From Förberedande kurs i matematik 2

Revision as of 13:35, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_2e-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_3_2e-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (13:21, 30 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we treat the expression <math>w=\frac{z+i}{z-i}</math> as an unknown, we have the equation
-	<center> [[Bild:3_3_2e-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>w^2=-1\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:3_3_2e-2(2).gif]] </center>	+	We know already that this equation has roots
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>w=\left\{\begin{align}
		+	-i\,,&\\[5pt]
		+	i\,,&
		+	\end{align}\right.</math>}}
		+
		+	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>}}
		+
		+	We solve these equations one by one.
		+
		+
		+	*<math>(z+i)/(z-i)=-i</math>:
		+
		+	:Multiply both sides by <math>z-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,
		+
		+	{{Displayed math||<math>z+iz=-1-i\,\textrm{.}</math>}}
		+
		+	:This gives
		+
		+	{{Displayed math||<math>z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.}</math>}}
		+
		+
		+	*<math>(z+i)/(z-i)=i</math>:
		+
		+	:Multiply both sides by <math>z-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,
		+
		+	{{Displayed math||<math>z-iz=1-i\,\textrm{.}</math>}}
		+
		+	:This gives
		+
		+	{{Displayed math||<math>z = \frac{1-i}{1-i} = 1\,\textrm{.}</math>}}
		+
		+
		+	The solutions are therefore <math>z=-1</math> and <math>z=1\,</math>.

---

Current revision

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\,.