Polynomial equations of higher degree.

In Chapter 6, Section thm fundamental, we prove the Fundamental Theorem of Algebra, which states that if $P(z)$ is a non constant polynomial over $\mathbb{C}$ , then $P(z)$ has a root.

Here are some examples of applications, based on the following theorems. Recall that a root of a polynomial $P(z)$ is a number $z_0$ such that $P(z_0)=0$ .

Theorem 1.7.1   Let $P(z)$ be a polynomial in one complex variable $z$ , with complex coefficients. The complex number $z_0$ is a root of $P(z)$ if, and only if, there exists a polynomial $Q(z)$ such that $P(z)=(z-z_0)Q(z)$ .

Proof. We denote

$$P(z)= \underset{k=0}{\overset{n}{\sum}} a_nz^n$$

where $a_i \in \mathbb{C}$ , for every $i \in \{ 0,1,2 \dots , n \}$ .

The ``if'' part is trivial.

For the ``only if'' part, suppose that $P(z_0)=0$ . Then we have:

$$P(z) = P(z)-\underbrace{P(z_0)}_{=0}$$

$$\quad = \underset{k=0}{\overset{n}{\sum}} a_nz^n - \underset{k=0}{\overset{n}{\sum}} a_nz_0^n$$

$$\quad = \underset{k=0}{\overset{n}{\sum}} a_n(z^n-z_0^n).$$

By section section algebraic form, all the terms $z^n-z_0^n$ have a common factor $z-z_0$ , whence the result. $\qedsymbol$

Example 1.7.2   Solve the equation $z^3-2z^2+(2+i)z-1-i=0$ .

An evident solution is 1 (why evident?).So we divide the polynomial on the left by $z-1$ . We have:

$$\forall z \in \mathbb{C}, z^3-2z^2+(2+i)z-1-i= (z-1)(z^2-z+1+i).$$

By the method of Section 6, we solve the equation $z^2-z+1+i=0$ , getting two solutions, namely $i$ and $1-i$ . In conclusion, the given cubic equation has three distinct solutions (as promised by 3.1): $1$ , $i$ and $1-i$ .

Theorem 1.7.3   Let $P(z)$ be a polynomial in one complex variable $z$ , with real coefficients. If the complex number $z_0$ is a root of $P(z)$ , then its complex conjugate $z_0$ (v.s. 1.5) is also a root of $P(z)$ .

Example 1.7.4   Let $P(z)=z^4+z^3+2z^2+z+1$ . We see easily that $P(i)=0$ .

By Theorem 7.3, we know that $-i$ is also a root of $P(z)$ .

By Theorem 7.1, there exists a polynomial $Q(z)$ (of degree 2) such that $P(z)=(z-i)(z+i)Q(z)$ . We have: $Q(z)=z^2+z+1$ . By the method of Section 6, we solve the equation $z^2+z+1=0$ . Finally the polynomial $P(z)$ has four distinct complex roots: $i$ , $-i$ , $-\frac 12 + i \frac {\sqrt{3}}{2}$ and $-\frac 12 - i \frac {\sqrt{3}}{2}$ . Remark that the third and the fourth solution are also conjugates.

The following corollary can be obtained either as a consequence of the Fundamental Theorem of Algebra thm fundamental, or as a consequence of the Intermediate Value Theorem in Calculus.

Corollary 1.7.5   A polynomial of odd degree over the reals has at least one real root.

Proof. Let $P(z)$ be a polynomial of odd degree $n$ with real coefficients. By the Fundamental Theorem of Algebra, it has exactly $n$ complex roots, counted with multiplicity. As the coefficients are real, the roots are organized by pairs of conjugate complex numbers. Suppose that $z_1$ is a root of $P(z)$ . If $z_1$ is real, we are done; otherwise, $\overline{z_1} \neq z_1$ is another root of $P(z)$ . Take now another root $z_2$ . If If $z_2$ is real, we are done; otherwise, $\overline{z_2} \neq z_2$ is another root of $P(z)$ . It is not important to know whether $z_1=z_2$ or $z_1 \neq z_2$ . Iterate this process until we discover the first (maybe the only) real root of $P(z)$ . As roots are organized by pairs, the maximum number of roots (either distinct or not) which can be involved in the process is even, and it is equal to $n-1$ . The last root $z_n$ cannot be equal to a previous one, as in such a case its conjugate should appear also now, and this is impossible. Thus, $z_n$ must be equal to its conjugate, i.e. $z_n$ is a real number. $\qedsymbol$

Remark 1.7.6   A consequence of these theorems is a method for factorizing polynomials of higher degree in one real variable, despite the fact that they have no real root. Let us see an example.

Let $p(x)=x^4+1$ . This polynomial has no real root, but it has complex conjugate roots:

$$\forall z \in \mathbb{C}, \; p(z)= (z^2+i)(z^2-i)$$

Moreover, by one of the methods described in subsection 6, we have:

$$\forall z \in \mathbb{C}, \; z^2+i = \left[ z- \frac {\sqrt{2}}{2... ...qrt{2}}{2} \right] \left[ z+ \frac {\sqrt{2}}{2}- i \frac {\sqrt{2}}{2} \right]$$

and

$$\forall z \in \mathbb{C}, \; z^2-i = \left[ z- \frac {\sqrt{2}}{2... ...qrt{2}}{2} \right] \left[ z+ \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right]$$

It follows:

$$\forall z \in \mathbb{C}, p(z) = \left[ z- \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right] \l... ...qrt{2}}{2} \right] \left[ z+ \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right]$$

$$\quad \quad$$

$$\quad = \underbrace{ \left[ z- \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{... ...eft[ z+ \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right] }_{=z^2+z\sqrt{2}+1}$$

Thus:

$$\forall x \in \mathbb{R}, \; x^4+1 = (x^2-x\sqrt{2}+1)(x^2+x\sqrt{2}+1).$$