[Math] Doubt about the statement of the Fundamental Theorem of Algebra.

abstract-algebraanalysiscalculuscomplex numbers

I have a doubt, not about the proof, but about the nomenclature of a theorem.

I'm researching some interesting applications of the Inverse Function Theorem and found a version of the Fundamental Theorem of Algebra, the proof of which applies that theorem. The version is as follows:

Fundamental Theorem of Algebra: If $p:\mathbb{C} \to \mathbb{C}$ is the nonconstant polynomial defined by
$$p(z) = a_0 + a_1z + \ldots + a_nz^n,$$
where $a_n \neq 0$ and $n \geq 1$, then $p$ is surjective. In particular, there exists $z_0 \in \mathbb{C}$ such that $p(z_0) = 0$.

I found this version a bit "strange" as it looks weaker than the following:

Fundamental Theorem of Algebra: If $p:\mathbb{C} \to \mathbb{C}$ is a nonconstant complex polynomial defined by
$$p(z) = a_0 + a_1z + \ldots + a_nz^n,$$
where $a_n \neq 0$ and $n \geq 1$, then the equation $p(z) = 0$ has $n$ solutions, not necessarily distinct.

Is the first version I listed correct? Are they equivalent? To me they don't seem to be, the first seems to be weaker than the second.

Taking advantage of the question, I would like to ask for suggestions for the application of this theorem (of Inverse Function in the case).

Best Answer

While the second version is logically stronger than the first version, it's not hard at all to get the second from the first: divide $p(z)$ by $z-z_0$, and it can be shown that the quotient is still a polynomial and then apply the first version to the quotient, until the quotient is a constant. This step is completely trivial compared to the hardness of establishing the first version.

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