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Theorem (Fundamental Theorem of Algebra).
Every polynomial of degree \(n\) with complex coefficients has exactly \(n\) complex roots, counted with multiplicity.
For a given polynomial \(p(x) = x^n + p_{n-1} x^{n-1} + \cdots + p_0\), we consider the family of polynomials \(p_t = t p + (1-t) q\) for \(t \in \mathbb{C}\).
Where \(p_0 = q\) is a polynomial with \(n\) distinct roots, e.g. \(q = x^n - 1\).
Let \(X \subset \IC\) be the set of parameters \(t\) where \(p_t\) has at least one root.
By construction \(0 \in X\).
We will show that \(1 \in X\), by deforming the roots of \(q\) along \(p_t\) using Resultant techniques and the Implicit Function theorem, and leveraging the Cauchy bound on root location to show stability of roots under limits with bounded coefficients.
This shows that \(p\) has a single root \(r\), by splitting off a linear factor \(x-r\) and induction on the degree of \(p\) we conclude that \(p\) has \(n\) roots.
Proposition.
\(X \subset \mathbb{C}\) is closed.
Proof.
Let \(t_k\) be a sequence in \(X\) converging to a point \(t \in \mathbb{C}\). We need to show that \(t \in X\).
As \(t_k\) is bounded, we know that the coefficients of \(p_{t_k}\) are bounded by a constant \(M\).
By the Cauchy bound, all roots of \(p_{t_k}\) lie within the disk of radius \(1 + \max_i |p_{t_k,i}| \leq M + 1\).
Let \(r_k\) be a convergent sequence of roots of \(p_{t_k}\), and let \(r = \lim_{k \to \infty} r_k\).
Then \(\lim_{k \to \infty} p_{t_k}(r_k) = p_t(r)\) by continuity of \(p_t\), but \(p_{t_k}(r_k) = 0\), so \(p_t(r) = 0\). \(\square\)
Now consider \(\Delta(t) = Res(p_t, p_t')\) the discriminant of the polynomial \(p_t\).
Recall that the Resultant \(Res(p,q)\) is a polynomial in the coefficients of the polynomials \(p\) and \(q\)
and that \(Res(p,q) = 0\) if and only if the polynomials \(p\) and \(q\) have a common factor (\(gcd(p,q) \neq 1\)).
In particular \(\Delta(t)\) is a polynomial in \(t\), with \(\Delta(0) \neq 0\) as \(q\) has \(n\) distinct roots.
Let \(D = \{ t \in \mathbb{C} \mid \Delta(t) = 0 \}\), and let \(U = \mathbb{C} \setminus D\).
As complement of a finite set, \(U\) is connected open and dense in \(\IC\).
Proposition.
The set \(X \cap U \subset \IC\) is open.
Proof.
Let \(t \in X \cap U\), we need to show that there is a neighborhood \(N\) of \(t\) such that \(N \subset X\).
Since \(t \in X\) we know that \(p_t\) has at least one root \(r\).
As \(t \in U\) we know that \(p_t'(r) \neq 0\), since otherwise \(p'\) and \(p_t'\) would have a common factor.
By the implicit function theorem applied to \(F(x,t) = p_t(x)\), there exists a neighborhood \(N\) of \(t\) and a function \(r(\tau)\) defined on \(N\), with \(r(t) = r\) and \(p_\tau(r(\tau)) = 0\) for all \(\tau \in N\).
This shows that \(N \subset X\). \(\square\)
We conclude that \(X \cap U\) is open and closed in \(U\). As \(U\) is connected, it follows that \(X \cap U = U\).
Thus, \(X\) contains the open dense set \(U\), which implies that \(X\) is all of \(\IC\), and in particular \(1 \in X\).
QED.