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Let $f$ be an analytic function in the upper half-plane with $|f(z)| <1$. Now if $f( \iota )=0$ then find the maximum possible value of $|f(2 \iota)|$. Clearly Reflection principle is not working as real values at real is not given. Then how do you find its maximum value? No idea is working. I think the function will be constant. Please suggest how to find this value?

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  • $\begingroup$ Transform the domain to the unit circle and use Schwarz's Lemma. $\endgroup$ – Chappers May 18 '15 at 11:18
  • $\begingroup$ how to transform ? can you help a little more. $\endgroup$ – neelkanth May 18 '15 at 11:20
  • $\begingroup$ There are bi linear transformations that transform the upper to unit disc.. $\endgroup$ – neelkanth May 18 '15 at 11:38
  • $\begingroup$ then how to use the given function? $\endgroup$ – neelkanth May 18 '15 at 11:47
  • $\begingroup$ please some one help me to solve this problem... $\endgroup$ – neelkanth May 18 '15 at 13:01
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The transformation $T(z) = \frac{z-i}{z+i}$ sends the upper half-plane $H$ to the unit disc $D$. Hence $g := f \circ T^{-1} : D \to D $. Also, $g(0) = f(i) = 0$, so $g$ satisfies the conditions of Schwarz's lemma. Hence $\lvert g(w) \rvert \leqslant \lvert w \rvert $, for all $w \in D$. Since $T$ is a bijection, it is easy to see that $T(2i)=1/3$ and hence $$ \lvert f(2i) \rvert = \lvert f(T^{-1}(1/3)) \rvert = \lvert g(1/3) \rvert \leqslant 1/3. $$ (You can show equality occurs by setting $f$ up so that $g(w) = w$.)

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  • $\begingroup$ nicely explained....thanking you.... $\endgroup$ – neelkanth May 18 '15 at 13:31
  • $\begingroup$ @ chappers thanking you.... $\endgroup$ – neelkanth May 18 '15 at 13:33

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