Aus Online Mathematik Brückenkurs 2

Version vom 15:13, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.4:4 moved to Solution 3.4:4: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:32, 28. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	Because
-	<center> [[Image:3_4_4.gif]] </center>	+	<math>z=\text{1}-\text{2}i</math>
-	{{NAVCONTENT_STOP}}	+	should be a root of the equation, we can substitute
		+	<math>z=\text{1}-\text{2}i</math>
		+	in and the equation should be satisfied:
		+
		+
		+	<math>\left( \text{1}-\text{2}i \right)^{3}+a\left( \text{1}-\text{2}i \right)+b=0</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,
		+
		+
		+	<math>-11+2i+a\left( \text{1}-\text{2}i \right)+b=0</math>
		+
		+
		+	and collect together the real and imaginary parts:
		+
		+
		+	<math>\left( -11+a+b \right)+\left( 2-2a \right)i=0</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.
		+
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	-11+a+b=0 \\
		+	2-2a=0 \\
		+	\end{array} \right.</math>
		+
		+
		+	This gives
		+	<math>a=\text{1}</math>
		+	and
		+	<math>b=\text{1}0</math>.
		+
		+	The equation is thus
		+
		+
		+	<math>z^{2}+z+10=0</math>
		+
		+
		+	and has the prescribed root
		+	<math>z=\text{1}-\text{2}i</math>.
		+
		+	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=\text{1}+\text{2}i</math>.
		+
		+	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
		+
		+
		+	<math>\left( z-\left( 1-2i \right) \right)\left( z-\left( 1+2i \right) \right)=z^{2}-2z+5</math>
		+
		+
		+	and this means that we can write
		+
		+
		+	<math>z^{3}+z+10=\left( z-A \right)\left( z^{2}-2z+5 \right)</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
		+
		+
		+	<math>\begin{align}
		+	& z-A=\frac{z^{3}+z+10}{z^{2}-2z+5} \\
		+	& =\frac{z^{3}-2z^{2}+5z+2z^{2}-5z+z+10}{z^{2}-2z+5} \\
		+	& =\frac{z\left( z^{2}-2z+5 \right)+2z^{2}-4z+10}{z^{2}-2z+5} \\
		+	& =z+\frac{2z^{2}-4z+10}{z^{2}-2z+5} \\
		+	& =z+\frac{2\left( z^{2}-2z+5 \right)}{z^{2}-2z+5} \\
		+	& =z+2 \\
		+	\end{align}</math>
		+
		+
		+	Thus, the remaining root is
		+	<math>z=-\text{2}</math>.

---

Version vom 13:32, 28. Okt. 2008

Because \displaystyle z=\text{1}-\text{2}i should be a root of the equation, we can substitute \displaystyle z=\text{1}-\text{2}i in and the equation should be satisfied:

\displaystyle \left( \text{1}-\text{2}i \right)^{3}+a\left( \text{1}-\text{2}i \right)+b=0

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

\displaystyle -11+2i+a\left( \text{1}-\text{2}i \right)+b=0

and collect together the real and imaginary parts:

\displaystyle \left( -11+a+b \right)+\left( 2-2a \right)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.

\displaystyle \left\{ \begin{array}{*{35}l} -11+a+b=0 \\ 2-2a=0 \\ \end{array} \right.

This gives \displaystyle a=\text{1} and \displaystyle b=\text{1}0.

The equation is thus

\displaystyle z^{2}+z+10=0

and has the prescribed root \displaystyle z=\text{1}-\text{2}i.

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

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

\displaystyle \left( z-\left( 1-2i \right) \right)\left( z-\left( 1+2i \right) \right)=z^{2}-2z+5

and this means that we can write

\displaystyle z^{3}+z+10=\left( z-A \right)\left( z^{2}-2z+5 \right)

where \displaystyle 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} \\ & =\frac{z^{3}-2z^{2}+5z+2z^{2}-5z+z+10}{z^{2}-2z+5} \\ & =\frac{z\left( z^{2}-2z+5 \right)+2z^{2}-4z+10}{z^{2}-2z+5} \\ & =z+\frac{2z^{2}-4z+10}{z^{2}-2z+5} \\ & =z+\frac{2\left( z^{2}-2z+5 \right)}{z^{2}-2z+5} \\ & =z+2 \\ \end{align}

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