Lösung 3.2:1b

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{{NAVCONTENT_START}}
 
We can easily calculate <math>z+u</math> and <math>z-u</math>,
We can easily calculate <math>z+u</math> and <math>z-u</math>,
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<math>\begin{align}z+u&=2+i+(-1-2i)=2-1+(1-2)i=1-i,\\
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{{Displayed math||<math>\begin{align}
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z-u&=2+i-(-1-2i)=2+1+(1+2)i=3+3i,\end{align}</math>
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z+u &= 2+i+(-1-2i) = 2-1+(1-2)i = 1-i,\\[5pt]
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z-u &= 2+i-(-1-2i) = 2+1+(1+2)i = 3+3i,
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\end{align}</math>}}
and then mark them on the complex plane.
and then mark them on the complex plane.
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[[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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or interpret <math>z-u</math> from the vector relation
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<math>z=(z-u)+u</math>
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{{Displayed math||<math>z=(z-u)+u\,,</math>}}
i.e. <math>z-u</math> is the vector we add to <math>u</math> to arrive at <math>z</math>.
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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{{NAVCONTENT_STOP}}
 

Version vom 09:21, 29. Okt. 2008

We can easily calculate \displaystyle z+u and \displaystyle z-u,

\displaystyle \begin{align}

z+u &= 2+i+(-1-2i) = 2-1+(1-2)i = 1-i,\\[5pt] z-u &= 2+i-(-1-2i) = 2+1+(1+2)i = 3+3i, \end{align}

and then mark them on the complex plane.

An alternative is to view \displaystyle z and \displaystyle u as vectors and \displaystyle z+u as a vector addition of \displaystyle z and \displaystyle u.

We can either view the vector subtraction \displaystyle z-u as \displaystyle z+(-u),

or interpret \displaystyle z-u from the vector relation

\displaystyle z=(z-u)+u\,,

i.e. \displaystyle z-u is the vector we add to \displaystyle u to arrive at \displaystyle z.