Showing $f(z) = \frac{z-a}{1-\bar{a}z}$ is a Moebius Transformation Given Conditions

Question

As the title suggests, I want to show that $f(z) = \frac{z-a}{1-\bar{a}z}$ where $z,a \in \mathbb{C}$ is a Moebius transformation with $|a|<1$ and $1-a\bar{a} \neq 0$

I'm uncertain as to how to start.
Any help would be appreciated.
Thanks.

Answer 1

A Möbius transformation $T$ is most simply defined as a nonconstant fractional linear transformation: the latter means $$ T(z) = \frac{az+b}{cz+d}. $$ To see what nonconstant needs, let $z$ and $w$ be distinct points. Then a calculation shows $$ T(z) - T(w) = \frac{(ad-bc)(z-w)}{(cz+d)(cw+d)}, $$ so if $z \neq w$, $T(z) \neq T(w)$ if and only if $ad-bc \neq 0$. So in fact, $T$ is injective if and only if it is nonconstant.

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So, since your function is already of the form $(az+b)/(cz+d)$, you just have to check that $ad-bc \neq 0$.