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I have the following equation:

$f(s)=g(s)f(s)$

holds for all complex number $s$.

I am little confused about this case: If $f(s)≠0$ then I can deduce directely that $g(s)=1$ as the unique case. However, when $f(s)=0$ for one and isolated $s$ (root) then $g(s)$ would be any function. This is the point. Is this case have some relations with the analytic continuation principle.

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Do you want $f(s)$ and $g(s)$ both to be analytic? –  user17762 Dec 30 '12 at 17:45
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1 Answer

up vote 1 down vote accepted

Claim: If $f(s)$ and $h(s)$ are both entire functions, then $f(s) h(s) = 0 \implies f(s) = 0 \,\,\,\,\, \forall s$ or $h(s) = 0 \,\,\,\,\, \forall s$.

Proof If $f(s) = 0$, $\forall s$ is not true, then there is a $s_1 \in \mathbb{C}$ such that $f(s_1) \neq 0$. Now since $f(s)$ is analytic, there exists a open $\epsilon$-ball around $s_1$ such that $f(s) \neq 0$ for all $s\in B(s_1, \epsilon)$. Since $f(s)h(s) = 0$, we have that $h(s) = 0$ for all $s\in B(s_1, \epsilon)$. Now we have $h$ to be analytic and $h(s) = 0$, in an $\epsilon$ ball. Hence, by analytic continuation, we have $h(s) = 0$ for all $s \in \mathbb{C}$.

In your problem, take $h(s)$ as $g(s) - 1$.

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@ Marvis: Yes, it was but I am talking about this case: "when $f(s)=0$ for one and isolated $s$ (root)". Please remark that $(x+iy)(u+iv)= ux-vy+i(uy+vx)=0$ if $ux-vy=0$ and $uy+vx=0$ without the implication of $x=y=0,u=v=0$ –  ZE1 Dec 30 '12 at 18:15
    
@RH1 The answer above is general, it doesn't matter if $s$ is a isolated root or not. –  user17762 Dec 30 '12 at 18:17
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