# Question on maximal value of $f(0.4 + 0.5i)$ subject to certain constraints. Use of maximum principle?

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!

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

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 $g(z)\equiv 6$.
• 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. Commented Mar 10, 2015 at 8:30