Lösning 1.8.1b - Versionshistorik

Samuel den 20 juni 2012 kl. 12.58

Samuel — Wed, 20 Jun 2012 12:58:42 GMT

← Äldre version		Versionen från 20 juni 2012 kl. 12.58
Rad 1:		Rad 1:
-	<math>\~~displaystyle~~ (3-2i)(4+i-(6-2i)) = (3-2i)(-2+3i)=~~</math>~~	+	<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	+

Sass — Tue, 12 Jun 2012 14:06:29 GMT

← Äldre version		Versionen från 12 juni 2012 kl. 14.06
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

Sass — Tue, 12 Jun 2012 13:58:46 GMT

← Äldre version		Versionen från 12 juni 2012 kl. 13.58
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> = 2-\frac{i}{4}+4i-\frac{i}{2}	+	<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>
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		+	<math>\displaystyle = 2-\frac{i}{4}+4i+\frac{1}{2} = \frac{5}{2} + \frac{15 i} {4}</math>

Sass — Tue, 12 Jun 2012 13:53:50 GMT

← Äldre version		Versionen från 12 juni 2012 kl. 13.53
Rad 1:		Rad 1:
-	<math>\displaystyle (1+2i)(2-\frac{i}{4}) = 1\cdot 2 + 1 \cdot \frac{i]{4}+2i\cdot 2+ 2i\cdot \frac{i}{4}</math>	+	<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> = 2-\frac{i}{4}+4i-\frac{i}{2}

Sass — Tue, 12 Jun 2012 13:52:27 GMT

Ny sida: <math>\displaystyle (1+2i)(2-\frac{i}{4}) = 1\cdot 2 + 1 \cdot \frac{i]{4}+2i\cdot 2+ 2i\cdot \frac{i}{4}</math>

Ny sida

<math>\displaystyle (1+2i)(2-\frac{i}{4}) = 1\cdot 2 + 1 \cdot \frac{i]{4}+2i\cdot 2+ 2i\cdot \frac{i}{4}</math>