Show that $f(z) = \varphi(z)^n$ locally around $0$

Question

Let $U$ be an open subset of $\mathbb{C}$ which contains $0$ and $f \, : \, U \, \rightarrow \, \mathbb{C}$ an holomorphic function on $U$ such that $f(0)=0$. I want to show that :

There exist a function $\varphi$, holomorphic around $0$ and $n \in \mathbb{N}^{\ast}$ such that $f = \varphi^n$.

My attempt :

Because $f$ is holomorphic on $U$, it is analytic around $z=0$ and that means that there exist $r > 0$ such that $D(0,r) \subset U$ and

$$ \forall z \in D(0,r), \; f(z) = \sum_{k=0}^{+\infty} a_k z^k $$

where $(a_n)_{n \geq 0} \in \mathbb{C}^{\mathbb{N}}$. Let $k_0 = \min \lbrace k \geq 1, \; a_k \neq 0 \rbrace$. Then :

$$ \forall z \in D(0,r), \; f(z) = z^{k_0} g(z) $$

where :

$$ \forall z \in D(0,r), \; g(z) = \sum_{k=k_0}^{+\infty} a_k z^{k- k_0}. $$

$g$ is holomorphic on $D(0,r)$ and $g(0) = a_{k_0} \neq 0$. Since $g$ is continuous, there exist $\eta > 0$ such that $g \neq 0$ on $D(0,\eta) \subset D(0,r)$. Therefore :

$$ \forall z \in D(0,\eta), \; f(z) = z^{k_0} g(z) $$

and $g$ is nonzero on $D(0,\eta)$. Which complex logarithm shall I consider to conclude ? As long as $z \in \mathbb{C} \smallsetminus ]-\infty,0]$, I can write $z^{k_0} = \exp\big( k_{0} \operatorname{Log}(z) \big)$.

Answer 1

Note that $f(z)=a_{k_0}z^{k_0}h(z)$, with$$h(z)=1+\frac{a_{k+1}}{a_k}z+\frac{a_{k+2}}{a_k}z^2+\cdots.$$Besides, $a_{k_0}\neq0$, and therefore $a_{k_0}=e^\omega$, for some $\omega$. Now, you consider$$\log(z)=z-\frac{z^2}2+\frac{z^3}3-\frac{z^4}4+\cdots$$on a neighborood of of $1$ so small that its image under $\log$ is contained in $D(0,\eta)$. Then$$f(z)=z^{k_0}\exp\left(\omega+\log\bigl(h(z)\bigr)\right).$$

Answer 2

The way I am inclined to look at this one is to note that since $g(z)$ is holomorphic in $D(0, \eta)$, so is $g'(z)$, and since

$g(z) \ne 0, \; z \in D(0,\eta) \tag 1$

the function

$\dfrac{g'(z)}{g(z)} \tag 2$

is also holomorphic in $D(0,\eta)$. For $z \in D(0, \eta)$ and $\gamma:[0, 1] \to D(0, \eta)$ any path in $D(0, \eta)$ with $\gamma(0) = 0$ and $\gamma(1) = z$, we may define the function

$G(z) = \displaystyle \int_{0, \gamma(t)}^z \dfrac{g'(z)}{g(z)}dz; \tag 3$

that is, $G(z)$ is given by the path integral of $g'(z) / g(z)$ along $\gamma(t)$; then $G(z)$ is holomorphic in $D(0, \eta)$ as well, and we have

$(g(z)\exp(-G(z))' = g'(z)\exp-(G(z)) + g(z)(-\dfrac{g'(z)}{g(z)}) \exp(-G(z))$ $= g'(z) \exp(-G(z)) - g'(z) \exp(-G(z)) = 0; \tag 4$

it follows that $g(z) \exp(-G(z))$ is constant in $D(0, \eta)$; hence there is some $c \in \Bbb C$ with

$g(z) \exp(-G(z)) = c, \tag 5$

or

$g(z) = c \exp(G(z)); \tag 6$

from (3) we see that

$G(0) = 0, \tag 7$

whence

$g(0) = c \exp(G(0)) = c \exp(0) = c(1) = c; \tag 8$

thus (6) becomes

$g(z) = g(0) \exp(G(z)); \tag 9$

now since $g(0) \ne 0$ there is a unique $\theta \in [0, 2\pi)$ with

$g(0) = \vert g(0) \vert \exp(i\theta); \tag{10}$

from (10) it follows that, setting

$\rho_j = \vert g(0) \vert^{1/k_0} \exp \left(i\dfrac{\theta + 2 \pi j}{k_0} \right )= \exp \left (\dfrac{\ln \vert g(0) \vert}{k_0} \right )\exp \left(i\dfrac{\theta + 2 \pi j}{k_0} \right ), \; 0 \le j < k_0, \tag{11}$

that

$\rho_j^{k_0} = \left ( \exp \left (\dfrac{\ln \vert g(0) \vert}{k_0} \right )\exp \left(i\dfrac{\theta + 2 \pi j}{k_0} \right ) \right )^{k_0}$ $= \exp \left (k_0\dfrac{\ln \vert g(0) \vert}{k_0} \right )\exp \left(i k_0 \dfrac{\theta + 2 \pi j}{k_0} \right )$ $= \exp( \ln \vert g(0) \vert) \exp \left(i (\theta + 2 \pi j \right )) = \vert g(0) \vert \exp(i \theta) \exp(2 j \pi i) = g(0), \; 0 \le j < k_0, \tag{12}$

since $\exp(2 j \pi i) = 1$. If we thus define, for $0 \le j < k_0$,

$h_j(z) = \rho_j \exp \left (\dfrac{G(z)}{k_0} \right ), \tag{13}$

then

$h_j^{k_0}(z) = \rho_j^{k_0} \left( \exp \left (\dfrac{G(z)}{k_0} \right ) \right )^{k_0} = g(0) \exp \left ( k_0 \dfrac{G(z)}{k_0} \right ) = g(0) \exp (G(z)) = g(z). \tag{14}$

Setting

$\phi_j(z) = z h_j(z), \tag{15}$

we have

$f(z) = z^{k_0} g(z) = z^{k_0} h_j^{k_0}(z) = \phi_j^{k_0}(z). \tag{16}$

We have thus established the existence of $k_0$ holomorphic functions $\phi_j(z)$ on $D(0, \eta)$, each satisfying

$f(z) = \phi_j^{k_0}(z). \tag{17}$

We now observe that we have defined the $\phi_j(z)$ without explicit reference to the complex logarithm function $\ln(z)$, having circumvented the need to do so via the construction of $G(z)$, (3)-(9); since $G(z)$ is single-valued in $D(0, \eta)$, the question of which branch of $\ln z$ to adopt is instead (perhaps tacitly) addressed by allowing all roots $\rho_j$ of $g(0)$ to come into play, each defining a similar but distinct $h_j(z)$ on $D(0, \eta)$; we thus see there are $k_0$ functions $\phi_j(z)$ all satisfying (17). If one takes the present path, the need to select a branch of $\ln z$ may be side-stepped in favor of the selection of one of the $\rho_j$, $0 \le j < k_0$.