Fundamental Theorem of Algebra §

(Argand, 1806)

The Fundamental Theorem of Algebra is a foundational result in the field of complex analysis, but it has wide-reaching implications for polynomial equations in general. The theorem essentially guarantees that every non-constant polynomial has at least one root in the complex number system.

Statement §

For every non-constant polynomial $p(z)$ with complex coefficients, there exists a complex number $c$ such that $p(c)=0$.

In other words, every polynomial of degree $n\ge 1$ has at least one complex root. Moreover, with multiplicity considered, it has exactly $n$ complex roots.

Proof §

The formal proof of the Fundamental Theorem of Algebra is beyond the scope of a simple introduction due to its reliance on advanced concepts from complex analysis. However, a general sketch can be provided.

Proof Sketch: §

1. Boundedness of ( p ): For a non-constant polynomial, it can be shown that there exists some radius $R$ such that $|p(z)|\to \infty$ as $|z|\to \infty$, for all $|z|>R$. This means that the polynomial’s magnitude grows without bound outside of some large circle in the complex plane.

2. Achievement of Minimum Magnitude: The function $|p(z)|$ achieves its minimum value over the closed disc of radius $R$ (due to compactness). If this minimum value is zero, we have found a root. If not…

3. Rolle’s Theorem for Complex Polynomials: If $p(z)$ doesn’t have a root in the disc of radius $R$, then neither does its derivative. This contradicts the fact that the derivative is of lower degree and, by induction, should have a root. Thus, $p(z)$ must have had a root in the disc.

Through a combination of these elements, we can establish the Fundamental Theorem of Algebra.

Implications §

1. Complex Roots Come in Pairs: If a polynomial has real coefficients and one non-real complex root, then its complex conjugate is also a root.

2. Degree of Polynomial: A polynomial of degree $n$ will have exactly $n$ roots in the complex numbers, considering multiplicity.

3. Roots and Factors: If $c$ is a root of $p(z)$, then $(z-c)$ is a factor of $p(z)$. Using synthetic division or polynomial long division, we can express $p(z)$ as $p(z)=(z-c)q(z)$ where $q(z)$ is a polynomial of degree $n-1$.

Practice Problems §

1. Show that the polynomial $p(z)=z^2+1$ has no real roots but has two complex roots.
2. Find all roots, real or complex, of the polynomial $p(z)=z^3-6z^2+11z-6$.

For those interested in a deeper dive, books on complex analysis or advanced algebra will provide rigorous proofs and further discussion on the Fundamental Theorem of Algebra.

Note that this also links with Galois Theory, and how it can prove that no formulae exist for polynomials above order 4.