Let $\Bbb D = \left \{z \in \Bbb C\ :\ |z| < 1 \right \}$ and $f : \Bbb D \longrightarrow \Bbb D$ be a holomorphic function having a zero of order $N$ at the origin. Then show that $|f(z)| \leq |z|^N$ for all $z \in \Bbb D.$
My attempt $:$ Since $f$ is a holomorphic function having a zero of order $N$ at the origin $\exists$ a holomorphic function $g : \Bbb D \longrightarrow \Bbb C$ with $g(0) \neq 0$ such that $f(z) = z^N g(z)$ for all $z \in \Bbb D.$ Let $h : \Bbb D \longrightarrow \Bbb C$ be a function defined by $h(z) = z g(z),$ $z \in \Bbb D.$ Then clearly $h$ is holomorphic since $g$ is so and $f(z) = z^{N - 1} h(z)$ for all $z \in \Bbb D.$ Also we have $h(0) = 0.$ So if we can show that $|h(z)| \leq 1$ for all $z \in \Bbb D$ then we are through by Schwarz lemma.
But I find it difficult to show that $|h(z)| = |zg(z)| \leq 1$ for all $z \in \Bbb D.$ Any help in this regard will be highly appreciated.
Thank you very much for your valuable time for reading.
Let $0 <t<1$. $|h(z)|=|\frac {f{(z)}} {z^{N-1}}| \leq \frac 1 {t^{N-1}}$ whenever $|z| = t$, hence for $|z| \leq t$ by MMP. From this can you conclude that $|h(z)| \leq 1$ for all $z$?