3.4 Komplexe Polynome
Aus Online Mathematik Brückenkurs 2
| Theorie | Übungen |
Inhalt:
- Factor theorem.
- Polynomial division
- Fundamental theorem of algebra
Lernziele:
Nach diesem Abschnitt sollst Du folgendes können to:
- Perform polynomial division.
- Understand the relationship between factors and zeros of polynomials.
- Know that a polynomial equation of degree n has n roots (including multiplicity).
- Know that real polynomial equations have complex conjugate roots.
Polynomials and equations
An expression of the form
where
Polynomials are essential for a large part of mathematics and have many properties that display great similarities with integers expressed using our "Arabic" (actually originally Indian) place value system. This means that we can often do calculations with polynomials in a similar way as with integers.
Beispiel 1
Compare the following integer written using a base 10,
 103+3102+510+3 |
with the polynomial in
| x3+3x2+5x+3 |
and then the following divisions,
111353=123 as1353=123 ,11
x+1x3+3x2+5x+3=x2+2x+3 sincex3+3x2+5x+3=(x2+2x+3)(x+1) .
If
As the above example showed, polynomials can be divided very like integers. Polynomial division, like integer division, is usually not exact. If, for example,
The calculation can also be written as 5+2
Similarly, if
or
Clearly, a division is exact if the remainder is zero. For polynomials this is expressed as follows: If
or
Polynomial division
If
Beispiel 2
Perform polynomial divisionen for
The first step is that we add and subtract an appropriate
The reason why we do this is that the sub-expression
Then we add and subtract an appropriate
The last step is that we add and subtract a constant
Thus ending up with
The quotient is
The connection between factors and zeros
If
Since 0=0
Factor theorem:
\displaystyle (x-a) is a divisor of a polynomial \displaystyle p(x) if and only if \displaystyle x=a is a zero of \displaystyle p(x).
Please note that the theorem applies in both directions, i.e. if we know that \displaystyle x=a is a zero of \displaystyle p(x) we automatically would know that \displaystyle p(x) is divisible by \displaystyle (x-a).
Beispiel 3
The polynomial \displaystyle p(x) = x^2-6x+8 can be factorised as
| \displaystyle x^2-6x+8 = (x-2)(x-4) |
and has therefore zeros at \displaystyle x=2 and \displaystyle x=4 (and no others). It is precisely these that are obtained if one solves the equation \displaystyle \ x^2-6x+8 = 0\,.
Beispiel 4
- Factorise the polynomial \displaystyle \ x^2-3x-10\,.
By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation \displaystyle \ x^2-3x-10=0\ has the solutions\displaystyle x= \frac{3}{2} \pm \sqrt{\Bigl(\frac{3}{2}\Bigr)^2 - (-10)} = \frac{3}{2} \pm \frac{7}{2}\,\mbox{,} i.e. \displaystyle x=-2 and \displaystyle x=5. This means that \displaystyle \ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,.
- Factorise the polynomial \displaystyle \ x^2+6x+9\,.
This polynomial has a repeated root\displaystyle x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3 and thus \displaystyle \ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,.
- Factorise the polynomial \displaystyle \ x^2 -4x+5\,.
In this case, the polynomial has two complex roots\displaystyle x= 2 \pm \sqrt{2^2 -5} = 2\pm \sqrt{-1} = 2\pm i and when factorised will be \displaystyle \ (x-(2-i))(x-(2+i))\,.
Beispiel 5
Determine a cubic polynomial having zeros, \displaystyle 1, \displaystyle -1 and \displaystyle 3.
According to the factor theorem, the polynomial must have factors \displaystyle (x-1), \displaystyle (x+1) and \displaystyle (x-3). Multiplying these factors, we get a cubic polynomial
| \displaystyle (x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.} |
Fundamental theorem of algebra
At the beginning of this chapter, we introduced the complex numbers to enable us to solve the quadratic equation \displaystyle x^2=-1 and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or whether we need to invent more types of numbers in order to solve other complicated polynomials. The answer to that question is that we need not; the complex numbers are enough. The German mathematician Carl Friedrich Gauss proved in the year 1799 the fundamental theorem of algebra which says the following:
Every polynomial of degree \displaystyle n\ge1 with complex coefficients has at least one zero which is a complex number.
As every zero according to the the factor theorem is matched by a factor, we can now also state the following theorem:
Every polynomial of degree \displaystyle n\ge1 has exactly \displaystyle n zeros if each zero is counted up to its multiplicity.
(By multiplicity is meant that a double zero is counted twice, a triple zero 3 times, etc.)
Note that these theorems only say that there exist complex roots of a polynomial, but not how to determine them. In general, there is no simple method to write a formula for the roots, and for polynomials of higher degree, we must use various devices to obtain a solution. If we restrict ourselves to polynomial with real coefficients, one of the devices that can help us is the knowledge that the complex roots of such polynomials always come in complex conjugate pairs.
Beispiel 6
Show that the polynomial \displaystyle p(x)=x^4-4x^3+6x^2-4x+5 has zeros \displaystyle x=i and \displaystyle x = 2-i. Thus determine the other zeros.
We have
| \displaystyle \begin{align*} p(i) &= i^4- 4i^3 +6i^2-4i+5 = 1+4i-6-4i+5=0\,\mbox{,}\\ p(2-i) &= (2-i)^4 -4(2-i)^3 + 6(2-i)^2 - 4(2-i) +5\,\mbox{.}\end{align*} |
In order to calculate the last term, we need to determine
| \displaystyle \begin{align*} (2-i)^2 &= 4-4i+i^2 = 3-4i\,\mbox{,}\\ (2-i)^3 &= (3-4i)(2-i) = 6-3i-8i+4i^2 = 2-11i\,\mbox{,}\\ (2-i)^4 &= (2-11i)(2-i) = 4-2i-22i+11i^2= -7-24i\,\mbox{.}\end{align*} |
This gives that
| \displaystyle \begin{align*} p(2-i) &= -7-24i-4(2-11i)+6(3-4i) -4(2-i) +5\\ &= -7-24i-8+44i+18-24i-8+4i+5=0\,\mbox{,}\end{align*} |
which proves that \displaystyle i and \displaystyle 2-i are zeros of this polynomial.
Since the polynomial has real coefficients, we can immediately say that the other two zeros are the complex conjugates of the first two zeros, i.e. the other two roots are \displaystyle z=-i and \displaystyle z=2+i.
One consequence of the fundamental theorem of algebra (and the factor theorem) is that all polynomials can be factored into a product of complex linear factors. This also applies to polynomials with real coefficients, but for such polynomials it is possible to multiply together the pair of factors belonging to complex conjugate roots. In this case the factorisation will consist of linear and quadratic real factors.
Beispiel 7
Show that \displaystyle x=1 is a zero of \displaystyle p(x)= x^3+x^2-2. Then first factorise \displaystyle p(x) into polynomials having real coefficients and then factorise \displaystyle p(x) completely into linear factors.
We have that \displaystyle \ p(1)= 1^3 + 1^2 -2 = 0\ which shows that \displaystyle x=1 is a zero of the polynomial. According to the factor theorem, this means that \displaystyle x-1 is a factor of \displaystyle p(x), i.e. \displaystyle p(x) is divisible by \displaystyle x-1. We therefore divide the polynomial with \displaystyle x-1 to get the remaining factors after \displaystyle x-1 is factored out of the polynomial
| \displaystyle \begin{align*} \frac{x^3+x^2-2}{x-1} &= \frac{x^2(x-1)+2x^2-2}{x-1} = x^2 + \frac{2x^2-2}{x-1} = x^2 + \frac{2x(x-1) +2x -2}{x-1}\\[4pt] &= x^2 + 2x + \frac{2x-2}{x-1} = x^2 + 2x + \frac{2(x-1)}{x-1} = x^2 + 2x + 2\,\mbox{.}\end{align*} |
So we have \displaystyle \ p(x)= (x-1)(x^2+2x+2)\, which is the first part of the problem.
It now remains to factorise \displaystyle x^2+2x+2. The equation \displaystyle x^2+2x+2=0 has the solutions
| \displaystyle x=-1\pm \sqrt{\smash{(-1)^2 -2}\vphantom{i^2}} = -1 \pm \sqrt{-1} = -1\pm i |
and therefore the polynomial has the following factorization into complex linear factors.
| \displaystyle \begin{align*} x^3+x^2-2 = (x-1)(x^2+2x+2) &= (x-1)(x-(-1+i))(x-(-1-i))\\ &= (x-1)(x+1-i)(x+1+i)\,\mbox{.}\end{align*} |
