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Is is possible to generalize the fundamental theorem of algebra to allow for any function $f(x)$ that is entire in the complex plane (or for polynomials of infinite order)?

I am interested in the following equation:

$$f(x) = 0$$

How many roots in the complex plane does this have? If $f(x)$ is entire, then can I represent it as a convergent series in $x$ like this:

$$f(x) = \sum_{n=0}^\infty a_n x^n = 0$$

Can I then view $f(x)$ as a polynomial of (countably) infinite order? So then can I use the standard fundamental theorem of algebra to say this has a countably infinite number of roots in the complex plane?

QuantumDot
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    http://math.stackexchange.com/questions/27546/zeroes-of-holomorphic-function – Hoot Jul 02 '16 at 17:13
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    How many roots has $e^x$? – quid Jul 02 '16 at 17:14
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    @quid how about an infinite number of them, all at $x=-\infty$? :D – QuantumDot Jul 02 '16 at 17:16
  • This is an excellent opportunity for you to look again at the proof(s) of the fundamental theorem of algebra, and see why they do not work for non-polynomials. – Greg Martin Jul 02 '16 at 18:01
  • @QuantumDot $;x=-\infty;$ is not a complex number , so that doesn't work. Perhaps you could some some kind of generalization to the complex plane union the point $;-\infty;$ , but maybe some things would get weird there... – DonAntonio Jul 02 '16 at 23:44

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