Lösung 3.4:2
Aus Online Mathematik Brückenkurs 2
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| - | {{ | + | If the equation has the root |
| - | < | + | <math>z=\text{1}</math>, this means, according to the factor rule, that the equation mustcontain the |
| - | {{ | + | <math>z=\text{1}</math>, i.e. the polynomial on the left-hand side can be written as |
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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 other unknown factor using polynomial division: | ||
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| + | <math>\begin{align} | ||
| + | & z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right) \\ | ||
| + | & z^{2}+Az+B=\frac{z^{3}-3z^{2}+4z-2}{z-1} \\ | ||
| + | & =\frac{z^{3}-z^{2}+z^{2}-3z^{2}+4z-2}{z-1} \\ | ||
| + | & =\frac{z^{2}\left( z-1 \right)-2z^{2}+4z-2}{z-1} \\ | ||
| + | & =z^{2}+\frac{-2z^{2}+4z-2}{z-1} \\ | ||
| + | & =z^{2}+\frac{-2z^{2}+2z-2z+4z-2}{z-1} \\ | ||
| + | & =z^{2}+\frac{-2z\left( z-1 \right)+2z-2}{z-1} \\ | ||
| + | & =z^{2}-2z+\frac{2z-2}{z-1} \\ | ||
| + | & =z^{2}-2z+\frac{2\left( z-1 \right)}{z-1} \\ | ||
| + | & =z^{2}-2z+2 \\ | ||
| + | \end{align}</math> | ||
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| + | Thus, the equation can be written as | ||
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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 | ||
| + | <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-\text{1}</math> | ||
| + | or | ||
| + | <math>z^{2}-2z+2</math> | ||
| + | is zero, and we see directly that | ||
| + | <math>z-\text{1}</math> | ||
| + | is zero only when | ||
| + | <math>z=\text{1}</math>. | ||
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| + | Hence, we determine the roots by solving the equation | ||
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| + | <math>z^{2}-2z+2=0</math> | ||
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| + | Completing the square gives | ||
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| + | <math>\begin{align} | ||
| + | & \left( z-\text{1} \right)^{2}-1^{2}+2=0 \\ | ||
| + | & \left( z-\text{1} \right)^{2}=-1 \\ | ||
| + | \end{align}</math> | ||
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| + | |||
| + | and taking the root gives that | ||
| + | <math>z-\text{1}=\pm i</math> | ||
| + | i.e. | ||
| + | <math>z=1-i</math> | ||
| + | and | ||
| + | <math>z=1+i</math>. | ||
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| + | The equation's other roots are | ||
| + | <math>z=1-i</math> | ||
| + | and | ||
| + | <math>z=1+i</math>. | ||
| + | |||
| + | As an extra check, we investigate whether | ||
| + | <math>z-\text{1}=\pm i</math> | ||
| + | really are roots of the equation. | ||
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| + | <math>\begin{align} | ||
| + | & z=1+i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\ | ||
| + | & =\left( \left( 1+i-3 \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\ | ||
| + | & =\left( \left( -2+i \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\ | ||
| + | & =\left( -2+i-2i-1+4 \right)\left( 1+i \right)-2 \\ | ||
| + | & =\left( 1-i \right)\left( 1+i \right)-2 \\ | ||
| + | & =1^{2}-i^{2}-2=1+1-2=0 \\ | ||
| + | \end{align}</math> | ||
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| + | <math>\begin{align} | ||
| + | & z=1-i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\ | ||
| + | & =\left( \left( 1-i-3 \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\ | ||
| + | & =\left( \left( -2-i \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\ | ||
| + | & =\left( -2-i+2i-1+4 \right)\left( 1-i \right)-2 \\ | ||
| + | & =\left( 1+i \right)\left( 1-i \right)-2 \\ | ||
| + | & =1^{2}-i^{2}-2=1+1-2=0 \\ | ||
| + | \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. | ||
Version vom 09:25, 28. Okt. 2008
If the equation has the root
z2+Az+B
z−1
for some constants
z2+Az+Bz−1z2+Az+B=z−1z3−3z2+4z−2=z−1z3−z2+z2−3z2+4z−2=z−1z2z−1−2z2+4z−2=z2+z−1−2z2+4z−2=z2+z−1−2z2+2z−2z+4z−2=z2+z−1−2zz−1+2z−2=z2−2z+z−12z−2=z2−2z+z−12z−1=z2−2z+2
Thus, the equation can be written as
z−1z2−2z+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
Hence, we determine the roots by solving the equation
Completing the square gives
z−12−12+2=0z−12=−1
and taking the root gives that
i
The equation's other roots are
As an extra check, we investigate whether
i
z−3z+4z−2=1+i−31+i+41+i−2=−2+i1+i+41+i−2=−2+i−2i−1+41+i−2=1−i1+i−2=12−i2−2=1+1−2=0
z−3z+4z−2=1−i−31−i+41−i−2=−2−i1−i+41−i−2=−2−i+2i−1+41−i−2=1+i1−i−2=12−i2−2=1+1−2=0
NOTE: Writing
z−3z+4z−2
is known as the Horner scheme and is used to reduce the amount of the arithmetical work.
