# How to show a function is holomorphic

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!

• $g$ is given by a convergent power series in $\{z:0<|z|<1\}$. This is enough to say that $g$ is holomorphic in $D$. Commented Mar 14 at 7:15
• If $\mathbb{D} = \{z\in\mathbb{C} : |z| < 1\}$, then from what I can tell, the result is false by looking at the function $f:\mathbb{D} \to \mathbb{D}$ defined by $f(z) = z$. How did you show that $|g(z)| < 1$ for all $z\in\mathbb{D}$? Commented Mar 14 at 7:18
• @AlexNguyen If $f$ is defined in that way, then you have $f(z) = z\cdot g(z)$ for each $z\in\mathbb{D}$ which implies that $g(z) = 1$ for each $z\in\mathbb{D}$. But then it is impossible that $|g(z)| < 1$ for each $z\in\mathbb{D}$. Commented Mar 14 at 7:25
• @AlexNguyen You can show that the power series of $g$ at $0$ has the same radius of convergence as the power series of $f$ at $0$. Commented Mar 14 at 7:31
• $g(z)=f(z)/z^{k}$. Doesn't that prove that $g$ is given by a convergent power series? Commented Mar 14 at 7:35

$$g$$ 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.

• It supposes to be $f(z)=z^kg(z)$, so your evaluation is false.
– user1277306
Commented Mar 14 at 8:03
• I made a correction. Commented Mar 14 at 8:36

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.