Aus Online Mathematik Brückenkurs 2

Version vom 15:06, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.3:2e moved to Solution 3.3:2e: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 11:09, 24. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	If we treat the expression
-	<center> [[Image:3_3_2e-1(2).gif]] </center>	+	<math>w=\frac{z+i}{z-i}</math>
-	{{NAVCONTENT_STOP}}	+	as an unknown, we have the equation
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_3_2e-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	<math>w^{2}=-1</math>
		+
		+
		+	We know already that this equation has roots
		+
		+
		+	<math>w=\left\{ \begin{array}{*{35}l}
		+	-i \\
		+	i \\
		+	\end{array} \right.</math>
		+
		+
		+	so
		+	<math>z\text{ }</math>
		+	should satisfy one of the equation's
		+
		+
		+	<math>\frac{z+i}{z-i}=-i</math>
		+	or
		+	<math>\frac{z+i}{z-i}=i</math>
		+
		+
		+	We solve these equations one by one.
		+
		+
		+	<math>\underline{\underline{\frac{z+i}{z-i}=-i}}</math>
		+
		+
		+	Multiply both sides by
		+	<math>z-i</math>:
		+
		+
		+	<math>z+i=-i\left( z-i \right)</math>
		+
		+
		+	Move all the
		+	<math>z</math>
		+	-terms over to the left-hand side and all the constants to the right-hand side,
		+
		+
		+	<math>z+iz=-1-i</math>
		+
		+
		+	This gives
		+
		+
		+	<math>z=\frac{-1-i}{1+i}=\frac{-\left( 1+i \right)}{1+i}=-1</math>
		+
		+
		+
		+
		+	<math>\underline{\underline{\frac{z+i}{z-i}=i}}</math>
		+
		+
		+	Multiply both sides by
		+	<math>z-i</math>:
		+
		+
		+	<math>z+i=i\left( z-i \right)</math>
		+
		+
		+	Move all the z-terms over to the left-hand side and all the constants to the right-hand side,
		+
		+
		+	<math>z-iz=1-i</math>
		+
		+
		+	This gives
		+
		+
		+	<math>z=\frac{1-i}{1-i}=1</math>
		+
		+
		+	The solutions are therefore
		+	<math>z=-\text{1}</math>
		+	and
		+	<math>z=\text{1}</math>.

---

Version vom 11:09, 24. Okt. 2008

If we treat the expression \displaystyle w=\frac{z+i}{z-i} as an unknown, we have the equation

\displaystyle w^{2}=-1

We know already that this equation has roots

\displaystyle w=\left\{ \begin{array}{*{35}l} -i \\ i \\ \end{array} \right.

so \displaystyle z\text{ } should satisfy one of the equation's

\displaystyle \frac{z+i}{z-i}=-i or \displaystyle \frac{z+i}{z-i}=i

We solve these equations one by one.

\displaystyle \underline{\underline{\frac{z+i}{z-i}=-i}}

Multiply both sides by \displaystyle z-i:

\displaystyle z+i=-i\left( z-i \right)

Move all the \displaystyle z -terms over to the left-hand side and all the constants to the right-hand side,

\displaystyle z+iz=-1-i

This gives

\displaystyle z=\frac{-1-i}{1+i}=\frac{-\left( 1+i \right)}{1+i}=-1

\displaystyle \underline{\underline{\frac{z+i}{z-i}=i}}

Multiply both sides by \displaystyle z-i:

\displaystyle z+i=i\left( z-i \right)

Move all the z-terms over to the left-hand side and all the constants to the right-hand side,

\displaystyle z-iz=1-i

This gives

\displaystyle z=\frac{1-i}{1-i}=1

The solutions are therefore \displaystyle z=-\text{1} and \displaystyle z=\text{1}.