From Förberedande kurs i matematik 2

Revision as of 08:09, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (11:57, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We calculate what the expression will be
-	<center> [[Image:3_2_4b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>(2-i)+(5+3i) = 2+5+(-1+3)i = 7+2i</math>}}
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		+	and then calculate the magnitude,
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		+	{{Displayed math||<math>|7+2i| = \sqrt{7^2+2^2} = \sqrt{49+4} = \sqrt{53}\,\textrm{.}</math>}}
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		+	Note: It is not possible to calculate the magnitude of the terms individually
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		+	{{Displayed math||<math>|(2-i)+(5+3i)| \ne |2-i| + |5+3i|\,\textrm{.}</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{.}