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Lösung 3.3:2e

Aus Online Mathematik Brückenkurs 2

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

If we treat the expression w=z−iz+i as an unknown, we have the equation

w2=−1.

We know already that this equation has roots

w=−ii 

so z should satisfy one of the equation's

z−iz+i=−i or z−iz+i=i.

We solve these equations one by one.


  • (z+i)(z−i)=−i:
Multiply both sides by z−i,
z+i=−i(z−i).
Move all the z-terms over to the left-hand side and all the constants to the right-hand side,
z+iz=−1−i.
This gives
z=1+i−1−i=1+i−(1+i)=−1.


  • (z+i)(z−i)=i:
Multiply both sides by z−i,
z+i=i(z−i).
Move all the z-terms over to the left-hand side and all the constants to the right-hand side,
z−iz=1−i.
This gives
z=1−i1−i=1.


The solutions are therefore z=−1 and z=1.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:2e“