Aus Online Mathematik Brückenkurs 2

Version vom 15:10, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.3:5d moved to Solution 3.3:5d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 11:32, 26. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	Let us first divide both sides by
-	<center> [[Image:3_3_5d-1(5).gif]] </center>	+	<math>4+i</math>, so that the coefficient in front of
-	{{NAVCONTENT_STOP}}	+	<math>z^{2}</math>
-	{{NAVCONTENT_START}}	+	becomes
-	<center> [[Image:3_3_5d-2(5).gif]] </center>	+	<math>\text{1}</math>,
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_3_5d-3(5).gif]] </center>	+	<math>z^{2}+\frac{1-21i}{4+i}z=\frac{17}{4+i}</math>
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_3_5d-4(5).gif]] </center>	+	The two complex quotients become
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_3_5d-5(5).gif]] </center>	+	<math>\begin{align}
-	{{NAVCONTENT_STOP}}	+	& \frac{\left( 1-21i \right)\left( 4-i \right)}{\left( 4+i \right)\left( 4-i \right)}=\frac{4-i-84i+21i^{2}}{4^{2}-i^{2}} \\
		+	& =\frac{-17-85i}{16+1}=\frac{-17-85i}{17}=-1-5i \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \frac{17}{4+i}=\frac{17\left( 4-i \right)}{\left( 4+i \right)\left( 4-i \right)}=\frac{17\left( 4-i \right)}{4^{2}-i^{2}} \\
		+	& =\frac{17\left( 4-i \right)}{17}=4-i \\
		+	\end{align}</math>
		+
		+
		+	Thus, the equation becomes
		+
		+
		+	<math>z^{2}-\left( 1+5i \right)z=4-i</math>
		+
		+
		+	Now, we complete the square of the left-hand side:
		+
		+
		+	<math>\begin{align}
		+	& \left( z-\frac{1+5i}{2} \right)^{2}-\left( \frac{1+5i}{2} \right)^{2}=4-i \\
		+	& \left( z-\frac{1+5i}{2} \right)^{2}-\left( \frac{1}{4}+\frac{5}{2}i+\frac{25}{4}i^{2} \right)=4-i \\
		+	& \left( z-\frac{1+5i}{2} \right)^{2}-\frac{1}{4}-\frac{5}{2}i+\frac{25}{4}=4-i \\
		+	& \left( z-\frac{1+5i}{2} \right)^{2}=-2+\frac{3}{2}i \\
		+	\end{align}</math>
		+
		+
		+	If we set
		+	<math>w=z-\frac{1+5i}{2}</math>, we have a binomial equation in
		+	<math>w</math>,
		+
		+
		+	<math>w^{2}=-2+\frac{3}{2}i</math>
		+
		+
		+	which we solve by putting
		+	<math>w=x+iy\text{ }</math>,
		+
		+
		+	<math>\left( x+iy \right)^{2}=-2+\frac{3}{2}i</math>
		+
		+
		+	or, if the left-hand side is expanded,
		+
		+
		+	<math>x^{2}-y^{2}+2xyi=-2+\frac{3}{2}i</math>
		+
		+
		+	If we identify the real and imaginary parts on both sides, we get
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-y^{2}=-2 \\
		+	& 2xy=\frac{3}{2} \\
		+	\end{align}</math>
		+
		+
		+	and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:
		+
		+
		+	<math>x^{2}+y^{2}=\sqrt{\left( -2 \right)^{2}+\left( \frac{3}{2} \right)^{2}}=\frac{5}{2}</math>
		+
		+
		+	Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.
		+
		+	Together, the three relations constitute the following system of equations:
		+
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	\ x^{2}-y^{2}=2 \\
		+	\ 2xy=-\frac{3}{2} \\
		+	\ x^{2}+y^{2}=\frac{5}{2} \\
		+	\end{array} \right.</math>
		+
		+
		+
		+	From the first and the third equations, we can relatively easily obtain the values that
		+	<math>x</math>
		+	and
		+	<math>y</math>
		+	can take.
		+
		+	Add the first and third equations,
		+
		+	EQ1
		+
		+	which gives that
		+	<math>x=\pm \frac{1}{2}</math>.
		+
		+	Then, subtract the first equation from the third equation,
		+
		+	EQ13
		+
		+	i.e.
		+	<math>y=\pm \frac{3}{2}</math>.
		+
		+	This gives four possible combinations,
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{1}{2} \\
		+	y=\frac{3}{2} \\
		+	\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
		+	x=\frac{1}{2} \\
		+	y=-\frac{3}{2} \\
		+	\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
		+	x=-\frac{1}{2} \\
		+	y=\frac{3}{2} \\
		+	\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
		+	x=-\frac{1}{2} \\
		+	y=-\frac{3}{2} \\
		+	\end{array} \right.</math>
		+
		+
		+	of which only two also satisfy the second equation.
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{1}{2} \\
		+	y=\frac{3}{2} \\
		+	\end{array} \right.\quad \quad \text{and}\quad \quad \left\{ \begin{array}{*{35}l}
		+	x=-\frac{1}{2} \\
		+	y=-\frac{3}{2} \\
		+	\end{array} \right.</math>
		+
		+
		+	This means that the binomial equation has the two solutions,
		+
		+
		+
		+	<math>w=\frac{1}{2}+\frac{3}{2}i</math>
		+	and
		+	<math>w=\frac{1}{-2}-\frac{3}{2}i</math>
		+
		+
		+	and that the original equation has the solutions
		+
		+
		+	<math>z=1+4i</math>
		+	and
		+	<math>z=i</math>
		+
		+	according to the relation
		+	<math>w=z-\frac{1+5i}{2}</math>.
		+
		+	Finally, we check that the solutions really do satisfy the equation.
		+
		+
		+
		+	<math>\begin{align}
		+	& 1+4i:\quad \left( 4+i \right)z^{2}+\left( 1-21i \right)z \\
		+	& =\left( 4+i \right)\left( 1+4i \right)^{2}+\left( 1-21i \right)\left( 1+4i \right) \\
		+	& =\left( 4+i \right)\left( 1+8i+16i^{2} \right)+\left( 1+4i-21i-84i^{2} \right) \\
		+	& =\left( 4+i \right)\left( -15+8i \right)+1-17i+84 \\
		+	& =-60+32i-15i+8i^{2}+1-17i+84 \\
		+	& =-60+32i-15i-8+1-17i+84=17 \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& z=i:\quad \left( 4+i \right)z^{2}+\left( 1-21i \right)z \\
		+	& =\left( 4+i \right)i^{2}+\left( 1-21i \right)i \\
		+	& =\left( 4+i \right)\left( -1 \right)+i-21i^{2} \\
		+	& =-4-i+i+21=17 \\
		+	\end{align}</math>

---

Version vom 11:32, 26. Okt. 2008

Let us first divide both sides by \displaystyle 4+i, so that the coefficient in front of \displaystyle z^{2} becomes \displaystyle \text{1},

\displaystyle z^{2}+\frac{1-21i}{4+i}z=\frac{17}{4+i}

The two complex quotients become

\displaystyle \begin{align} & \frac{\left( 1-21i \right)\left( 4-i \right)}{\left( 4+i \right)\left( 4-i \right)}=\frac{4-i-84i+21i^{2}}{4^{2}-i^{2}} \\ & =\frac{-17-85i}{16+1}=\frac{-17-85i}{17}=-1-5i \\ \end{align}

\displaystyle \begin{align} & \frac{17}{4+i}=\frac{17\left( 4-i \right)}{\left( 4+i \right)\left( 4-i \right)}=\frac{17\left( 4-i \right)}{4^{2}-i^{2}} \\ & =\frac{17\left( 4-i \right)}{17}=4-i \\ \end{align}

Thus, the equation becomes

\displaystyle z^{2}-\left( 1+5i \right)z=4-i

Now, we complete the square of the left-hand side:

\displaystyle \begin{align} & \left( z-\frac{1+5i}{2} \right)^{2}-\left( \frac{1+5i}{2} \right)^{2}=4-i \\ & \left( z-\frac{1+5i}{2} \right)^{2}-\left( \frac{1}{4}+\frac{5}{2}i+\frac{25}{4}i^{2} \right)=4-i \\ & \left( z-\frac{1+5i}{2} \right)^{2}-\frac{1}{4}-\frac{5}{2}i+\frac{25}{4}=4-i \\ & \left( z-\frac{1+5i}{2} \right)^{2}=-2+\frac{3}{2}i \\ \end{align}

If we set \displaystyle w=z-\frac{1+5i}{2}, we have a binomial equation in \displaystyle w,

\displaystyle w^{2}=-2+\frac{3}{2}i

which we solve by putting \displaystyle w=x+iy\text{ },

\displaystyle \left( x+iy \right)^{2}=-2+\frac{3}{2}i

or, if the left-hand side is expanded,

\displaystyle x^{2}-y^{2}+2xyi=-2+\frac{3}{2}i

If we identify the real and imaginary parts on both sides, we get

\displaystyle \begin{align} & x^{2}-y^{2}=-2 \\ & 2xy=\frac{3}{2} \\ \end{align}

and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:

\displaystyle x^{2}+y^{2}=\sqrt{\left( -2 \right)^{2}+\left( \frac{3}{2} \right)^{2}}=\frac{5}{2}

Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.

Together, the three relations constitute the following system of equations:

\displaystyle \left\{ \begin{array}{*{35}l} \ x^{2}-y^{2}=2 \\ \ 2xy=-\frac{3}{2} \\ \ x^{2}+y^{2}=\frac{5}{2} \\ \end{array} \right.

From the first and the third equations, we can relatively easily obtain the values that \displaystyle x and \displaystyle y can take.

Add the first and third equations,

EQ1

which gives that \displaystyle x=\pm \frac{1}{2}.

Then, subtract the first equation from the third equation,

EQ13

i.e. \displaystyle y=\pm \frac{3}{2}.

This gives four possible combinations,

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{1}{2} \\ y=\frac{3}{2} \\ \end{array} \right.\quad \left\{ \begin{array}{*{35}l} x=\frac{1}{2} \\ y=-\frac{3}{2} \\ \end{array} \right.\quad \left\{ \begin{array}{*{35}l} x=-\frac{1}{2} \\ y=\frac{3}{2} \\ \end{array} \right.\quad \left\{ \begin{array}{*{35}l} x=-\frac{1}{2} \\ y=-\frac{3}{2} \\ \end{array} \right.

of which only two also satisfy the second equation.

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{1}{2} \\ y=\frac{3}{2} \\ \end{array} \right.\quad \quad \text{and}\quad \quad \left\{ \begin{array}{*{35}l} x=-\frac{1}{2} \\ y=-\frac{3}{2} \\ \end{array} \right.

This means that the binomial equation has the two solutions,

\displaystyle w=\frac{1}{2}+\frac{3}{2}i and \displaystyle w=\frac{1}{-2}-\frac{3}{2}i

and that the original equation has the solutions

\displaystyle z=1+4i and \displaystyle z=i

according to the relation \displaystyle w=z-\frac{1+5i}{2}.

Finally, we check that the solutions really do satisfy the equation.

\displaystyle \begin{align} & 1+4i:\quad \left( 4+i \right)z^{2}+\left( 1-21i \right)z \\ & =\left( 4+i \right)\left( 1+4i \right)^{2}+\left( 1-21i \right)\left( 1+4i \right) \\ & =\left( 4+i \right)\left( 1+8i+16i^{2} \right)+\left( 1+4i-21i-84i^{2} \right) \\ & =\left( 4+i \right)\left( -15+8i \right)+1-17i+84 \\ & =-60+32i-15i+8i^{2}+1-17i+84 \\ & =-60+32i-15i-8+1-17i+84=17 \\ \end{align}

\displaystyle \begin{align} & z=i:\quad \left( 4+i \right)z^{2}+\left( 1-21i \right)z \\ & =\left( 4+i \right)i^{2}+\left( 1-21i \right)i \\ & =\left( 4+i \right)\left( -1 \right)+i-21i^{2} \\ & =-4-i+i+21=17 \\ \end{align}