**introduction**:

o*fundamental theorem of algebra*is the statement 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 if all repeated roots are counted individually.

This theorem has a long and tortuous history. In 1608 Peter Roth wrote that a polynomial equation of degree $n$ with real coefficients $n$ can have solutions, but he gave no proof. Leibniz and Nikolaus Bernoulli claimed that quartic polynomials of the form $x^4 + a^4$, where $a$ is real and nonzero, could not be resolved or factored into linear or quadratic factors, but Euler later noted $ $x ^4 + a^4 = (x^2 + a \sqrt{2} x + a^2) (x^2 – a \sqrt{2} x + a^2).$$ Finally they convinced the mathematicians 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 wrong or incomplete. The first complete and rigorous proof was given by Argand in 1806. For more historical information on the Fundamental Theorem of Algebra, see hereWikipedia article.

A proof of the fundamental theorem of algebra is usually presented in a college-level course in complex analysis, but only after a comprehensive foundation of the underlying theory, such as Cauchy's theorem, the principle of argumentation, and Liouville's theorem. Only a small percentage of college students take those courses, and even many of those who take those courses never really understand the essential idea behind the Fundamental Theorem of Algebra because of the way it is usually presented in textbooks (this one was certainly the case). . editor's first experiences with the fundamental theorem of algebra). All of this is generally accepted as an unpleasant but unavoidable quality of mathematics pedagogy.

Here we present a proof of the fundamental theorem of algebra. We note that both are the case*elemental*mi*independent*– is based solely on a well-known completeness axiom and simple reasoning with estimates and inequalities. It should be understandable to anyone with a solid knowledge of high school algebra, trigonometry, and complex numbers, although some experience with limits and estimation, such as those typically taught in freshman math courses, may also be helpful. high school or college.

**test summary**:

For readers familiar with Newton's method of solving equations, start with an approximation fairly close to a root, and then adjust the approximation by approximating in an appropriate direction. We will use the same strategy here and show that if we assume that the argument in which the polynomial function reaches its minimum absolute value is not a root, there is a close argument in which the polynomial function has an even smaller absolute value, which contradicts the assumption that the absolute minimum value argument is not a root.

**Definitions and axioms**:

Then $p(z)$ denotes the polynomial $n$th degree $p(z) = p_0 + p_1 z + p_2 z^2 + \cdots + p_n z^n$, where the coefficients $p_i$ are any complex Numbers where neither $p_0$ nor $p_n$ are equal to zero (otherwise the polynomial is equivalent to one degree less). We will use a fundamental property of completeness of real and complex numbers, that is, a continuous function on a closed set reaches its minimum at some point in the domain. This can be taken as an axiom or easily proved by applying other well-known completeness axioms, such as the Cauchy sequence axiom or the nested interval axiom. watch thisWikipedia articlefor a more detailed discussion of the completeness axioms.

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

**court hearing**: Suppose that $p(z)$ has no roots in the complex plane. First note that for large $z$ you need $|z| say > 2 \max_i |p_i/p_n|$ the $z^n$ term of $p(z)$ is greater than the sum of all other terms. Then given $B > 0$ then for $s$ large enough we have $|p(z)| > B$ for all $z$ with $|z| \ge s$. We take $B = 2 |p(0)| = 2 |p_0|$. Since $|p(z)|$ is continuous inside and in the limit of the circle of radius $s$, it follows from the axiom of completeness mentioned above that $|p(z)|$ reaches its minimum value at some point $ t$ in this circle (possibly in the limit). But since $|p(0)| < 1/2 \cdot |p(z)|$ for every $z$ on the circumference of the circle it follows that $|p(z)|$ reaches its minimum at some point on the circle $t$*Internal*of the circle

Now rewrite the polynomial $p(z)$ by translating the argument $z$ to $t$, giving a new polynomial $$q(z) = p(z + t) \; =\; q_0 + q_1 z + q_2 z^2 + \cdots + q_n z^n,$$ and translate the circle described above in the same way. Presumably, the polynomial $q(z)$, defined on a circle with center of origin (whose circle is contained in the circle above), has an absolute minimum value $M > 0$ at $z = 0$. Note that $M = |q(0)| = |q_0|$.

Our proof strategy is to construct a point $x$ near the origin such that $|q(x)| < |q(0)|,$, which contradicts the assumption that $|q(z)|$ has a non-zero minimum value at $z = 0$. If our method gives us only one direction in the complex plane for which the value of the function decreases in magnitude (a*descent direction*), so by moving a bit in this direction we hope to achieve our goal of constructing a complex $x$ such that $|q(x)| < |q(0)|$. That is the strategy we will follow.

**Construction of $x$ with $|q(x)| < |q(0)|$**:

Let $q(z)$ be the first non-zero coefficient after $q_0$, $q_m$ such that $q(z) = q_0 + q_m z^m + q_{m+1} z^{m + 1 } + \cdots + q_n z^n$. We choose $x$ as the complex number $$x = r \left(\frac{- q_0}{ q_m}\right)^{1/m},$$ where $r$ is a small positive real value we will do a below, and where $(-q_0/q_m)^{1/m}$ denotes one of the $m$th roots of $(-q_0/q_m)$.

**Comment**: By the way, note that in the complex number system, unlike real numbers, the existence of $m$th roots of a real or complex number is always guaranteed: 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.

**Prove 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 The terms $E$ can be restricted as follows. Suppose $q_0 \leq q_m$ (results in a very similar expression for $|E|$ in the case of $q_0 \geq q_m$) and define $s = r(|q_0/q_m|)^{1/m } $ . . . . Then, using the well-known formula for the sum of a geometric series, we can get $$|E| write \leq r^{m+1} \max_i |q_i| \left|\frac{q_0}{q_m}\right|^{(m+1)/m} (1 + s + s^2 + \cdots + s^{n-m-1}) \leq \frac{r ^{m+1}\max_i |q_i|}{1 – s} \left|\frac{q_0}{q_m}\right|^{(m+1)/m}.$$ Then $|E| $ can be arbitrarily smaller than $|q_0 r^m| become = |q_0|r^m$ choose $r$ sufficiently small. For example, select $r$ so that $|E| < |q _0|r^m / 2$. So for such $r$ 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 value distinct from zero at $z = 0$.

**SET 2**: Every polynomial of degree $n$ with real or complex coefficients has exactly $n$ complex roots if all repeated roots are counted individually.

**court hearing**: If $\alpha$ is a real or complex root of the polynomial $p(z)$ of degree $n$ with real or complex coefficients, divide this polynomial by $(z – \alpha)$ using the well-known process of the division of polynomials we obtain $p(z) = (z – \alpha) q(z) + r$, where $q(z)$ has degree $n – 1$ and $r$ is a constant. Note, however, that $p(\alpha) = r = 0$, so $p(z) = (z – \alpha) q(z)$. Continuing with the induction, we conclude that the original polynomial $p(z)$ has exactly $n$ complex zeros, although some may be repeated.

For more evidence in this series, see the list below.Simple proofs of great theorems.