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Lösung 3.4:4

Aus Online Mathematik Brückenkurs 2

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Because
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Because <math>z=1-2i</math> should be a root of the equation, we can substitute
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<math>z=\text{1}-\text{2}i</math>
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<math>z=1-2i</math> in and the equation should be satisfied,
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should be a root of the equation, we can substitute
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<math>z=\text{1}-\text{2}i</math>
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in and the equation should be satisfied:
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{{Displayed math||<math>(1-2i)^3 + a(1-2i) + b = 0\,\textrm{.}</math>}}
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<math>\left( \text{1}-\text{2}i \right)^{3}+a\left( \text{1}-\text{2}i \right)+b=0</math>
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We will therefore adjust the constants <math>a</math> and <math>b</math> so that the relation above holds. We simplify the left-hand side,
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{{Displayed math||<math>-11+2i+a(1-2i)+b=0</math>}}
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We will therefore adjust the constants
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and collect together the real and imaginary parts,
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<math>a</math>
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and
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<math>b</math>
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so that the relation above holds. We simplify the left-hand side,
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<math>-11+2i+a\left( \text{1}-\text{2}i \right)+b=0</math>
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and collect together the real and imaginary parts:
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<math>\left( -11+a+b \right)+\left( 2-2a \right)i=0</math>
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{{Displayed math||<math>(-11+a+b)+(2-2a)i=0\,\textrm{.}</math>}}
If the left-hand side is to equal the right-hand side, the left-hand side's real and imaginary parts must be equal to zero, i.e.
If the left-hand side is to equal the right-hand side, the left-hand side's real and imaginary parts must be equal to zero, i.e.
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{{Displayed math||<math>\left\{\begin{align}
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-11+a+b &= 0\,,\\[5pt]
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2-2a &= 0\,\textrm{.}
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\end{align}\right.</math>}}
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This gives <math>a=1</math> and <math>b=10</math>.
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<math>\left\{ \begin{array}{*{35}l}
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-11+a+b=0 \\
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2-2a=0 \\
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\end{array} \right.</math>
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This gives
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<math>a=\text{1}</math>
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and
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<math>b=\text{1}0</math>.
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The equation is thus
The equation is thus
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{{Displayed math||<math>z^3+z+10=0</math>}}
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<math>z^{2}+z+10=0</math>
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and has the prescribed root <math>z=1-2i</math>.
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What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root <math>z=1+2i</math>.
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and has the prescribed root
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<math>z=\text{1}-\text{2}i</math>.
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What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root
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<math>z=\text{1}+\text{2}i</math>.
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Hence, we know two of the equation's three roots and we can obtain the third root with help of the factor theorem. According to the factor theorem, the equation's left-hand side contains the factor
Hence, we know two of the equation's three roots and we can obtain the third root with help of the factor theorem. According to the factor theorem, the equation's left-hand side contains the factor
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{{Displayed math||<math>\bigl(z-(1-2i)\bigr)\bigl(z-(1+2i)\bigr)=z^2-2z+5</math>}}
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<math>\left( z-\left( 1-2i \right) \right)\left( z-\left( 1+2i \right) \right)=z^{2}-2z+5</math>
 
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and this means that we can write
and this means that we can write
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{{Displayed math||<math>z^3+z+10 = (z-A)(z^2-2z+5)</math>}}
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<math>z^{3}+z+10=\left( z-A \right)\left( z^{2}-2z+5 \right)</math>
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where <math>z-A</math> is the factor which corresponds to the third root <math>z=A</math>. Using polynomial division, we obtain the factor
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where
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<math>z-A</math>
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is the factor which corresponds to the third root
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<math>z=A</math>. Using polynomial division, we obtain the factor
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<math>\begin{align}
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& z-A=\frac{z^{3}+z+10}{z^{2}-2z+5} \\
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& =\frac{z^{3}-2z^{2}+5z+2z^{2}-5z+z+10}{z^{2}-2z+5} \\
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& =\frac{z\left( z^{2}-2z+5 \right)+2z^{2}-4z+10}{z^{2}-2z+5} \\
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& =z+\frac{2z^{2}-4z+10}{z^{2}-2z+5} \\
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& =z+\frac{2\left( z^{2}-2z+5 \right)}{z^{2}-2z+5} \\
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& =z+2 \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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z-A
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&= \frac{z^3+z+10}{z^2-2z+5}\\[5pt]
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&= \frac{z^3-2z^2+5z+2z^2-5z+z+10}{z^2-2z+5}\\[5pt]
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&= \frac{z(z^2-2z+5)+2z^2-4z+10}{z^2-2z+5}\\[5pt]
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&= z + \frac{2z^2-4z+10}{z^2-2z+5}\\[5pt]
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&= z + \frac{2(z^2-2z+5)}{z^2-2z+5}\\[5pt]
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&= z+2\,\textrm{.}
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\end{align}</math>}}
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Thus, the remaining root is
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Thus, the remaining root is <math>z=-2</math>.
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<math>z=-\text{2}</math>.
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Version vom 13:45, 31. Okt. 2008

Because z=1−2i should be a root of the equation, we can substitute z=1−2i in and the equation should be satisfied,

(1−2i)3+a(1−2i)+b=0.

We will therefore adjust the constants a and b so that the relation above holds. We simplify the left-hand side,

−11+2i+a(1−2i)+b=0

and collect together the real and imaginary parts,

(−11+a+b)+(2−2a)i=0.

If the left-hand side is to equal the right-hand side, the left-hand side's real and imaginary parts must be equal to zero, i.e.

−11+a+b2−2a=0=0. 

This gives a=1 and b=10.

The equation is thus

z3+z+10=0

and has the prescribed root z=1−2i.

What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root z=1+2i.

Hence, we know two of the equation's three roots and we can obtain the third root with help of the factor theorem. According to the factor theorem, the equation's left-hand side contains the factor

z−(1−2i)z−(1+2i)=z2−2z+5 

and this means that we can write

z3+z+10=(z−A)(z2−2z+5)

where z−A is the factor which corresponds to the third root z=A. Using polynomial division, we obtain the factor

z−A=z2−2z+5z3+z+10=z2−2z+5z3−2z2+5z+2z2−5z+z+10=z2−2z+5z(z2−2z+5)+2z2−4z+10=z+z2−2z+52z2−4z+10=z+z2−2z+52(z2−2z+5)=z+2.

Thus, the remaining root is z=−2.

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