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The function $f(z)$ is analytic in the unit disk $U = {z:|z|<1}$ and continuous in the closed unit disk. Suppose that $\frac{f(z)}{z^2}$ can be extended to be analytic in the (open) unit disk U (also at the origin). If $|f(z)| \leq 6$ in the closed unit disk , what is the maximal possible value for $f(0.4 + 0.5i)$ ?

I guess I should somehow be using the maximum modulus principle or the Cauchy estimates, but I'm not sure how. Grateful for any help with this!

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  • $\begingroup$ Is there a typo with the arguments for $f$? It currently has 3 arguments. $\endgroup$
    – msteve
    Commented Mar 9, 2015 at 20:43
  • $\begingroup$ Not quite sure what you mean... $\endgroup$
    – Jarvi79
    Commented Mar 9, 2015 at 20:48
  • $\begingroup$ What complex number are you plugging into $f$? Currently, your input is the tuple $(0,4+0,5i)$. $\endgroup$
    – msteve
    Commented Mar 9, 2015 at 22:19
  • $\begingroup$ Well, I want to plug in the complex number 0.4 + 0.5i. $\endgroup$
    – Jarvi79
    Commented Mar 9, 2015 at 22:24
  • $\begingroup$ Sorry, in Finland we use , instead of . Any thoughts on the problem? $\endgroup$
    – Jarvi79
    Commented Mar 9, 2015 at 22:26

1 Answer 1

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Yes, you should apply the maximum principle to the function $g(z) = f(z)/z^2$. Since $|g|\le 6$ on the boundary, $|g(0.4+0.5i)|\le 6$. Put this back in terms of $f$ to get the desired upper bound.

To demonstrate its sharpness, arrange $f$ so that $g(z)\equiv 6$.

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  • $\begingroup$ Okey, is $|g| \leq 6$ on the boundary because by maximum principle $f(z)$ (and $g(z)$) takes its maximum value on the boundary and $|g(z)| = |\frac {f(z)}{z^2}| = \frac {|f(z)|}{|z|^2} \leq \frac {6}{1^2} = 6$? I get: $|g(0.4 + 0.5i)| \leq 6 \rightarrow \frac {|f(0.4 + 0.5i)|}{|z|^2} \leq 6 \rightarrow |f(0.4 + 0.5i)| \leq 6|z|^2 \rightarrow |f(0.4 + 0.5i)| \leq 6(0.4^2 + 0.5^2) = 2.46$ Is that right? Very grateful for feedback. $\endgroup$
    – Jarvi79
    Commented Mar 10, 2015 at 8:30

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