Let $f$ be an analytic function that maps upper half plane to itself. Also, let $f(i)=i, \ f(2i)=\frac{i}{2}$. Then, what can be said about $|f(1+i)|$?
Now, by using the Schwarz-pick lemma and its corollary ineqaulity, we may observe that $f$ is a Mobius transformation with positive determinant. But then, the problem arises as to whether the value of $|f(1+i)|$ can be determined precisely. This is because, three points are at the least required to completely determine the Mobius transformations. So how do we determine $|f(1+i)|$? Thanks beforehand.
Best Answer
Consider $g=\phi \circ f \circ \phi^{-1}$ where $\phi(z)= \frac{z-i}{z+i}$ is an isomorphism $\Bbb{H\to D}$ sending $i$ to $0$.
So $g$ maps $\Bbb{D\to D}$ to itself and $g(0)=0, g(1/3)=g(\phi(2i)=\phi(f(2i))=\phi(i/2)=-1/3$.
Schwarz lemma gives that $g(z)=-z$.
Therefore $f(z) = -1/z$.