Lösung 3.2:1b

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<center> [[Image:3_2_1b-1(2).gif]] </center>
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We can easily calculate <math>z+u</math> and <math>z-u</math>,
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<center> [[Image:3_2_1b-2(2).gif]] </center>
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<math>\begin{align}z+u&=2+i+(-1-2i)=2-1+(1-2)i=1-i,\\
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z-u&=2+i-(-1-2i)=2+1+(1+2)i=3+3i,\end{align}</math>
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and then mark them on the complex plane.
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An alternative is to view <math>z</math> and <math>u</math> as vectors and <math>z+u</math> as a vector addition of <math>z</math> and <math>u</math>.
[[Image:3_2_1_b1.gif|center]]
[[Image:3_2_1_b1.gif|center]]
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We can either view the vector subtraction <math>z-u</math> as <math>z+(-u)</math>,
[[Image:3_2_1_b2.gif|center]]
[[Image:3_2_1_b2.gif|center]]
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or interpret <math>z-u</math> from the vector relation
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<math>z=(z-u)+u</math>
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i.e. <math>z-u</math> is the vector we add to <math>u</math> to arrive at <math>z</math>.
[[Image:3_2_1_b3.gif|center]]
[[Image:3_2_1_b3.gif|center]]
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Version vom 10:30, 3. Okt. 2008