Let $\phi$ be a holomorphic function from the unit disk onto the upper
half-plane such that $\phi(0)=\alpha$. Give a method to find an upper
bound for $\lvert\phi ′(0)\rvert$?
To apply Schwarz's lemma, don't I just need to find a Möbius transformation that can be composed with $\phi$ so that the image of $\alpha$ under this Möbius transformation is $0$? (or am i wrong??)
Is there a general form for such Möbius transformations from the upper half-plane $H$ to the disc $D$ that sends $\alpha$ to $0$?
Best Answer
Combine the hint above with this case when $\alpha = i$:
Consider the Cayley transform $W(z) = \frac{z-i}{z+i}$ which maps the upper half plane conformally onto the unit disk. Now, $W \circ \phi (0) = W(\alpha) = 0$. Hence, $W \circ \phi$ is a conformal map of the unit disk to the unit disk that fixes $0$, so Schwarz applies. Thus $|(W \circ \phi)'(0)| \leq 1$. By the chain rule, $(W \circ \phi)' (0)= W'(\phi(0)) \phi'(0) = W'(i) \phi'(0)$. Now $W'(z) = \frac{2i}{(z+i)^2}$, so $W'(i) = -i/2$. Thus
$$|\phi′(0)| = | \frac{(W \circ \phi )' (0) }{ W'(i)} | \leq 2.$$