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Lösning 1.8.1b

Förberedande kurs i matematik

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Rad 1: Rad 1:
-
<math>\displaystyle (1+2i)\left( 2-\frac{i}{4} \right) = 1\cdot 2 - 1 \cdot \frac{i}{4}+2i\cdot 2- 2i\cdot \frac{i}{4}=</math>
+
<math>\displaystyle (3-2i)(4+i-(6-2i)) = (3-2i)(-2+3i)=</math>
-
<math>\displaystyle = 2-\frac{i}{4}+4i+\frac{1}{2} = \frac{5}{2} + \frac{15 i} {4}</math>
+
<math>\displaystyle =(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i = -6 +9i + 4i +6=</math>
 +
 
 +
<math>\displaystyle =13i

Versionen från 12 juni 2012 kl. 14.06

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

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

=13i