From Förberedande kurs i matematik 2

Revision as of 14:59, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.2:4d moved to Solution 3.2:4d: Robot: moved page) ← Previous diff		Current revision (12:16, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	For magnitudes of quotients, we have the arithmetical rule
-	<center> [[Image:3_2_4d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\left|\frac{z}{w}\right| = \frac{|z|}{|w|}\,\textrm{.}</math>}}
		+
		+	We can therefore take the magnitude of the numerator and denominator separately and then divide the magnitudes by each other,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\left|\frac{3-4i}{3+2i}\right|
		+	&= \frac{|3-4i|}{|3+2i|}
		+	= \frac{\sqrt{3^2+(-4)^2}}{\sqrt{3^2+2^2}}
		+	= \frac{\sqrt{9+16}}{\sqrt{9+4}}
		+	= \frac{\sqrt{25}}{\sqrt{13}}
		+	= \frac{5}{\sqrt{13}}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

For magnitudes of quotients, we have the arithmetical rule

\displaystyle \left|\frac{z}{w}\right| = \frac{|z|}{|w|}\,\textrm{.}

We can therefore take the magnitude of the numerator and denominator separately and then divide the magnitudes by each other,

\displaystyle \begin{align} \left|\frac{3-4i}{3+2i}\right| &= \frac{|3-4i|}{|3+2i|} = \frac{\sqrt{3^2+(-4)^2}}{\sqrt{3^2+2^2}} = \frac{\sqrt{9+16}}{\sqrt{9+4}} = \frac{\sqrt{25}}{\sqrt{13}} = \frac{5}{\sqrt{13}}\,\textrm{.} \end{align}