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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 $|p(0)| 0$ 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.