Aus Online Mathematik Brückenkurs 2

Version vom 15:09, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.3:5c moved to Solution 3.3:5c: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 10:01, 26. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	As usual we begin by completing the square,
-	<center> [[Image:3_3_5c-1(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	<math>\begin{align}
-	<center> [[Image:3_3_5c-2(4).gif]] </center>	+	& \left( z-\frac{1+3i}{2} \right)^{2}-\left( \frac{1+3i}{2} \right)^{2}-4+3i=0 \\
-	{{NAVCONTENT_STOP}}	+	& \left( z-\frac{1+3i}{2} \right)^{2}-\left( \frac{1}{4}+\frac{3}{2}i+\frac{9}{4}i^{2} \right)-4+3i=0 \\
-	{{NAVCONTENT_START}}	+	& \left( z-\frac{1+3i}{2} \right)^{2}-\frac{1}{4}-\frac{3}{2}i+\frac{9}{4}-4+3i=0 \\
-	<center> [[Image:3_3_5c-3(4).gif]] </center>	+	& \left( z-\frac{1+3i}{2} \right)^{2}-2+\frac{3}{2}i=0 \\
-	{{NAVCONTENT_STOP}}	+	\end{align}</math>
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_3_5c-4(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	and if we treat as unknown, we have the equation's
		+
		+
		+	<math>w=z-\frac{1+3i}{2}</math>
		+
		+
		+	Up until now, we have solved binomial equations of this type by going over to polar form, but if we were to do that in this case, we would have a problem with determining the exact value of the argument of the right-hand side. Instead, we put
		+	<math>w=x+iy</math>
		+	and try to obtain
		+	<math>x</math>
		+	and
		+	<math>y</math>
		+	from the equation.
		+
		+	With
		+	<math>w=x+iy</math>, the equation becomes
		+
		+
		+	<math>\left( x+iy \right)^{2}=2-\frac{3}{2}i</math>
		+
		+
		+	and, with the left-hand side expanded,
		+
		+
		+	<math>x^{2}-y^{2}+2xyi=2-\frac{3}{2}i</math>
		+
		+
		+	If we set the real and imaginary part of both sides equal, we obtain the equation system
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x^{2}-y^{2}=2 \\
		+	2xy=-\frac{3}{2} \\
		+	\end{array} \right.</math>
		+
		+
		+	We would very well be able to solve this system, but there is a further relation that we can obtain which will simplify the calculations. If we go back to the relation
		+
		+
		+	<math>\left( x+iy \right)^{2}=2-\frac{3}{2}i</math>
		+
		+
		+	and take the modulus of both sides, we obtain
		+
		+
		+	<math>x^{2}+y^{2}=\sqrt{2^{2}+\left( -\frac{3}{2} \right)^{2}}</math>
		+
		+
		+
		+	i.e.
		+
		+
		+	<math>x^{2}+y^{2}=\frac{5}{2}</math>
		+
		+
		+	We add this relation to our two other 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>
		+
		+
		+	Now, add the first and third equations
		+
		+	EQU1
		+
		+	which gives that
		+	<math>y=\pm \frac{3}{2}</math>. Then, subtracting the first equation from the third,
		+
		+	EQU2
		+
		+
		+	we get
		+	<math>y=\pm \frac{1}{2}</math>. This gives potentially four solutions to the equation system,
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{3}{2} \\
		+	y=\frac{1}{2} \\
		+	\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
		+	x=\frac{3}{2} \\
		+	y=-\frac{1}{2} \\
		+	\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
		+	x=-\frac{3}{2} \\
		+	y=\frac{1}{2} \\
		+	\end{array} \right.\quad \left\{ \begin{array}{*{35}l}
		+	x=-\frac{3}{2} \\
		+	y=-\frac{1}{2} \\
		+	\end{array} \right.</math>
		+
		+
		+	but only two of these satisfy the second equation
		+	<math>2xy=-\frac{3}{2}</math>, namely
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=\frac{3}{2} \\
		+	y=-\frac{1}{2} \\
		+	\end{array} \right.\quad \quad \text{and}\quad \left\{ \begin{array}{*{35}l}
		+	x=-\frac{3}{2} \\
		+	y=\frac{1}{2} \\
		+	\end{array} \right.</math>
		+
		+
		+	Hence, we have that the solutions are
		+
		+
		+	<math>w=\frac{3-i}{2}</math>
		+	and
		+	<math>w=\frac{-3+i}{2}</math>
		+
		+
		+	or, expressed in
		+	<math>z</math>,
		+
		+
		+	<math>z=2+i</math>
		+	and
		+	<math>z=-1+2i</math>
		+
		+
		+	Because the calculation has been rather long, there is the risk that we have calculated incorrectly somewhere and we therefore check that the solutions satisfy the equation in the exercise:
		+
		+
		+	<math>\begin{align}
		+	& z=2+i:\quad z^{2}-\left( 1+3i \right)z-4+3i=\left( 2+i \right)^{2}-\left( 1+3i \right)\left( 2+i \right)-4+3i \\
		+	& =4+4i+i^{2}-\left( 2+i+6i+3i^{2} \right)-4+3i \\
		+	& =4+4i-1-2-7i+3-4+3i=0 \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& z=-1+2i:\quad z^{2}-\left( 1+3i \right)z-4+3i=\left( -1+2i \right)^{2}-\left( 1+3i \right)\left( -1+2i \right)-4+3i \\
		+	& =\left( -1 \right)^{2}-4i+4i^{2}-\left( -1+2i-3i+6i^{2} \right)-4+3i \\
		+	& =1-4i-4+1+i+6-4+3i=0 \\
		+	\end{align}</math>

---

Version vom 10:01, 26. Okt. 2008

As usual we begin by completing the square,

\displaystyle \begin{align} & \left( z-\frac{1+3i}{2} \right)^{2}-\left( \frac{1+3i}{2} \right)^{2}-4+3i=0 \\ & \left( z-\frac{1+3i}{2} \right)^{2}-\left( \frac{1}{4}+\frac{3}{2}i+\frac{9}{4}i^{2} \right)-4+3i=0 \\ & \left( z-\frac{1+3i}{2} \right)^{2}-\frac{1}{4}-\frac{3}{2}i+\frac{9}{4}-4+3i=0 \\ & \left( z-\frac{1+3i}{2} \right)^{2}-2+\frac{3}{2}i=0 \\ \end{align}

and if we treat as unknown, we have the equation's

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

Up until now, we have solved binomial equations of this type by going over to polar form, but if we were to do that in this case, we would have a problem with determining the exact value of the argument of the right-hand side. Instead, we put \displaystyle w=x+iy and try to obtain \displaystyle x and \displaystyle y from the equation.

With \displaystyle w=x+iy, the equation becomes

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

and, with the left-hand side expanded,

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

If we set the real and imaginary part of both sides equal, we obtain the equation system

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

We would very well be able to solve this system, but there is a further relation that we can obtain which will simplify the calculations. If we go back to the relation

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

and take the modulus of both sides, we obtain

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

i.e.

\displaystyle x^{2}+y^{2}=\frac{5}{2}

We add this relation to our two other 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.

Now, add the first and third equations

EQU1

which gives that \displaystyle y=\pm \frac{3}{2}. Then, subtracting the first equation from the third,

EQU2

we get \displaystyle y=\pm \frac{1}{2}. This gives potentially four solutions to the equation system,

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

but only two of these satisfy the second equation \displaystyle 2xy=-\frac{3}{2}, namely

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

Hence, we have that the solutions are

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

or, expressed in \displaystyle z,

\displaystyle z=2+i and \displaystyle z=-1+2i

Because the calculation has been rather long, there is the risk that we have calculated incorrectly somewhere and we therefore check that the solutions satisfy the equation in the exercise:

\displaystyle \begin{align} & z=2+i:\quad z^{2}-\left( 1+3i \right)z-4+3i=\left( 2+i \right)^{2}-\left( 1+3i \right)\left( 2+i \right)-4+3i \\ & =4+4i+i^{2}-\left( 2+i+6i+3i^{2} \right)-4+3i \\ & =4+4i-1-2-7i+3-4+3i=0 \\ \end{align}

\displaystyle \begin{align} & z=-1+2i:\quad z^{2}-\left( 1+3i \right)z-4+3i=\left( -1+2i \right)^{2}-\left( 1+3i \right)\left( -1+2i \right)-4+3i \\ & =\left( -1 \right)^{2}-4i+4i^{2}-\left( -1+2i-3i+6i^{2} \right)-4+3i \\ & =1-4i-4+1+i+6-4+3i=0 \\ \end{align}