# Why is $\log(f(z))$ entire if $f(z)\ne0$ and $f$ is entire? [duplicate]

Let $$f:\mathbb{C}\to\mathbb{C}\setminus\{0\}$$ be an entire function. Why is $$\log(f(z))$$ entire?

I don't understand the answer because if we have log with any branch $$B=\{Re^{i\theta}:R\geq0\}$$, (and assume we choose for example the principal branch, $$\theta=0$$), then it may be that $$f(z_0)\in B$$ for some $$z_0\ne0$$ and then $$\log(f(z_0))$$ is not defined.

• There is a well-known theorem that a non-zero holomorphic function on a simply-connected domain has a holomorphic logarithm. Jul 6 '19 at 9:09
• I dont think it can be holomorphic in $\Bbb C\setminus\{0\}$, the complex logarithm have a half-line of discontinuity in the complex plane, for any chosen branch cut Jul 6 '19 at 9:09
• @MartinR In what domain is the log function holomorphic? Jul 6 '19 at 9:11
• Or this one:math.stackexchange.com/q/1862115/42969. Jul 6 '19 at 9:13
• I would like to address the downvotes - neither of the questions linked as duplicates addresses OP's problem, which appears to be confusion regarding what the notation $\log(f(z))$ is supposed to mean. Jul 6 '19 at 9:16

There is a certain subtlety regarding what $$\log(f(z))$$ means. It is not true that for any nonvanishing $$f$$ there is a branch of logarithm for which $$\log(f(z))$$ is defined everywhere (indeed, for e.g. $$f(z)=e^z$$, which is onto $$\mathbb C\setminus\{0\}$$, that would imply existence of a branch of logarithm defined everywhere apart from zero, which you probably know is impossible.

What is true, and what that statement implicitly means, is that for a nonvanishing $$f$$ there exists a function $$g(z)$$, which we can denote by $$\log(f(z))$$, such that $$e^{g(z)}=f(z)$$.

• one more subtle thing is that $\arg{f}=\Im{g}$ is an unbounded (both positively and negatively by Liouville) harmonic function in this case (assuming $f$ non constant of course), so actually $\log(f)$ goes through infinitely many branches of the general complex logarithm multi-valued function (or the general argument multi-valued function), making choices by analytic continuation Jul 6 '19 at 12:10

You have good reasons to find the question unclear. However, here is anothor way of stating it:

Let $$f\colon\mathbb C\longrightarrow\mathbb C\setminus\{0\}$$ be an entire function. Why is there an entire function $$g\colon\mathbb C\longrightarrow\mathbb C$$ such that$$(\forall z\in\mathbb C):e^{g(z)}=f(z)?$$

Just take a primitive $$h$$ of $$\frac{f'}f$$. It is not hard to prove that $$\frac{e^h}f$$ is constant. So, there is a $$k\in\mathbb C$$ such that $$(\forall z\in\mathbb C):\frac{e^{h(z)}}{f(z)}=e^k$$, and therefore $$(\forall z\in\mathbb C):f(z)=e^{h(z)-k}$$.

• If $f$ is entire without zeros then there exists $g$ such that $\exp g=f$. That is, there exists a holomorphic logarithm of $f$. For a proof of this fact you may consult Theorem 2.2.g on page 202 of Conway's book. However, $\log |f|$ is not a holomorphic function! The modulus makes it real and $\log$ is the real logarithm. The function $\log |f|$ is harmonic and it is the real part of the holomorphic function $\log f$. The fact that $\log |f|$ is harmonic can be checked by direct calculations based on the fact $f$ satifies the Cauchy-Riemann equations. Apr 27 '20 at 11:51