its ok for me to leave my answer for part a as is, though if anyone has any hints on finding 4 functions for this then by all means don't hesitate to tell me to keep looking.

b) This is where I'm having trouble.

Translations don't change the geometric structure so I'm not going to discuss them.

Inversions send z to z_bar / |z|^2 ~ Ie: Flips z over the real axis and then scales them to 1/|z|.

I can see how this may send a straight line to a half circle.. though I don't see how it could possibly encompass an entire disk.

Rotation-Dilation rotates z by some theta and then scales it by some radius.

Basically I'm a little lost on how to start on these, any hints you can provide would be appreciated.
 
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Your part a) is fine. When c=0, you can't decompose the mobius transformation in the way they suggest. For b) what part don't you get? You seem to have the right idea for translations and rotation-dilations. It's just inversions, right?
I can see how this may send a straight line to a half circle.. though I don't see how it could possibly encompass an entire disk.
Why?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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