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Lösning 1.8.1a

Förberedande kurs i matematik

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(Ny sida: <math>\displaystyle (3-2i)(4+i-(6-2i)) = (3-2i)(-2+3i)=</math> <math>\displaystyle =(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i = -6 +9i + 4i +6=</math> <math>/displaystyle =13i)
m
Rad 3: Rad 3:
<math>\displaystyle =(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i = -6 +9i + 4i +6=</math>
<math>\displaystyle =(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i = -6 +9i + 4i +6=</math>
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<math>/displaystyle =13i
+
<math>\displaystyle =13i

Versionen från 12 juni 2012 kl. 14.05

(32i)(4+i(62i))=(32i)(2+3i)=

=(3(2)+33i2i(2)2i3i=6+9i+4i+6=

=13i