Algebra of the complex plane

Question

Suppose $a$,$b$,$c$,$d$ be the points of set of all complex no$(C)$. with $c$ not equal to $0$ and $ad$ not equal to $bc$.$f$ be a function such that $f(z)$=$(az+b)$/$(cz+d)$. how to prove that $f$ defines a bijection between $C$-{d/c} and $C$-{a/c}? how to show that it takes circles into circles?What about circles through $(-d/c)$?

Answer 1

We can use the fact that every Mobius map can be expressed as the composition of four operations: inversion, rotation, scaling, translation, each of which is bijective, and therefore invertible --rotate by inverse amount, rescale by $1/k$ (opposite of original scaling factor), invert again (using the fact that $\frac {1}{1/z}=z$).

EDIT: More specifically, every Mobius map $$\frac {az+b}{cz+d} $$ , can be expressed as a composition of four operations (in this order; note that in some cases, any of these may be trivial, e.g., rotating by $0$ and stretching by a factor of $1$):

1)Translation: $z+d/c$

2)Inversion: $1/z$

3)Stretching/Rotating : $(\frac {bc-ad}{c^2})z$

4)Another translation : $z+a/c$

Now, this shows bijectivity: each operation is invertible:

4') Translate by $-a/c$

3')Stretch and rotate by inverse amounts.

2') $\frac {1}{1/z}=z$

1')Translate by $-d/c$.

You can also see each of these operations sends circles to circles:

I)Clear for translations; just shifts each point by the same amount.

ii) Inversion $1/z$ sends $e^{i\theta} \rightarrow e^{-i\theta}$ ;

iii)Stretching by $k$ sends $re^{i\theta} \rightarrow rke^{i \theta}$ , and,

iv) Rotation by $\theta'$ just sends : $ re^{ i \theta} \rightarrow e^{i (\theta+ \theta')}$

Answer 2

The fraction $f(z)=\frac{az+b}{cz+d}$ can be written as

$$f(z)=\frac{\frac{-ad}{c^2}}{z+\frac{d}{c}}+\frac{a}{c}$$ So we have $$f=t_{\frac{-a}{c}} \circ r_{\arg{(\frac{-ad}{c^2})}} \circ d_{|\frac{ad}{c^2}|} \circ v \circ t_{\frac{-d}{c}}$$ with the following meaning: $$t_u(z)=z-u$$ $$v(z)=\frac{1}{z}$$ $$d_\lambda(z)=\lambda z, \lambda \in \mathbb{R^+}$$ $$r_\varphi(z)=ze^{i\varphi}, \varphi \in \left[0,2\pi\right[$$

The translation $t$, the rotation $r$ and the dilation $d$ are bijections on $\mathbb{C}$, the inversion $v$ is a bijection on $\mathbb{C \setminus \{0\}}$. All of them take circles into circles.

$v$ takes lines through the origin into lines though the origin. All others takes lines into lines. So lines through $\frac{-d}{c}$ are taken into lines through $\frac{-a}{c}$