Fundamental theorem of algebra

Complex numbers: Complex polynomials

Fundamental theorem of algebra

Let $p(z)$ be a complex polynomial of degree $n$ .

The sum of the multiplicities of all solutions to $p(z)$ is equal to $n$ .

The proof of this theorem is beyond the scope of this course.

The theorem was previously proven for polynomials of type $z^n-w$ .

This theorem does not apply for real numbers: $x^2+1$ is a real polynomial of degree 2 with no zeros in $\mathbb{R}$ . As a complex polynomial $x^2+1$ has exactly two zeros: $x=\ii$ and $x=-\ii$ .

A version of the fundamental theorem of algebra that applies to the real numbers, is formulated in the theory Fundamental Theorem of Algebra. We will later show that the real version is a result of the version given here.

According to the statement, any complex polynomial $p(z)$ with degree $n$ with leading coefficient $1$ can be written as the product

$\left(z-w_1\right)\cdot\left(z-w_2\right)\cdots \left(z-w_n\right)\tiny,$ in which $w_1,w_2,\ldots,w_n$ are zeros of $p(z)$ ; each zero point is equally common as its multiplicity. As a consequence any irreducible complex polynomial is linear.

Polynomials of degree 1

For polynomials of degree 1, the zero can always be found directly: such a polynomial can be written as $a\cdot z+b$ , in which $a$ and $b$ are complex numbers with $a\ne0$ . If

$az+b=0\tiny,$
then follows $z=-\frac{b}{a}$ .

Polynomials of degree 2 We now consider polynomials of degree 2:

$az^2+bz+c\tiny,$ in which $a$ , $b$ , and $c$ are complex numbers with $a\ne0$ . The zeros of this quadratic polynomial satisfy the following version of the abc-formula

$z=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2a}\tiny.$ In which $\sqrt{b^2-4\cdot a\cdot c}$ has to be read as the solutions $w$ of the equation $w^2={b^2-4ac}$ (the two solutions are each others opposite).

We rewrite the polynomial as follows:

$\begin{array}{rcl}az^2+bz+c&=& a\cdot\left(z^2+\frac{b}{a}\cdot z\right)+c\\ &=&a\left(z+\frac{b}{2 a}\right)^2+\frac{4a\cdot c- b^2}{4a}\\ &=&\frac{\left(2a\cdot\left(z+\frac{b}{2 a}\right)\right)^2-\left(b^2-4a\cdot c\right)}{4a}\end {array}$ No write $w={2a}\cdot\left(z+\frac{b}{2 a}\right)$ . Then we find the quadratic equation

$w^2={b^2-4ac}$ with unknown $w$ . According to the statement of Higher power roots this equation has two solutions (unless $b^2-4a\cdot c=0$ ), from which the two solutions of $z$ directly follow.

The above technique is called completing the square.

In the abc-formula the root of the complex number $b^2-4a\cdot c$ usually speaking is only defined if it is a non-negative real number (in the theory Imaginary numbers). Because $\pm$ is in front of the root, the lack of clarity about $\sqrt{b^2-4a\cdot c}$ is removed. A solution $w$ of $w^2={b^2-4a\cdot c}$ is indeed unique, apart from the sign.

If we write $w=\pm\sqrt{b^2-4a\cdot c}$ for the two solutions $w$ , then the quadratic formula

$z=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2a}$ follows by substituting for $w$ the expression ${2a}\cdot\left(z+\frac{b}{2 a}\right)$ , dividing both sides by $2a$ , and finally, subtracting $\frac{b}{2 a}$ from both sides.

It is still possible to give algorithms for polynomials of grade 3 and grade 4 which lead to the three or four zeros of these polynomials. For polynomials of a degree higher than four such algorithms do not exist, and it is therefore fairly accidental if we can find these zeros exactly. Of course, there are algorithms that numerically approximate the zeros.

Solve the equation

$z^2 + \left(2+4\cdot \ii\right)\cdot z +\ii=0$ by means of completing the square.

$\begin{array}{rcl} z& =&-1 + \sqrt[4]{18}\cos \left({{3\cdot \pi}\over{8}}\right) + \left(-2 + \sqrt[4]{18}\sin \left({{3\cdot \pi}\over{8}}\right)\right)\cdot\ii\\& \lor&\\ z &=&-1 - \sqrt[4]{18}\cos \left({{3\cdot \pi}\over{8}}\right) + \left(-2 - \sqrt[4]{18}\sin \left({{3\cdot \pi}\over{8}}\right)\right)\cdot\ii\end{array}$

After all, completing the square gives

$(z+(1+2 \ii))^2=-3+3\ii\,$
such that, according to the theorem of Higher power roots,

$\begin{array}{rcl} z+1+2\ii & =&\sqrt[4]{18}\cdot\left (\cos \left(\dfrac{3\pi}{8}\right)+\sin\left (\dfrac{3\pi}{8}\right)\right)\cdot\ii \\ &\text{of}&\\ z+1+2\ii & =&\sqrt[4]{18}\cdot\left (\cos \left(\dfrac{11\pi}{8}\right)+\sin\left( \dfrac{11\pi}{8}\right)\right)\cdot\ii \\ \end{array}$
The answer follows by subtracting $1+2\ii$ from both sides.

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