This is a complex analisys exercise I came across while studying.
Let $\Omega \subset \mathbb{C}$ open such that $\overline{B_1(0)} := \{ z \in \mathbb{C} : |z| \le 1\} \subseteq \Omega$ and $f$ holomorphic on $\Omega$ such that: $f(0) = 1$, $ f(z) \not = 0$ for $z : |z| = 1$ and $|f(z)| \le 5$ for $z \in \overline{B_1 (0)}$. Find an upperbound to the number of zeros of $f$ in $B_{1/2}$ in such a way that it is optimal.
What I thought is that I want to find an $m \in \mathbb{N}$ such that for $f$ with above hypothesis $n(1/2) \le m$ and there exists an $f_0$ as above with exactly $m$ zeros.
My solution:
Without pasting the whole calculations, I found an upperbound to be $\frac{\ln(5)}{\ln(2)}$ as done on this answer which is a more general form of the problem. In particular, $$1 < \frac{\ln(5)}{\ln(2)} < 2$$ which means that the number of zeros in $\overline{B_{1/2}(0)}$ is $$n(1/2) \le 1 .$$ So This should be the best approximation if we are able to find an $f_0$ as above with exactly $1$ zero. In particular it would look like $f_0(z) = 4z + 1$ does the job: satisfies the hypotesys and has exactly one root which coincidentally is $-1/4 \in \overline{B_{1/2} (0)}$.
I posted this question mainly to check that I understood the assignment well and also to ask if the condition $ f(z) \not = 0$ for $z : |z| = 1$ was used at all.
