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If I have define a Möbius transformation as "a map on the extended complex plane, $\bar{\mathbb{C}} \rightarrow \bar{\mathbb{C}}$, given by $\omega = \frac{az + b}{cz + d}$ where $a,b,c,d \in \mathbb{C}$ and $ad - bc \neq 0$", then can I define an affine Möbius transormation as

"A map in the extended complex plane, $\bar{\mathbb{C}} \rightarrow \bar{\mathbb{C}}$, given by $\omega = az + d$, i.e

$$\left\{ \infty \rightarrow \infty, a \neq 0 : \omega = \frac{1z + b}{cz + 1}\right\}$$

Is this correct?

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The mapping $z \mapsto \frac{1 \cdot z+b}{c \cdot z+1}=:w$ is in general no affine one (for example take $b:=0$, $c:=1$, then $w = \frac{z}{z+1}$ and this is obviously not an affine mapping.)

The transformation $w = \frac{a \cdot z+b}{c \cdot z+d}$ is affine iff $c=0$, because in this case $$w= \frac{a}{d} \cdot z+ \frac{b}{d}$$ (and $d \not= 0$ since $a \cdot d - 0 \not=0$). This shows that affine mappings are a special case of Möbius transformations.

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$z\mapsto w=az+b$ is a Moebius transformation with only $\infty$ as fixed point. It is called in general a similarity (it is a composition of rotation, expansion/contraction and translation, not necessarily all of them).

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  • $\begingroup$ That's the definition of an affine Mobius transformation? $\endgroup$ – Kaish Dec 29 '12 at 13:02
  • $\begingroup$ I don't know if it is called affine but this is just a name. $\endgroup$ – Sigur Dec 29 '12 at 13:04

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