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

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

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:

$\displaystyle P(z)$ $\displaystyle = P(z)-\underbrace{P(z_0)}_{=0}$    
$\displaystyle \quad$ $\displaystyle = \underset{k=0}{\overset{n}{\sum}} a_nz^n - \underset{k=0}{\overset{n}{\sum}} a_nz_0^n$    
$\displaystyle \quad$ $\displaystyle = \underset{k=0}{\overset{n}{\sum}} a_n(z^n-z_0^n).$    

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:

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

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:

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

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

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

and

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

It follows:

$\displaystyle \forall z \in \mathbb{C}, p(z)$ $\displaystyle = \left[ z- \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right] \l...
...qrt{2}}{2} \right] \left[ z+ \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right]$    
$\displaystyle \quad$ $\displaystyle \quad$    
$\displaystyle \quad$ $\displaystyle = \underbrace{ \left[ z- \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{...
...eft[ z+ \frac {\sqrt{2}}{2}+ i \frac {\sqrt{2}}{2} \right] }_{=z^2+z\sqrt{2}+1}$    

Thus:

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

Noah Dana-Picard 2007-12-24