Inverse of Möbius transform - w(z)?

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Homework Statement



I want to know how I could extract T from Tz in matrix form so I can
get T^{-1} to get the inverse of w(z).

w(z)=Tz=\frac{az+b}{cz+d}

Homework Equations





The Attempt at a Solution

 
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You can't. The moebius transform is not linear and only linear transforms can be written as matrices.
 
Thanks.

So how how do I solve it if I don't want to remember the answer z(w)=\frac{-dw+b}{cw-a}?
 
You invert it like you invert any other function. If w=(az+b)/(cz+d), solve that equation for z using algebra.
 
Of course, I am to tired now I should go to sleep :) Thank you.
 
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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