Simple proofs: The fundamental theorem of algebra

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The fundamental theorem of algebra is the assertion that every polynomial with real or complex coefficients has at least one complex root. An immediate extension of this result is that every polynomial of degree $n$ with real or complex coefficients has exactly $n$ complex roots, when counting individually any repeated roots.

This theorem has a long, tortuous history. In 1608, Peter Roth wrote that a polynomial equation of degree $n$ with real coefficients may have $n$ solutions, but offered no proof. Leibniz and Nikolaus Bernoulli both asserted that quartic polynomials of the form $x^4 + a^4$, where $a$ is real and nonzero, could not be solved or factored into linear or quadratic factors, but Euler later pointed out that $$x^4 + a^4 = (x^2 + a \sqrt{2} x + a^2) (x^2 – a \sqrt{2} x + a^2).$$ Eventually mathematicians became convinced that something like the fundamental theorem must be true. Numerous mathematicians, including d’Alembert, Euler, Lagrange, Laplace and Gauss, published proofs in the 1600s and 1700s, but each was later found to be flawed or incomplete. The first complete and fully rigorous proof was by Argand in 1806. For additional historical background on the fundamental theorem of algebra, see this Wikipedia article.

A proof of the fundamental theorem of algebra is typically presented in a college-level course in complex analysis, but only after an extensive background of underlying theory such as Cauchy’s theorem, the argument principle and Liouville’s theorem. Only a tiny percentage of college students ever take such coursework, and even many of those who do take such coursework never really grasp the essential idea behind the fundamental theorem of algebra, because of the way it is typically presented in textbooks (this was certainly the editor’s initial experience with the fundamental theorem of algebra). All this is generally accepted as an unpleasant but unavoidable feature of mathematical pedagogy.

We present here a proof of the fundamental theorem of algebra. We emphasize that it is both elementary and self-contained — it relies only a well-known completeness axiom and some straightforward reasoning with estimates and inequalities. It should be understandable by anyone with a solid high school background in algebra, trigonometry and complex numbers, although some experience with limits and estimates, as is commonly taught in high school or first-year college calculus courses, would also help.

Gist of the proof:
For readers familiar with Newton’s method for solving equations, one starts with a reasonably close approximation to a root, then adjusts the approximation by moving closer in an appropriate direction. We will employ the same strategy here, showing that if one assumes that the argument where the polynomial function achieves its minimum absolute value is not a root, then there is a nearby argument where the polynomial function has an even smaller absolute value, contradicting the assumption that the argument of the minimum absolute value is not a root.

Definitions and axioms:
In the following $p(z)$ will denote the $n$-th degree polynomial $p(z) = p_0 + p_1 z + p_2 z^2 + \cdots + p_n z^n$, where the coefficients $p_i$ are any complex numbers, with neither $p_0$ nor $p_n$ equal to zero (otherwise the polynomial is equivalent to one of lesser degree). We will utilize a fundamental completeness property of real and complex numbers, namely that a continuous function on a closed set achieves its minimum at some point in the domain. This can be taken as an axiom, or can be easily proved by applying other well-known completeness axioms, such as the Cauchy sequence axiom or the nested interval axiom. See this Wikipedia article for more discussion on completeness axioms.

THEOREM 1: Every polynomial with real or complex coefficients has at least one complex root.

Proof: Suppose that $p(z)$ has no roots in the complex plane. First note that for large $z$, say $|z| > 2 \max_i |p_i/p_n|$, the $z^n$ term of $p(z)$ is greater in absolute value than the sum of all the other terms. Thus given some $B > 0$, then for any sufficiently large $s$, we have $|p(z)| > B$ for all $z$ with $|z| \ge s$. We will take $B = 2 |p(0)| = 2 |p_0|$. Since $|p(z)|$ is continuous on the interior and boundary of the circle with radius $s$, it follows by the completeness axiom mentioned above that $|p(z)|$ achieves its minimum value at some point $t$ in this circle (possibly on the boundary). But since $|p(0)| < 1/2 \cdot |p(z)|$ for all $z$ on the circumference of the circle, it follows that $|p(z)|$ achieves its minimum at some point $t$ in the interior of the circle.

Now rewrite the polynomial $p(z)$, translating the argument $z$ by $t$, thus producing a new polynomial $$q(z) = p(z + t) \; = \; q_0 + q_1 z + q_2 z^2 + \cdots + q_n z^n,$$ and similarly translate the circle described above. Presumably the polynomial $q(z)$, defined on some circle centered at the origin (which circle is contained within the circle above), has a minimum absolute value $M > 0$ at $z = 0$. Note that $M = |q(0)| = |q_0|$.

Our proof strategy is to construct some point $x$, close to the origin, such that $|q(x)| < |q(0)|,$ thus contradicting the presumption that $|q(z)|$ has a minimum nonzero value at $z = 0$. If our method gives us merely a direction in the complex plane for which the function value decreases in magnitude (a descent direction), then by moving a small distance in that direction, we hope to achieve our goal of constructing a complex $x$ such that $|q(x)| < |q(0)|$. This is the strategy we will pursue.

Construction of $x$ such that $|q(x)| < |q(0)|$:
Let the first nonzero coefficient of $q(z)$, following $q_0$, be $q_m$, so that $q(z) = q_0 + q_m z^m + q_{m+1} z^{m+1} + \cdots + q_n z^n$. We will choose $x$ to be the complex number $$x = r \left(\frac{- q_0}{ q_m}\right)^{1/m},$$ where $r$ is a small positive real value we will specify below, and where $(-q_0/q_m)^{1/m}$ denotes any of the $m$-th roots of $(-q_0/q_m)$.

Comment: As an aside, note that unlike the real numbers, in the complex number system the $m$-th roots of a real or complex number are always guaranteed to exist: if $z = z_1 + i z_2$, with $z_1$ and $z_2$ real, then the $m$-th roots of $z$ are given explicitly by $$\{R^{1/m} \cos ((\theta + 2k\pi)/m) + i R^{1/m} \sin ((\theta+2k\pi)/m), \, k = 0, 1, \cdots, m-1\},$$ where $R = \sqrt{z_1^2 + z_2^2}$ and $\theta = \arctan (z_2/z_1)$. The guaranteed existence of $m$-th roots, a feature of the complex number system, is the key fact behind the fundamental theorem of algebra.

Proof that $|q(x)| < |q(0)|$:
With the definition of $x$ given above, we can write
$$q(x) = q_0 – q_0 r^m + q_{m+1} r^{m+1} \left(\frac{-q_0} {q_m}\right)^{(m+1)/m} + \cdots + q_n r^n \left(\frac{-q_0} {q_m}\right)^{n/m}$$ $$= q_0 – q_0 r^m + E,$$ where the extra terms $E$ can be bounded as follows. Assume that $q_0 \leq q_m$ (a very similar expression is obtained for $|E|$ in the case $q_0 \geq q_m$), and define $s = r(|q_0/q_m|)^{1/m}$. Then, by applying the well-known formula for the sum of a geometric series, we can write $$|E| \leq r^{m+1} \max_i |q_i| \left|\frac{q_0}{q_m}\right|^{(m+1)/m} (1 + s + s^2 + \cdots + s^{n-m-1}) \leq \frac{r^{m+1}\max_i |q_i|}{1 – s} \left|\frac{q_0}{q_m}\right|^{(m+1)/m}.$$ Thus $|E|$ can be made arbitrarily smaller than $|q_0 r^m| = |q_0|r^m$ by choosing $r$ small enough. For instance, select $r$ so that $|E| < |q _0|r^m / 2$. Then for such an $r$, we have $$|q(x)| = |q_0 - q_0 r^m + E| < |q_0 - q_0 r^m / 2| = |q_0|(1 - r^m / 2)< |q_0| = |q(0)|,$$ which contradicts the original assumption that $|q(z)|$ has a minimum nonzero value at $z = 0$.

THEOREM 2: Every polynomial of degree $n$ with real or complex coefficients has exactly $n$ complex roots, when counting individually any repeated roots.

Proof: If $\alpha$ is a real or complex root of the polynomial $p(z)$ of degree $n$ with real or complex coefficients, then by dividing this polynomial by $(z – \alpha)$, using the well-known polynomial division process, one obtains $p(z) = (z – \alpha) q(z) + r$, where $q(z)$ has degree $n – 1$ and $r$ is a constant. But note that $p(\alpha) = r = 0$, so that $p(z) = (z – \alpha) q(z)$. Continuing by induction, we conclude that the original polynomial $p(z)$ has exactly $n$ complex roots, although some might be repeated.

For other proofs in this series see the listing at Simple proofs of great theorems.

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