The given function is $$f(z)=e^{i\psi}\cdot{{z-a}\over {1-\bar az}}$$. Now this is a mobious transformation. I know every mobious transformation maps circles to circles. But the question is to show that it maps the unit disk $\{z\in \mathbb C:|z|\le 1\}$ onto itself. Now by searching this site I found out that The Mobious transformation that maps the unit disk onto itself is of the form $f(z)=e^{i\psi}\cdot{{z-a}\over {1-\bar az}}$with $|a|\le 1,\psi\in[0,2\pi].$ and that is proved too.
But how do I proceed when I have to go the other way round , that is the map is given and its image is to be found.One thing I understand is that if I show that the $|f(z)|\le 1$ then its image is proven the unit disk.
Now $$|f(z)|\le1\\\implies |f(z)|^2\le 1\\\implies f(z)\cdot\bar{f(z)}\le 1\\\implies {{z-a}\over {1-\bar az}}\cdot{\bar{{z-a}\over {1-\bar a z}}}\le 1$$ Got stuck here. With the information that both $|z|\le 1,|a|\le 1$ I have to prove that the last inequality holds.Do I explicitely have to consider $a=b+ic,z=x+iy$ and proceed? I think there is another shorter way of doing that last equation. Help pls. Thanks.
EDIT: The other post that has been linked to here is the one I mentioned reading before in my post. That one find out the function from the given characteristic. But I have been given the function to find out some characteristics.