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Lösung 3.2:1b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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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>,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\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 = 1-i,\\[5pt]
z-u &= 2+i-(-1-2i) = 2+1+(1+2)i = 3+3i,
z-u &= 2+i-(-1-2i) = 2+1+(1+2)i = 3+3i,
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or interpret <math>z-u</math> from the vector relation
or interpret <math>z-u</math> from the vector relation
-
{{Displayed math||<math>z=(z-u)+u\,,</math>}}
+
{{Abgesetzte Formel||<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]]

Version vom 13:07, 10. Mär. 2009

We can easily calculate z+u and zu,

z+uzu=2+i+(12i)=21+(12)i=1i=2+i(12i)=2+1+(1+2)i=3+3i

and then mark them on the complex plane.

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

We can either view the vector subtraction zu as z+(u),

or interpret zu from the vector relation

z=(zu)+u

i.e. zu is the vector we add to u to arrive at z.