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Lösung 3.4:4

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 2: Zeile 2:
<math>z=1-2i</math> in and the equation should be satisfied,
<math>z=1-2i</math> in and the equation should be satisfied,
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{{Displayed math||<math>(1-2i)^3 + a(1-2i) + b = 0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>(1-2i)^3 + a(1-2i) + b = 0\,\textrm{.}</math>}}
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,
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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{{Abgesetzte Formel||<math>-11+2i+a(1-2i)+b=0</math>}}
and collect together the real and imaginary parts,
and collect together the real and imaginary parts,
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{{Displayed math||<math>(-11+a+b)+(2-2a)i=0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<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}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
-11+a+b &= 0\,,\\[5pt]
-11+a+b &= 0\,,\\[5pt]
2-2a &= 0\,\textrm{.}
2-2a &= 0\,\textrm{.}
Zeile 23: Zeile 23:
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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{{Abgesetzte Formel||<math>z^3+z+10=0</math>}}
and has the prescribed root <math>z=1-2i</math>.
and has the prescribed root <math>z=1-2i</math>.
Zeile 31: Zeile 31:
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>}}
+
{{Abgesetzte Formel||<math>\bigl(z-(1-2i)\bigr)\bigl(z-(1+2i)\bigr)=z^2-2z+5</math>}}
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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{{Abgesetzte Formel||<math>z^3+z+10 = (z-A)(z^2-2z+5)</math>}}
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
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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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z-A
z-A
&= \frac{z^3+z+10}{z^2-2z+5}\\[5pt]
&= \frac{z^3+z+10}{z^2-2z+5}\\[5pt]

Version vom 13:15, 10. Mär. 2009

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 \displaystyle z=A. Using polynomial division, we obtain the factor

\displaystyle \begin{align}

z-A &= \frac{z^3+z+10}{z^2-2z+5}\\[5pt] &= \frac{z^3-2z^2+5z+2z^2-5z+z+10}{z^2-2z+5}\\[5pt] &= \frac{z(z^2-2z+5)+2z^2-4z+10}{z^2-2z+5}\\[5pt] &= z + \frac{2z^2-4z+10}{z^2-2z+5}\\[5pt] &= z + \frac{2(z^2-2z+5)}{z^2-2z+5}\\[5pt] &= z+2\,\textrm{.} \end{align}

Thus, the remaining root is \displaystyle z=-2.

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