Lösung 3.2:1c

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By calculation, we obtain
By calculation, we obtain
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<math>\begin{align}2z+w &= 2(2+i)+(2+3i)=2\cdot 2 + 2 + (2+3)i\\
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{{Displayed math||<math>2z+w = 2(2+i)+(2+3i) = 2\cdot 2 + 2 + (2+3)i = 6+5i</math>}}
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&=6+5i\end{align}</math>
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and we can mark this point on the complex plane.
and we can mark this point on the complex plane.
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If we treat <math>z</math> and <math>w</math> as vectors, then <math>2z</math> is the vector which has the same direction as <math>z</math>, but is twice as long.
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If we treat <math>z</math> and <math>w</math> as vectors, then <math>2z</math> is the vector which has the same direction as <math>z</math>, but is twice as long.
[[Image:3_2_1_c1.gif|center]]
[[Image:3_2_1_c1.gif|center]]
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[[Image:3_2_1_c2.gif|center]]
[[Image:3_2_1_c2.gif|center]]
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Version vom 09:23, 29. Okt. 2008

By calculation, we obtain

\displaystyle 2z+w = 2(2+i)+(2+3i) = 2\cdot 2 + 2 + (2+3)i = 6+5i

and we can mark this point on the complex plane.

If we treat \displaystyle z and \displaystyle w as vectors, then \displaystyle 2z is the vector which has the same direction as \displaystyle z, but is twice as long.

We add \displaystyle w to this vector and get \displaystyle 2z+w.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:1c“