Aus Online Mathematik Brückenkurs 2

Version vom 13:21, 30. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 13:12, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	If we treat the expression <math>w=\frac{z+i}{z-i}</math> as an unknown, we have the equation		If we treat the expression <math>w=\frac{z+i}{z-i}</math> as an unknown, we have the equation
-	{{Displayed math||<math>w^2=-1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<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}	+	{{Abgesetzte Formel||<math>w=\left\{\begin{align}
	-i\,,&\\[5pt]		-i\,,&\\[5pt]
	i\,,&		i\,,&
Zeile 12:		Zeile 12:
	so <math>z</math> should satisfy one of the equation's		so <math>z</math> should satisfy one of the equation's
-	{{Displayed math||<math>\frac{z+i}{z-i}=-i\quad</math> or <math>\quad\frac{z+i}{z-i}=i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<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.
Zeile 21:		Zeile 21:
	:Multiply both sides by <math>z-i</math>,		:Multiply both sides by <math>z-i</math>,
-	{{Displayed math||<math>z+i=-i(z-i)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<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,		:Move all the <math>z</math>-terms over to the left-hand side and all the constants to the right-hand side,
-	{{Displayed math||<math>z+iz=-1-i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>z+iz=-1-i\,\textrm{.}</math>}}
	:This gives		:This gives
-	{{Displayed math||<math>z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.}</math>}}
Zeile 36:		Zeile 36:
	:Multiply both sides by <math>z-i</math>,		:Multiply both sides by <math>z-i</math>,
-	{{Displayed math||<math>z+i=i(z-i)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<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,		:Move all the <math>z</math>-terms over to the left-hand side and all the constants to the right-hand side,
-	{{Displayed math||<math>z-iz=1-i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>z-iz=1-i\,\textrm{.}</math>}}
	:This gives		:This gives
-	{{Displayed math||<math>z = \frac{1-i}{1-i} = 1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>z = \frac{1-i}{1-i} = 1\,\textrm{.}</math>}}
	The solutions are therefore <math>z=-1</math> and <math>z=1\,</math>.		The solutions are therefore <math>z=-1</math> and <math>z=1\,</math>.

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Version vom 13:12, 10. Mär. 2009

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\,.