# Möbius transform example explanation

If a Möbius transform is a map that goes from the extended complex plane to the extended complex plane, given by some $\omega = \frac{az + b}{cz + d}$, where $ad - bc \neq 0$. In my notes, underneath the definition, I have:

Ex: $\frac{1z + 2}{2z + 4} = \frac{1}{2}$.

How does this work? Isn't ad - bc = 0? Or have I made some kind of mistake in writing down the notes properly? Where does the fraction come from?

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The example illustrates why we want $ad - bc\neq 0$ since we would like these to not be constant functions. – Tobias Kildetoft Dec 13 '12 at 18:08
I think the "ex" is probably the reason that we require $ad-bc\neq 0$ - you get problematic cases if $ad-bc=0$, and this is an example of why it is problematic. (Another example, $c=d=0$ is obviously problematic.) – Thomas Andrews Dec 13 '12 at 18:08
Where does the 1/2 come into it then? – Kaish Dec 13 '12 at 18:10
$2z+4=2(1z+2)$, so $\frac{1z+2}{2z+4}=\frac{1}{2}$ – Thomas Andrews Dec 13 '12 at 18:10

As per your definition, the map is not a Möbius transform. As for how it's $1/2$? $2z+4=2(1z+2)$.