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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.

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  • $\begingroup$ There is a well-known theorem that a non-zero holomorphic function on a simply-connected domain has a holomorphic logarithm. $\endgroup$
    – Martin R
    Jul 6 '19 at 9:09
  • $\begingroup$ 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 $\endgroup$
    – Masacroso
    Jul 6 '19 at 9:09
  • $\begingroup$ @MartinR In what domain is the log function holomorphic? $\endgroup$
    – J. Doe
    Jul 6 '19 at 9:11
  • $\begingroup$ Or this one:math.stackexchange.com/q/1862115/42969. $\endgroup$
    – Martin R
    Jul 6 '19 at 9:13
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    $\begingroup$ 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. $\endgroup$
    – Wojowu
    Jul 6 '19 at 9:16
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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 $e^{g(z)}=f(z)$.

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  • $\begingroup$ 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 $\endgroup$
    – Conrad
    Jul 6 '19 at 12:10
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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 $(\forall z\in\mathbb C):f(z)=e^{h(z)-k}$.

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  • $\begingroup$ 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. $\endgroup$ Apr 27 '20 at 11:51

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