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Context, I was reading Joseph Bak and Donald J. Newman's Complex Analysis as a refresher on complex analysis (I have some prior experience but am washed) and I was attempting to understand theorem 16.13.

Suppose $f$ is an entire function of finite order. Then either $f$ has infinitely many zeroes or $$f (z) = Q(z)e^{P(z)}$$ where $Q$ and $P$ are polynomials.

In particular, the following step, let $f$ be entire and of finite order (bounded in magnitude by $A|z|^n$ for some integer $n$ and real constant $A$). The book goes on to say that if $f$ has finitely many zeros, then $f(z) = Q(z)g(z)$ where: $$Q(z):=(z-z_1)\cdots(z-z_n)$$ Where $z_1,\dots,z_n$ are the zeros of $f$ and $g$ is nonzero and holomorphic. Then it goes on to claim that $\ln(g)$ is entire. This is where the argument loses me. I understand that $g$ is nonzero, but that doesn't mean that the image of $g$ doesn't cross the branch cut of $\ln$. I'm guessing that they want to define a branch cut of $\ln$ such that this is true but I cannot understand what specific cut they are using.

Moreover, $g$, being holomorphic would attain every value (except $0$ by assumption) by Picard's Theorem, so how exactly are they defining this branch cut here? Or is there something else that I am missing? If this argument is indeed flawed, how can it be fixed?

Image of original text attached for context, page 237: page 237

Kenny Wong
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person
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1 Answers1

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You use the fundamental theorem of calculus/line integrals: Since $g$ doesn't vanish, the function $h:= g'/g$ is entire, and so there exists an entire $H$ such that $H'=h$ (by defining $H(z)= \int_0^z h(w)\, dw$, where this last is interpreted as a complex line integral over the line segment joining $0$ and $z$).

Now notice $(e^H/g)'= \frac{e^H H'g -e^H g'}{g^2}=0$, and so $e^H=cg$ for some constant $c\neq 0$ (in fact $c=1/g(0)$) that you can put in with $Q$ so that we can choose $H=P$.

Jose27
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