Let $A$ be a ragion: {$z: \lvert z-1-i \rvert <1$} $\setminus$ {$z: \lvert z \rvert \leq 1$} on complex plane. Let $f(z)=(1+i)\frac{z-i}{z-1}$ I know that $f$ map boundry of $A$ to the sum of half lines {$t \geq 0$} $\cup$ {$-it : t \geq 0$}. Moreover $f(1+i)=-i+1$. So intuitively $A$ should be mapped to the open fourth quadrant.
How to quickly prove it?
My professor says $f$ is continuos so it is obvious.
But I want to see formal mathematical argument which I am unable to provide. You can say other arguments than continuity.