# logarithm of an entire function

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 ?

• Possible duplicate of this question. Feb 17, 2014 at 2:16

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 $$\text{Log}\ f$$
• 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 ($\times$). Dec 21, 2022 at 11:17
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. :)