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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Share to your groups: 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.
I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.
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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