From Förberedande kurs i matematik 2

Revision as of 15:25, 22 October 2008 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (11:57, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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	We calculate what the expression will be		We calculate what the expression will be
		+	{{Displayed math||<math>(2-i)+(5+3i) = 2+5+(-1+3)i = 7+2i</math>}}
-	<math>\left( 2-i \right)+\left( 5+3i \right)=2+5+\left( -1+3 \right)i=7+2i</math>	+	and then calculate the magnitude,
-	and then take the magnitude:	+	{{Displayed math||<math>|7+2i| = \sqrt{7^2+2^2} = \sqrt{49+4} = \sqrt{53}\,\textrm{.}</math>}}
-	<math>\left| 7+2i \right|=\sqrt{7^{2}+2^{2}}=\sqrt{49+4}=\sqrt{53}</math>	+	Note: It is not possible to calculate the magnitude of the terms individually
-		+	{{Displayed math||<math>|(2-i)+(5+3i)| \ne |2-i| + |5+3i|\,\textrm{.}</math>}}
-	NOTE: Note that it is not possible to take the magnitude of the terms individually	+
-		+
-		+
-	<math>\left| \left( 2-i \right)+\left( 5+3i \right) \right|\ne \left| 2-i \right|+\left| 5+3i \right|</math>	+

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Current revision

We calculate what the expression will be

\displaystyle (2-i)+(5+3i) = 2+5+(-1+3)i = 7+2i

and then calculate the magnitude,

\displaystyle |7+2i| = \sqrt{7^2+2^2} = \sqrt{49+4} = \sqrt{53}\,\textrm{.}

Note: It is not possible to calculate the magnitude of the terms individually

\displaystyle |(2-i)+(5+3i)| \ne |2-i| + |5+3i|\,\textrm{.}