Domain such that $f(z) = \dfrac{z - a}{1 - \bar {a}z}$ is analytic

Question

GIven $a\in\mathbb{C}.\space$Is there a domain such that $f(z) = \dfrac{z - a}{1 - \bar {a}z}$ is analytic?

I have tried to show it using Cauchy-Riemann equations but got stuck with the differentiation. I was also thinking to show that there exists a domain in $\mathbb{C} -\{1 - \bar {a}z=0\}$ but couldn't figure how to show it formally.

I was wondering if there's an elegant way to show or disprove it using different tools.

Thanks for the help.

Answer 1

Being holomorphic on an open set is the same thing as being $\Bbb C$-differentiable there, the subtlety being that being $\Bbb C$-differentiable at a point does not imply that the function is $\Bbb C$-differentiable (hence holomorphic) in a neighborhood of such point. As a rational function, it will be holomorphic whenever the denominator is non-zero. So the domain you're looking for is $\{z \in \Bbb C \mid 1-\overline{a}z \neq 0\}$, and $$f'(z) = \frac{1-|a|^2}{(1-\overline{a}z)^2} $$by the quotient rule.