Aus Online Mathematik Brückenkurs 2

Version vom 11:09, 24. Okt. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 13:21, 30. Okt. 2008 (bearbeiten) (rückgängig) Tek (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	If we treat the expression	+	If we treat the expression <math>w=\frac{z+i}{z-i}</math> as an unknown, we have the equation
-	<math>w=\frac{z+i}{z-i}</math>	+
-	as an unknown, we have the equation	+
-		+
-		+
-	<math>w^{2}=-1</math>	+
		+	{{Displayed math||<math>w^2=-1\,\textrm{.}</math>}}
	We know already that this equation has roots		We know already that this equation has roots
		+	{{Displayed math||<math>w=\left\{\begin{align}
		+	-i\,,&\\[5pt]
		+	i\,,&
		+	\end{align}\right.</math>}}
-	<math>w=\left\{ \begin{array}{*{35}l}	+	so <math>z</math> should satisfy one of the equation's
-	-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>	+
		+	{{Displayed math||<math>\frac{z+i}{z-i}=-i\quad</math> or <math>\quad\frac{z+i}{z-i}=i\,\textrm{.}</math>}}
	We solve these equations one by one.		We solve these equations one by one.
-	<math>\underline{\underline{\frac{z+i}{z-i}=-i}}</math>	+	*<math>(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>	+
-		+
		+	:Multiply both sides by <math>z-i</math>,
-		+	{{Displayed math||<math>z+i=-i(z-i)\,\textrm{.}</math>}}
-	<math>\underline{\underline{\frac{z+i}{z-i}=i}}</math>	+
		+	:Move all the <math>z</math>-terms over to the left-hand side and all the constants to the right-hand side,
-	Multiply both sides by	+	{{Displayed math||<math>z+iz=-1-i\,\textrm{.}</math>}}
-	<math>z-i</math>:	+
		+	:This gives
-	<math>z+i=i\left( z-i \right)</math>	+	{{Displayed math||<math>z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.}</math>}}
-	Move all the z-terms over to the left-hand side and all the constants to the right-hand side,	+	*<math>(z+i)/(z-i)=i</math>:
		+	:Multiply both sides by <math>z-i</math>,
-	<math>z-iz=1-i</math>	+	{{Displayed math||<math>z+i=i(z-i)\,\textrm{.}</math>}}
		+	:Move all the <math>z</math>-terms over to the left-hand side and all the constants to the right-hand side,
-	This gives	+	{{Displayed math||<math>z-iz=1-i\,\textrm{.}</math>}}
		+	:This gives
-	<math>z=\frac{1-i}{1-i}=1</math>	+	{{Displayed math||<math>z = \frac{1-i}{1-i} = 1\,\textrm{.}</math>}}
-	The solutions are therefore	+	The solutions are therefore <math>z=-1</math> and <math>z=1\,</math>.
-	<math>z=-\text{1}</math>	+
-	and	+
-	<math>z=\text{1}</math>.	+

---

Version vom 13:21, 30. 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\,\textrm{.}

We know already that this equation has roots

\displaystyle w=\left\{\begin{align} -i\,,&\\[5pt] i\,,& \end{align}\right.

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

\displaystyle \frac{z+i}{z-i}=-i\quad or \displaystyle \quad\frac{z+i}{z-i}=i\,\textrm{.}

We solve these equations one by one.

• \displaystyle (z+i)/(z-i)=-i:

Multiply both sides by \displaystyle z-i,

\displaystyle z+i=-i(z-i)\,\textrm{.}

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\,\textrm{.}

This gives

\displaystyle z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.}

• \displaystyle (z+i)/(z-i)=i:

Multiply both sides by \displaystyle z-i,

\displaystyle z+i=i(z-i)\,\textrm{.}

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\,\textrm{.}

This gives

\displaystyle z = \frac{1-i}{1-i} = 1\,\textrm{.}

The solutions are therefore \displaystyle z=-1 and \displaystyle z=1\,.