Q: Suppose that $f(z)$ is analytic in the unit disc $D = \lbrace z: |z|<1 \rbrace$ and continuous in the corresponding closed disc $\overline{D}=\lbrace z: |z| \leq 1 \rbrace$. Suppose also that $f(z)/z^5$ can be extended to be analytic analytic in all of D (including the origin). If $|f(z)| \leq 4$ in $\overline{D}$, what is the maximal value that $|f(0.6i)|$ can assume under these conditions?
I am having some trouble with this problem. The answer i have gotten is $|f(0.6i)|\leq 4\cdot(0.6^5)$, but this is returned as incorrect. Am I missing a way to get a better bound with these conditions?
The question makes no sense ! Let $n \in \mathbb N$ and consider $f_n(z)=nz^5.$
Then $|f_n(0.6i)|=n (0.6)^5$
Consequence: $|f(0.6i)|$ can made arbitrayly big.