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Lösning 1.8.1b

Förberedande kurs i matematik

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Nuvarande version (20 juni 2012 kl. 12.58) (redigera) (ogör)
 
Rad 1: Rad 1:
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<math>\displaystyle (3-2i)(4+i-(6-2i)) = (3-2i)(-2+3i)=</math>
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<math>\begin{align} (3-2i)(4+i-(6-2i)) &= (3-2i)(-2+3i)=\\&=(3\cdot (-2) + 3 \cdot 3i -2i\cdot(-2) -2i\cdot 3i =\\&= -6 +9i + 4i+6=\\&=13i\end{align}
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<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
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Nuvarande version

(32i)(4+i(62i))=(32i)(2+3i)==(3(2)+33i2i(2)2i3i==6+9i+4i+6==13i