How to show a function is holomorphic

Question

Let $f:\mathbb{D}\to\mathbb{D}$ be a holomorphic function s.t $0$ is a zero of order $k\geq 1$. Prove that $f(z)=z^kg(z)$, in which $g:\mathbb{D}\to\mathbb{D}$ is holomorphic.

My attempt:
Since $f$ is holomorphic on $\mathbb{D}$, then by Taylor's expansion we have $$f(z)=\sum\limits_{n=0}^{\infty}c_nz^n,\quad c_n=\dfrac{f^{(n)}(0)}{n!}.$$ Since $0$ is zero of order $k\geq 1$, we have $c_1=...=c_{k-1}=0, c_k\neq 0$.
$\Rightarrow\quad f(z)=z^kg(z)$, in which $g(z)=c_k+c_{k+1}z+...$
My question is how can I show such $g$ is holomorphic on $\mathbb{D}$ and $|g(z)|<1$? Could someone help me? Thanks in advance!

Answer 1

$g$ is holomorphic because it has a power series. Holomorphic is equivalent to analytic. $c_k+c_{k+1}z+\dots $ is a power series.

The inequality $1\ge\mid f(z)\mid={\mid g(z)\mid}{\mid z^k\mid}\mid=\mid g(z)\mid$ holds for $\mid z\mid=1.$

Now apply the maximum modulus principle to get the inequality on the whole disk.

Answer 2

I will assume that $\mathbb{D} = \{z\in\mathbb{C} : |z| \leq 1\}$. This can be done by induction on $k$. Suppose first that $k=1$. Define the function $g:\mathbb{D} \to \mathbb{C}$ by

$$g(z) := \begin{cases} \frac{f(z)}{z} & \text{ if } z\in \mathbb{D} \setminus \{0\}, \\ f'(0) & \text{ if } z\in \{0\}. \end{cases}$$

Then $g$ is a holomorphic function, because if $f$ has a convergent power series expansion of the form $f(z) = \sum_{n=0}^{\infty}c_{n}z^{n}$, $g$ has a convergent power series expansion of the form $g(z) = \sum_{n=0}^{\infty}c_{n+1}z^{n}$.

By Schwarz's lemma, $|g(z)| \leq 1$ holds for all $z\in\text{int}(\mathbb{D})$, and by continuity it holds for all $z\in\mathbb{D}$. So it may be assumed that $g:\mathbb{D} \to \mathbb{D}$, and consequently $g$ is the desired function.

Now suppose the result has been proved for $k\in\mathbb{N}$. Suppose that $g$ has a zero of order $k+1$ at $0$. By the above argument, there is some holomorphic $h:\mathbb{D} \to \mathbb{D}$ such that $f(z) = z\cdot h(z)$ for all $z\in\mathbb{D}$. $h$ has a zero of order $k$, so by the induction assumption there is some holomorphic $g:\mathbb{D} \to \mathbb{D}$ such that $h(z) = z^{k}\cdot g(z)$ for all $z\in\mathbb{D}$. As $f(z) = z^{k+1} \cdot g(z)$ for all $z\in\mathbb{D}$, it follows that $g$ is the desired function. The desired result follows by induction.