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Let $f$ be an entire function (holomorphic over the complex plane). If $f$ has no zero point, then $\text{Log} f$ is also an entire function. How to prove this?

My idea: one branch of $\text{Log}f$ is well-defined on $\mathbb{C}\setminus \gamma$ for any $\gamma:[0,\infty)\longrightarrow \mathbb{C}$ such that $\gamma(\infty)=\infty$. Once it is well-defined, $\text{Log}f$ is holomorphic. By the open mapping theorem, $f(\mathbb{C})$ is open in $\mathbb{C}$. To show that $\text{Log}$ is well-defined on $f(\mathbb{C})$, we only need to show that $f(\mathbb{C})$ is simply-connected. But I do not know how to continue... Is a holomorphic function with no zero point a homeomorphism?

Since my way is inconvenient and problematic, is there any otehr ways to prove:

Let $f$ be an entire function (holomorphic over the complex plane). If $f$ has no zero point, then $\text{Log} f$ is also an entire function ?

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  • $\begingroup$ Possible duplicate of this question. $\endgroup$ Feb 17, 2014 at 2:16

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We can find $\text{Log}\ f$ explicitely. Consider the function $g:\Bbb C \to \Bbb C$ such that $g(z)=f'(z)/f(z)$ since $f$ is non-zero and entire this function is also entire. Take the integral $\int_{0}^{z}g(\zeta) d\zeta$ along any path from $0$ to $z$. This is well defined since $f'/f$ is entire (hence every closed integral is zero). This integral is a branch of $\text{Log}(f)$. Proof: $$I=e^{\int_{0}^{z}g(\zeta) d\zeta}$$ $$I'=f'/f\cdot I$$ $$(I'f-If')/f^2=0$$ $$I=\text{constant}\cdot f$$ So, (this integral - some constant) is $\text{Log}\ f$

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  • $\begingroup$ Avoid the use of $*$ to denote multiplication. That's a common practice in programming, not in Mathematics, where it has other meanings. Use \cdot ($\cdot$) or \times ($\times$). $\endgroup$
    – jjagmath
    Dec 21, 2022 at 11:17
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Hint: Integrate $d\operatorname{Log} f$; that is, find a holomorphic $1$-form on $\mathbf{C}$ that "should" be $d\operatorname{Log} f$.

($\exp$ is entire, never zero, and not a homeomorphism, so your strategy, though natural, is problematic. In fact, constructing $\operatorname{Log} f$ using branches is rather inconvenient. :)

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  • $\begingroup$ since my idea is problematic, how to prove the question? $\endgroup$
    – Shiquan
    Feb 17, 2014 at 2:06
  • $\begingroup$ There's a hint in my post, but as it happens, your question has been answered here. $\endgroup$ Feb 17, 2014 at 2:18

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