Are Möbius transformations onto?

Question

I know that Möbius transformations are one to one, but are they onto ? I am not sure about it as my professor said it's not always onto but I came to know that they are invertible I am confused help . Thnx and regards

Answer 1

$$\begin{cases}Z=f(z)=\dfrac{az+b}{cz+d}&(1)\\ \text{with} \ \ ad-bc \neq 0 & (2)\end{cases} \ \ \ \ \ \ \ \text{has two "problems":}$$

• $z=-\dfrac{d}{c}$ has no image, because it would give a zero denominator in (1).

• $\dfrac{a}{c}$ has no preimage $z$ because $\dfrac{az+b}{cz+d}=\dfrac{a}{c}$ would imply $ad=bc$, which is forbidden by (2).

These two issues are removed if you decide to work in $\tilde C=\mathbb{C}\cup \{\infty\}$:

• $f\left(-\dfrac{d}{c}\right)=\infty$ (intuitive explanation: the module of the limit of a ratio when the denominator tends to zero and the numerator tends to a finite value is $\infty$).

• $f\left(\infty\right)=\dfrac{a}{c}$ (intuitive explanation: the behavior at $\infty$ of a rational function is the behavior of the ratio or its dominant terms)

In this framework, $f$ is one-to-one onto (bijective).