Is this an equivalent statement to the Fundamental Theorem of Algebra? Is the following equivalent to the usual statement of the fundamental theorem of algebra:



Let $$f(z)=c_nz^n+\cdots+c_1z+c_0$$

be a polynomial with complex coefficients. For all but finitely many $w \in \mathbb C$, $f(z)-w$ has $n$ distinct roots in $\mathbb C$.



This seems different to just saying that $f(z)$ has, including multiplicities, $n$ roots. Because this statement does not leave the possibility that there could be finitely many points $w\in\mathbb C$ such that $f(z)=c_nz^n+\cdots+c_1z+(c_0-w)$ will not have $n$ roots. Any insight is appreciated!
A polynomial is separable (has distinct roots) if it shares no zeroes with its formal derivative. If $ P(X) - w $ is a polynomial of degree $ n $ for $ w $ a constant, then its formal derivative $P'(X) $ is a polynomial of degree at most $ n-1 $ and does not depend on $ w $. Let the distinct zeroes of the formal derivative be $ z_1, z_2, \ldots, z_r $ where $r \leq n-1 $; then $ P $ can only be inseparable if it has one of these as a root. But this requires $ w \in \{P(z_1), P(z_2), \ldots, P(z_r) \}$, and there are only finitely many elements of this set; which means $ P(X) - w $ has $ n $ distinct roots for all $ w $ outside of this set.

For the reverse implication, note that if $ P(X) $ was a polynomial with no roots, then $|P(X)| $ would have a nonzero minimum value (this is guaranteed by the growth lemma), say $ r $. In that case, we would have that $ |P(X) + q| \geq ||P(X)| - q| \geq r/2 $ for all $ 0 \leq q \leq r/2 $, meaning that none of these polynomials can have any roots in $ \mathbb{C} $. This violates our hypothesis, since there were only finitely many such polynomials which did not have $ n $ distinct roots.
ok, Let's look at the statemt that says


  For all but finitely many $w \in C$, $f(z)-w$ has $n$ distinct roots in $\mathbb C$.


As written, that statement doesn't say there are any roots at all when $w$ is one of those finitely many exceptions.  Thus it falls short of entailing the fundamental theorem, which says there is always at least one root.

You wrote "including multiplicities, $n$ dinstinct roots". Maybe you're missing the meaning of the word "distinct".  If any of the roots has multiplicity more than $1$, and the sum of the multiplicities is $n$, then the $n$ roots that you have (including multiplicities) are not distinct.

Regardless of the value of $w$, there will be $n$ roots if you count them by multiplicities, i.e. the sum of the multiplicities will always be $n$.  But for finitely many values of $w$, the roots will not be distinct.

You ask whether the statement is "equivalent" to the "fundamental theorem of algebra".  If "equivalent" means they are either both true or both false, then they are.  If equivalence is to be defined relative to a specified axiom system in which the axioms are not enough to decide whether either statement is true or false, then it can make sense to say what "equivalent" means.

But if "equivalent" means the deduction of either as a corollary of the other is quick and simple, then "equivalent" is not all that precisely defined.  Sometimes it means somebody has proved that either both are true or both are false, without saying which.

Obviously if there are $n$ roots, counted by multiplicities, then there is at least one root, and that's what the fundamental theorem says.

On the other hand, if the fundamental theorem is true (and it's not really a theorem of algebra as we usually understand that term today), then a simple theorem in algebra says we can write
$$
f(z) = (z-z_1) g(z)
$$
where $z_1$ is a zero of $f(z)$, and then applying the fundamental theorem and that simple theorem of algebra again, we get
$$
f(z) = (z-z_1)(z-z_2)h(z)
$$
and so on. So you get $n$ roots, counted by multiplicities.  You can make this into a precisely stated proof by mathematical induction.