### CSIR JUNE 2011 PART C QUESTION 81 SOLUTION (Analytic function with $f(0) = \frac{1}{2}$ and $f(\frac{1}{2}) = 0$)

Let $f$ be an analytic function from the unit disc $\Delta$ to $\Delta$ with $\Delta = \{z \in \Bbb C : |z| < 1\}$ satisfying $f(0) = \frac{1}{2}$ and $f(\frac{1}{2}) = 0$.  Which of the following is/are true?
1. $|f^{'}(0)| \le \frac{3}{4}$,
2. $|f^{'}(\frac{1}{2})| \le \frac{4}{3}$.
3. option 1 and 2 both are true.
4. $f$ is the identity function.

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Solution:
Schwarz's pick lemma:
Let $f$ be an analytic function from the unit disc $\Delta$ to $\Delta$. Further if $f(a) = b$ then we have $|f^{'}(a)| \le \frac{1-|f(a)|^2}{1-|a|^2}$.
Given that $f(0) = \frac{1}{2}$. This implies that $$|f^{'}(0)| \le \frac{1 - (\frac{1}{2})^2}{1-(0)^2} = \frac{3}{4}.$$
Given that $f(\frac{1}{2}) = 0$. This implies that $$|f^{'}(\frac{1}{2})| \le \frac{1 - (0)^2}{1-(\frac{1}{2})^2} = \frac{4}{3}.$$ So option 1, 2 and 3 are true.
option 4: (False) Let $f(z) =frac{az + b}{cz + d}$. Now, $f(0) = \frac{1}{2}$ implies that $2b = d$ and we have $f(z) = \frac{az+b}{cz+2b}$. Now, $f(\frac{1}{2}) = 0$ implies that $a = -2b$. Let us take $a = 2, b= -1, c= 1$ and $d = 2$, then all these conditions are met and $ad-bc \ne 0$. Therefore the function $$f(z) = \frac{2z - 1}{z - 2}$$ satisfies the given conditions of the problem which is different from the identity function.
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