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Lösung 3.4:2

Aus Online Mathematik Brückenkurs 2

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If the equation has the root
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If the equation has the root <math>z=1</math>, this means, according to the factor rule, that the equation must contain the factor <math>z-1</math>, i.e. the polynomial on the left-hand side can be written as
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<math>z=\text{1}</math>, this means, according to the factor rule, that the equation mustcontain the
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<math>z=\text{1}</math>, i.e. the polynomial on the left-hand side can be written as
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{{Displayed math||<math>z^3-3z^2+4z-2 = (z^2+Az+B)(z-1)</math>}}
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<math>z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right)</math>
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for some constants <math>A</math> and <math>B</math>. We can determine the second unknown factor using polynomial division,
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for some constants
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<math>A</math>
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and
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<math>B</math>. We can determine the other unknown factor using polynomial division:
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<math>\begin{align}
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& z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right) \\
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& z^{2}+Az+B=\frac{z^{3}-3z^{2}+4z-2}{z-1} \\
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& =\frac{z^{3}-z^{2}+z^{2}-3z^{2}+4z-2}{z-1} \\
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& =\frac{z^{2}\left( z-1 \right)-2z^{2}+4z-2}{z-1} \\
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& =z^{2}+\frac{-2z^{2}+4z-2}{z-1} \\
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& =z^{2}+\frac{-2z^{2}+2z-2z+4z-2}{z-1} \\
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& =z^{2}+\frac{-2z\left( z-1 \right)+2z-2}{z-1} \\
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& =z^{2}-2z+\frac{2z-2}{z-1} \\
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& =z^{2}-2z+\frac{2\left( z-1 \right)}{z-1} \\
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& =z^{2}-2z+2 \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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z^2+Az+B
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&= \frac{z^3-3z^2+4z-2}{z-1}\\[5pt]
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&= \frac{z^3-z^2+z^2-3z^2+4z-2}{z-1}\\[5pt]
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&= \frac{z^2(z-1)-2z^2+4z-2}{z-1}\\[5pt]
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&= z^2 + \frac{-2z^2+4z-2}{z-1}\\[5pt]
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&= z^2 + \frac{-2z^2+2z-2z+4z-2}{z-1}\\[5pt]
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&= z^2 + \frac{-2z(z-1)+2z-2}{z-1}\\[5pt]
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&= z^2 - 2z + \frac{2z-2}{z-1}\\[5pt]
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&= z^2 - 2z + \frac{2(z-1)}{z-1}\\[5pt]
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&= z^2 - 2z + 2\,\textrm{.}
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\end{align}</math>}}
Thus, the equation can be written as
Thus, the equation can be written as
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{{Displayed math||<math>(z-1)(z^2-2z+2) = 0\,\textrm{.}</math>}}
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<math>\left( z-1 \right)\left( z^{2}-2z+2 \right)=0</math>
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The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor
The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor
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<math>z^{2}-2z+2</math>. This is because the left-hand side is zero only when at least one of the factors
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<math>z^2-2z+2</math>. This is because the left-hand side is zero only when at least one of the factors <math>z-1</math> or <math>z^2-2z+2</math> is zero, and we see directly that <math>z-1</math> is zero only when <math>z=1\,</math>.
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<math>z-\text{1}</math>
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or
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<math>z^{2}-2z+2</math>
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is zero, and we see directly that
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<math>z-\text{1}</math>
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is zero only when
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<math>z=\text{1}</math>.
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Hence, we determine the roots by solving the equation
Hence, we determine the roots by solving the equation
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{{Displayed math||<math>z^2-2z+2 = 0\,\textrm{.}</math>}}
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<math>z^{2}-2z+2=0</math>
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Completing the square gives
Completing the square gives
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{{Displayed math||<math>\begin{align}
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(z-1)^2-1^2+2 &= 0\,,\\[5pt]
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(z-1)^2 &= -1\,,
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\end{align}</math>}}
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<math>\begin{align}
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and taking the root gives that <math>z-1=\pm i</math>, i.e. <math>z=1-i</math> and
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& \left( z-\text{1} \right)^{2}-1^{2}+2=0 \\
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<math>z=1+i\,</math>.
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& \left( z-\text{1} \right)^{2}=-1 \\
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\end{align}</math>
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The equation's other roots are <math>z=1-i</math> and <math>z=1+i\,</math>.
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and taking the root gives that
 
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<math>z-\text{1}=\pm i</math>
 
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i.e.
 
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<math>z=1-i</math>
 
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and
 
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<math>z=1+i</math>.
 
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The equation's other roots are
 
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<math>z=1-i</math>
 
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and
 
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<math>z=1+i</math>.
 
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As an extra check, we investigate whether
 
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<math>z-\text{1}=\pm i</math>
 
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really are roots of the equation.
 
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As an extra check, we investigate whether <math>z = 1 \pm i</math> really are roots of the equation.
<math>\begin{align}
<math>\begin{align}
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& z=1+i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\
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z = 1+i:\quad z^3-3z^2+4z-2
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& =\left( \left( 1+i-3 \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\
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&= \bigl((z-3)z+4\bigr)z-2\\[5pt]
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& =\left( \left( -2+i \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\
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&= \bigl((1+i-3)(1+i)+4\bigr)(1+i)-2\\[5pt]
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& =\left( -2+i-2i-1+4 \right)\left( 1+i \right)-2 \\
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&= \bigl((-2+i)(1+i)+4\bigr)(1+i)-2\\[5pt]
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& =\left( 1-i \right)\left( 1+i \right)-2 \\
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&= (-2+i-2i-1+4)(1+i)-2\\[5pt]
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& =1^{2}-i^{2}-2=1+1-2=0 \\
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&= (1-i)(1+i)-2\\[5pt]
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&= 1^2 - i^2 - 2\\[5pt]
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&= 1+1-2\\[5pt]
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&= 0\,,\\[10pt]
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z = 1-i:\quad z^3-3z^2+4z-2
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&= \bigl((z-3)z+4\bigr)z-2\\[5pt]
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&= \bigl((1-i-3)(1-i)+4\bigr)(1-i)-2\\[5pt]
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&= \bigl((-2-i)(1-i)+4\bigr)(1-i)-2\\[5pt]
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&= (-2-i+2i-1+4)(1-i)-2\\[5pt]
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&= (1+i)(1-i)-2\\[5pt]
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&= 1^2-i^2-2\\[5pt]
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&= 1+1-2\\[5pt]
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&= 0\,\textrm{.}
\end{align}</math>
\end{align}</math>
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Note: Writing
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<math>\begin{align}
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{{Displayed math||<math>z^3-3z^2+4z-2 = \bigl((z-3)z+4\bigr)z-2</math>}}
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& z=1-i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\
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& =\left( \left( 1-i-3 \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\
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& =\left( \left( -2-i \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\
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& =\left( -2-i+2i-1+4 \right)\left( 1-i \right)-2 \\
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& =\left( 1+i \right)\left( 1-i \right)-2 \\
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& =1^{2}-i^{2}-2=1+1-2=0 \\
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\end{align}</math>
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NOTE: Writing
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<math>z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2</math>
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is known as the Horner scheme and is used to reduce the amount of the arithmetical work.
is known as the Horner scheme and is used to reduce the amount of the arithmetical work.

Version vom 13:14, 31. Okt. 2008

If the equation has the root z=1, this means, according to the factor rule, that the equation must contain the factor z−1, i.e. the polynomial on the left-hand side can be written as

z3−3z2+4z−2=(z2+Az+B)(z−1)

for some constants A and B. We can determine the second unknown factor using polynomial division,

z2+Az+B=z−1z3−3z2+4z−2=z−1z3−z2+z2−3z2+4z−2=z−1z2(z−1)−2z2+4z−2=z2+z−1−2z2+4z−2=z2+z−1−2z2+2z−2z+4z−2=z2+z−1−2z(z−1)+2z−2=z2−2z+z−12z−2=z2−2z+z−12(z−1)=z2−2z+2.

Thus, the equation can be written as

(z−1)(z2−2z+2)=0.

The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor z2−2z+2. This is because the left-hand side is zero only when at least one of the factors z−1 or z2−2z+2 is zero, and we see directly that z−1 is zero only when z=1.

Hence, we determine the roots by solving the equation

z2−2z+2=0.

Completing the square gives

(z−1)2−12+2(z−1)2=0=−1

and taking the root gives that z−1=i, i.e. z=1−i and z=1+i.

The equation's other roots are z=1−i and z=1+i.


As an extra check, we investigate whether z=1i really are roots of the equation.

z=1+i:z3−3z2+4z−2z=1−i:z3−3z2+4z−2=(z−3)z+4z−2=(1+i−3)(1+i)+4(1+i)−2=(−2+i)(1+i)+4(1+i)−2=(−2+i−2i−1+4)(1+i)−2=(1−i)(1+i)−2=12−i2−2=1+1−2=0=(z−3)z+4z−2=(1−i−3)(1−i)+4(1−i)−2=(−2−i)(1−i)+4(1−i)−2=(−2−i+2i−1+4)(1−i)−2=(1+i)(1−i)−2=12−i2−2=1+1−2=0.


Note: Writing

z3−3z2+4z−2=(z−3)z+4z−2 

is known as the Horner scheme and is used to reduce the amount of the arithmetical work.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:2“