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Suppose $f,g$ are entire functions, such that $\frac{f}{g}$ is also entire. Can I conclude that the set of zeros of $g$ is contained in the set of zeros of $f$? I think the answer is yes, otherwise $\frac{f}{g}$ would have singularities. Do you agree with me?

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Yes.{}{}{}{}{}{} –  mrf Jan 10 '13 at 19:54

2 Answers 2

up vote 2 down vote accepted

Let us restate the condition "$\frac fg \textrm{is also entire}$": this means that there is a entire function $h$ such that $gh=f$. Now it is clear that every zero of $g$ is also a zero of $f$.

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Certainly,

Let $a$ be a zero of g, and expand $f$ and $g$ in talyor series centered at $a$. Then,

$g=(z-a)^k\cdot h$ for some $k>0$ such that $h$ is entire and $h(a)\not=0$

$f=b_0+b_1(z-a)+\cdots$

Now, $h\cdot \frac{f}{g}=\frac{b_0}{(z-a)^k}+\frac{b_1}{(z-a)^{k-1}}+\cdots$

We know $h\cdot\frac{f}{g}$ is entire so, $b_0=0,\ldots,b_{k-1}=0$

Thus, $f=b_k(z-a)^k+b_{k+1}(z-a)^{k+1}+\cdots=(z-a)^k\cdot \phi$ where $\phi$ is entire.

Overall if $g$ has a zero of order $k$ then $f$ must have a zero of at least order $k$.

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