A simple example related to Schwarz' lemma Let $f$ be a holomorphic function on $B(0,R)$, where $R>0$.  Assume that there exists an $M>0$ such that $\forall z\in B(0,R): |f(z)| \le M$ and a natural number $n$ such that
$$ 0 = f(0)=f'(0) = ... = f^{(n)}(0).$$
$1)$ Prove that $\forall z\in B(0,R) : |f(z)| \le M \left( \frac{|z|}{R} \right)^{n+1}$ with equality iff there exists an $\alpha \in \mathbb{C}$ such that $|\alpha|=1$ and $ f(z)=\alpha M \left( \frac{z}{R} \right)^{n+1}$.
$2)$ Assume that either $ |f(z_0)|=M \left(\frac{|z_0|}{R}\right)^{n+1}$ for some $z_0 \in B(0,R) \setminus \{0\}$ or $ |f^{(n+1)}(0)|=(n+1)!M / R^{n+1}$. Then $ f(z)=\alpha M \left(\frac{z}{R}\right)^{n+1}$ for an $\alpha \in \mathbb{C}$ such that $|\alpha|=1$.
Sorry for asking this problem, but I'd like to see some examples related to the Schwarz lemma. Thanks!
 A: Here's a solution:
1) Consider the function
$$
g(z) := \frac{f(z)}{z^{n+1}}.
$$
Due to the constraints on the derivative, we see from the Taylor series of $f$ that $g$ is a holomorphic function at $0$ (at other points, it's holomorphic anyway). Therefore, if $r < R$, we can apply the maximum modulus principle to $g$, obtaining
$$
|g(z)| \le \frac{M}{r^{n+1}} \Leftrightarrow |f(z)| \le M \left( \frac{|z|}{r} \right)^{n+1}
$$
and $r \to R$ gives the first part of the result.
Now if there is an $\alpha \in S^1$ such that $f(z) = \alpha M \left( \frac{z}{R} \right)^{n+1}$, then equality will hold. On the other hand, if equality holds at some point, $g$ attains a maximum at that point, which must be in the interior if $B_R(0)$ since $B_R(0)$ is open. Hence, the maximum modulus principle implies that $g$ is constant, and solving for $f$ gives the second part of the result, proving also the first point of 2).
2) We observe that $g(0) = \frac{f^{(n+1)}(0)}{(n+1)!}$ (the constant coefficient of the Taylor expansion), and then if $f^{(n+1)}(0) = (n+1)!M/R^{n+1}$, $g$ will attain a maximum at $0$; indeed, the above argument gives that $g \le M/R^{n+1}$ as $r \to R$. Hence, the maximum modulus principle will again imply that $g$ is constant.
