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3.4 Komplexe Polynome

Aus Online Mathematik Brückenkurs 2

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{{Info|
{{Info|
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'''Content:'''
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'''Contents:'''
* Factor theorem.
* Factor theorem.
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* Polynomial long division
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* Polynomial division
* Fundamental theorem of algebra
* Fundamental theorem of algebra
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After this section, you will have learned to:
After this section, you will have learned to:
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* Perform polynomial division.
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* Perform polynomial division.
* Understand the relationship between factors and zeros of polynomials.
* Understand the relationship between factors and zeros of polynomials.
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* Know that a polynomial equation of degree n has n roots (including multiplicity).
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* Know that a polynomial equation of degree ''n'' has ''n'' roots (including multiplicity).
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* Know that real polynomial equations have complex conjugate roots.
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* Know that real polynomial equations have complex conjugate roots.
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== Polynom and equations ==
== Polynom and equations ==
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Polynomials are essential for a large part of mathematics and have many properties which display great similarities with our integers, which means that we can do calculations with polynomials in a similar way as with integers.
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Polynomials are essential for a large part of mathematics and have many properties which display great similarities with our integers, which means that we can do calculations with polynomials in a similar way as with integers.
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If <math>p(x)</math> is a polynomial of degree <math>n</math> then <math>p(x)=0</math> is called a ''polynomial equation'' of degree <math>n</math>. If <math>x=a</math> is a number such that <math>p(a)=0</math> then <math>x=a</math> is called a ''root'', or solution to the equation. One also says that <math>x=a</math> is a ''zero'' of <math>p(x)</math>.
If <math>p(x)</math> is a polynomial of degree <math>n</math> then <math>p(x)=0</math> is called a ''polynomial equation'' of degree <math>n</math>. If <math>x=a</math> is a number such that <math>p(a)=0</math> then <math>x=a</math> is called a ''root'', or solution to the equation. One also says that <math>x=a</math> is a ''zero'' of <math>p(x)</math>.
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As the above example showed polynomial, can be divided just like integers. Polynomial division, just like integer division usually is not exact. If, for example, <math>37</math> is divided by <math>5</math>, one gets
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As the above example showed polynomials can be divided just like integers. Polynomial division, just like integer division is usually not exact. If, for example, <math>37</math> is divided by <math>5</math>, one gets
{{Fristående formel||<math>\frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}</math>}}
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== Polynomial long division ==
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== Polynomial division ==
If <math>p(x)</math> is a polynomial of higher degree than polynomial <math>q(x)</math> then one can divide <math>p(x)</math> by <math>q(x)</math>. For example, this may be done by gradually subtracting appropriate multiples of <math>q(x)</math> from <math>p(x)</math> until a remaining numerator is of lower degree than the denominator <math>q(x)</math>.
If <math>p(x)</math> is a polynomial of higher degree than polynomial <math>q(x)</math> then one can divide <math>p(x)</math> by <math>q(x)</math>. For example, this may be done by gradually subtracting appropriate multiples of <math>q(x)</math> from <math>p(x)</math> until a remaining numerator is of lower degree than the denominator <math>q(x)</math>.
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{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}</math>}}
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The reason why we do this is now the sub-expression <math>x^3+2x^2</math> can be written as <math>x^2(x+2)</math> and cancellation with the denominator can be done,
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The reason why we do this is that the sub-expression <math>x^3+2x^2</math> can be written as <math>x^2(x+2)</math> and cancellation with the denominator can be done,
{{Fristående formel||<math>\frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}</math>}}
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{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}</math>}}
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The quotient is <math>x^2 -x + 1</math> and the remainder is <math>2</math>. Since the remainder is not zero division is not exact, that is, <math>q(x)= x+2</math>is not a ''divisor'' of <math>p(x)=x^3 + x^2 -x +4</math>.
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The quotient is <math>x^2 -x + 1</math> and the remainder is <math>2</math>. Since the remainder is not zero division is not exact, that is, <math>q(x)= x+2</math> is not a ''divisor'' of <math>p(x)=x^3 + x^2 -x +4</math>.
</div>
</div>
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==The connection between factors and zeros ==
==The connection between factors and zeros ==
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If <math>q(x)</math> is a divisor of <math>p(x)</math> then <math>p(x)=k(x)\cdot q(x)</math>. We have thus ''factorised'' <math>p(x)</math> . One says that <math>q(x)</math> is a factor of <math>p(x)</math>. Especially, if a polynomial of first degree <math>(x-a)</math> is a dividor of <math>p(x)</math> then <math>(x-a)</math> is a factor of <math>p(x)</math> , i.e.
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If <math>q(x)</math> is a divisor of <math>p(x)</math> then <math>p(x)=k(x)\cdot q(x)</math>. We have thus ''factorised'' <math>p(x)</math> . One says that <math>q(x)</math> is a factor of <math>p(x)</math>. Especially, if a polynomial of first degree <math>(x-a)</math> is a divisor of <math>p(x)</math> then <math>(x-a)</math> is a factor of <math>p(x)</math> , i.e.
{{Fristående formel||<math>p(x)= q(x)\cdot (x-a)\,\mbox{.}</math>}}
{{Fristående formel||<math>p(x)= q(x)\cdot (x-a)\,\mbox{.}</math>}}
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</div>
</div>
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Please note that the theorem applies in both directions, i.e. if we know that <math>x=a</math> is a zeros of <math>p(x)</math> we automatically would know that <math>p(x)</math> is divisible by <math>(x-a)</math>.
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Please note that the theorem applies in both directions, i.e. if we know that <math>x=a</math> is a zero of <math>p(x)</math> we automatically would know that <math>p(x)</math> is divisible by <math>(x-a)</math>.
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<ol type="a">
<ol type="a">
<li> Factorise the polynomial <math>\ x^2-3x-10\,</math>.
<li> Factorise the polynomial <math>\ x^2-3x-10\,</math>.
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+
<br>
 +
<br>
By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation <math>\ x^2-3x-10=0\ </math> has the solutions
By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation <math>\ x^2-3x-10=0\ </math> has the solutions
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<li> Factorise the polynomial <math>\ x^2+6x+9\,</math>.
<li> Factorise the polynomial <math>\ x^2+6x+9\,</math>.
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<br>
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<br>
This polynomial has a repeated root
This polynomial has a repeated root
{{Fristående formel||<math>x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3</math>}}
{{Fristående formel||<math>x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3</math>}}
 +
and thus <math>\ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,</math>.
and thus <math>\ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,</math>.
</li>
</li>
<li>Factorise the polynomial <math>\ x^2 -4x+5\,</math>.
<li>Factorise the polynomial <math>\ x^2 -4x+5\,</math>.
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+
<br>
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<br>
In this case, the polynomial has two complex roots
In this case, the polynomial has two complex roots
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Determine a cubic equation having zeros, <math>1</math> , <math>-1</math> and <math>3</math>.
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Determine a cubic polynomial having zeros, <math>1</math>, <math>-1</math> and <math>3</math>.
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<br>
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<br>
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The polynomial according to the factor theorem, must have factors <math>(x-1)</math>, <math>(x+1)</math> and <math>(x-3)</math>. Multiplying these factors, we get a cubic equation
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The polynomial according to the factor theorem, must have factors <math>(x-1)</math>, <math>(x+1)</math> and <math>(x-3)</math>. Multiplying these factors, we get a cubic polynomial
{{Fristående formel||<math>(x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}</math>}}
{{Fristående formel||<math>(x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}</math>}}
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== Fundamental theorem of algebra ==
== Fundamental theorem of algebra ==
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We introduced at the beginning of this chapter, the complex numbers to enable us to solve quadratic equations <math>x^2=-1</math> and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or do 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 as 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:
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We introduced at the beginning of this chapter, the complex numbers to enable us to solve the quadratic equation <math>x^2=-1</math> and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or do 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 as 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:
<div class="regel">
<div class="regel">
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<div class="regel">
<div class="regel">
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very polynomial of degree <math>n\ge1</math> has exactly <math>n</math> zeros if each zero is counted up to its multiplicity.
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Every polynomial of degree <math>n\ge1</math> has exactly <math>n</math> zeros if each zero is counted up to its multiplicity.
</div>
</div>
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(By multiplicity is meant that a double zero is counted twice , a triple zero 3 times, etc.)
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(By multiplicity is meant that a double zero is counted twice, a triple zero 3 times, etc.)
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Note that these theorems only say that there ''exists'' complex roots of polynomial, but not ''how'' to determine them. In general, there is no simple method to write a formula for the roots, but 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 .
+
Note that these theorems only say that there ''exists'' 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.

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Contents:

  • Factor theorem.
  • Polynomial division
  • Fundamental theorem of algebra

Learning outcomes:

After this section, you will have learned 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.

Polynom and equations

An expression of the form

  1. REDIRECT Template:Abgesetzte Formel

where n is an integer, is called a polynomial of degree n in an unknown variable x. The number a1 is called the coefficient of x, a2 the coefficient of x2, etc. The constant a0 is called the constant term.


Polynomials are essential for a large part of mathematics and have many properties which display great similarities with our integers, which means that we can do calculations with polynomials in a similar way as with integers.


Example 1


Compare the following integer written using a base 10,

  1. REDIRECT Template:Abgesetzte Formel

with the polynomial in x

  1. REDIRECT Template:Abgesetzte Formel

and then the following divisions,

  • 111353=123 as  1353=12311,
  • x+1x3+3x2+5x+3=x2+2x+3 since  x3+3x2+5x+3=(x2+2x+3)(x+1).

If p(x) is a polynomial of degree n then p(x)=0 is called a polynomial equation of degree n. If x=a is a number such that p(a)=0 then x=a is called a root, or solution to the equation. One also says that x=a is a zero of p(x).

As the above example showed polynomials can be divided just like integers. Polynomial division, just like integer division is usually not exact. If, for example, 37 is divided by 5, one gets

  1. REDIRECT Template:Abgesetzte Formel

The calculation can also be written as  37=75+2. The number 7 is called the quotient and the number 2 the remainder. One says that dividing 37 by 5 gives the quotient 7 and the remainder 2.


If p(x) and q(x) are polynomials one similarly can divide p(x) with q(x) and unambiguously determine polynomials k(x) and r(x) such that

  1. REDIRECT Template:Abgesetzte Formel

or  p(x)=k(x)q(x)+r(x). One here says that polynomial division has resulted in a quotient k(x) and remainder r(x).


It is obvious that a division is exact if the remainder is zero. For polynomials this is expressed as follows: If r(x)=0 then p(x) is divisible by q(x), or, q(x) is a divisor of p(x). One writes

  1. REDIRECT Template:Abgesetzte Formel

or  p(x)=k(x)q(x).


Polynomial division

If p(x) is a polynomial of higher degree than polynomial q(x) then one can divide p(x) by q(x). For example, this may be done by gradually subtracting appropriate multiples of q(x) from p(x) until a remaining numerator is of lower degree than the denominator q(x).


Example 2


Perform polynomial divisionen for  x+2x3+x2−x+4.


The first step is that we add and subtract an appropriate x2-term in the numerator

  1. REDIRECT Template:Abgesetzte Formel

The reason why we do this is that the sub-expression x3+2x2 can be written as x2(x+2) and cancellation with the denominator can be done,

  1. REDIRECT Template:Abgesetzte Formel

Then we add and subtract an appropriate x-term so that the leading x2-term in the numerator can be cancelled,

  1. REDIRECT Template:Abgesetzte Formel

The last step is that we add and subtract a constant

  1. REDIRECT Template:Abgesetzte Formel

Thus ending up with

  1. REDIRECT Template:Abgesetzte Formel

The quotient is x2−x+1 and the remainder is 2. Since the remainder is not zero division is not exact, that is, q(x)=x+2 is not a divisor of p(x)=x3+x2−x+4.


The connection between factors and zeros

If q(x) is a divisor of p(x) then p(x)=k(x)q(x). We have thus factorised p(x) . One says that q(x) is a factor of p(x). Especially, if a polynomial of first degree (x−a) is a divisor of p(x) then (x−a) is a factor of p(x) , i.e.

  1. REDIRECT Template:Abgesetzte Formel

Since  p(a)=q(a)(a−a)=q(a)0=0  this means that x=a is a zero of p(x). This is exactly the content of the so-called factor theorem.

Factor theorem:

(x−a) is a divisor of a polynomial p(x) if and only if x=a is a zero of p(x).

Please note that the theorem applies in both directions, i.e. if we know that x=a is a zero of p(x) we automatically would know that p(x) is divisible by (x−a).


Example 3


The polynomial p(x)=x2−6x+8 can be factorised as

  1. REDIRECT Template:Abgesetzte Formel

and has therefore zeros at x=2 and x=4 (and no others). It is precisely these that are obtained if one solves the equation  x2−6x+8=0.

Example 4


  1. Factorise the polynomial  x2−3x−10.

    By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation  x2−3x−10=0  has the solutions
    1. REDIRECT Template:Abgesetzte Formel
    i.e. x=−2 and x=5. This means that  x2−3x−10=(x−(−2))(x−5)=(x+2)(x−5).
  2. Factorise the polynomial  x2+6x+9.

    This polynomial has a repeated root
    1. REDIRECT Template:Abgesetzte Formel
    and thus  x2+6x+9=(x−(−3))(x−(−3))=(x+3)2.
  3. Factorise the polynomial  x2−4x+5.

    In this case, the polynomial has two complex roots
    1. REDIRECT Template:Abgesetzte Formel
    and when factorised will be  (x−(2−i))(x−(2+i)).

Example 5


Determine a cubic polynomial having zeros, 1, −1 and 3.

The polynomial according to the factor theorem, must have factors (x−1), (x+1) and (x−3). Multiplying these factors, we get a cubic polynomial

  1. REDIRECT Template:Abgesetzte Formel


Fundamental theorem of algebra

We introduced at the beginning of this chapter, the complex numbers to enable us to solve the quadratic equation x2=−1 and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or do 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 as 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 n1 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 n1 has exactly 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 exists 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.


Example 6


Show that the polynomial p(x)=x4−4x3+6x2−4x+5 has zeros x=i and x=2−i. Thus determine the other zeros.


We have

  1. REDIRECT Template:Abgesetzte Formel

In order to calculate the last term, we need to determine

  1. REDIRECT Template:Abgesetzte Formel

This gives that

  1. REDIRECT Template:Abgesetzte Formel

which proves that i and 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 z=−i and 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 first order 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 first and second order real factors.


Example 7


Show that x=1 is a zero of p(x)=x3+x2−2. Then first factorise p(x) into polynomials having real coefficients and then factorise p(x) completely into first order factors.


We have that  p(1)=13+12−2=0  which shows that x=1 is a zero of the polynomial. According to the factor theorem, this means that x−1 is a factor of p(x), i.e. p(x) is divisible by x−1. We therefore divide the polynomial with x−1 to get the remaining factors after x−1 is factored out of the polynomial

  1. REDIRECT Template:Abgesetzte Formel

So we have  p(x)=(x−1)(x2+2x+2) which is the first part of the problem.


It now remains to factorise x2+2x+2. The equation x2+2x+2=0 has the solutions

  1. REDIRECT Template:Abgesetzte Formel

and therefore the polynomial has the following factorization into complex first order factors.

  1. REDIRECT Template:Abgesetzte Formel
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/3.4_Komplexe_Polynome“