Prove if $g$ is holomorphic on $V$ and non zero then it's $k$ -th power of other holomorphic function

Question

Prove if $g(z)$ is holomorphic on $V$(such that $V$ is some open subset of $\Bbb{C}$) and non-zero over it.

Then it's $k$ -th power of other holomorphic function for any $k\ge 1$.That is exist some holomorphic $h(z)$, such that $g(z) = (h(z))^k$ .

Answer 1

You cannot prove it, since it is false. Suppose that $V=\Bbb C\setminus\{0\}$, and that $g(z)=\frac1z$. Then there is no holomorphic function $h\colon V\longrightarrow\Bbb C$ such that $g=h^2$, because the order of $g$ at $0$ (which is $-1$) is twice the order of $h$ at $0$.

Answer 2

The correct statement is as follows:

Let $V$ be a simply connected domain and $f:V \rightarrow \mathbb{C^*}:=\mathbb{C} \setminus \{0 \}$ be a holomorphic function on $V$. Let $n\geq 1$. Then $\exists$ a holomorphic function $h:V \rightarrow \mathbb{C^*}$ such that $(h(z))^n=f(z)$.

Proof: Since $f$ is non-vanishing, $\frac{f'}{f}$ is holomorphic on $V$. By Cauchy's Theorem, for any closed curve $\gamma$ on $V$(which is simply connected), we have $\int_{\gamma} \frac{f'}{f} = 0$. So By the Fundamental theorem of Calculus, $\exists$ an antiderivative $g$ of $\frac{f'}{f}$ on $V$(g is unique upto variation by a constant).

Let $z_o \in V$ and $w_o \in \mathbb{C^*}$ be such that $f(z_o)=w_o$. Let $g$ be picked in such a manner that $e^{g(z_o)}=w_o$ (constant of g also fixed now).

Notice that $\dfrac{d}{dz} (\dfrac{e^{g(z)}}{f(z)}) =0$. This along with $e^{g(z_o)}=w_o$ gives us $e^{g(z)}=f(z)$.

Define $h(z)=e^{g(z)/n}$ which is a holomorphic function on $V$ (Since exp is an entire function and g is holomorphic on V). This satisfies $(h(z))^n=f(z)$ as required.