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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 $0$?

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Do you know a transform $f$ suitable for $\alpha=i$? Then $z\mapsto f(\frac{z-\Re\alpha}{\Im\alpha})$ should do the trick. –  Hagen von Eitzen Jan 16 '13 at 8:38

1 Answer 1

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

$$|\phi′(0)| = | \frac{(W \circ \phi )' (0) }{ W'(i)} | \leq 2.$$

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nice solution :) so the main strategy should be to play the game on $D\to D$ –  Une Femme Douce May 3 '13 at 8:24

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