From Förberedande kurs i matematik 2

Revision as of 14:59, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.2:4c moved to Solution 3.2:4c: Robot: moved page) ← Previous diff		Current revision (12:05, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	One way to determine the magnitude is to calculate the product <math>(3-4i)(3+2i)</math> and then to take the magnitude of the result, but for products we have that
-	<center> [[Image:3_2_4c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>|zw| = |z|\cdot |w|</math>}}
		+
		+	and we can take the magnitude of the factors <math>3-4i</math> and <math>3+2i</math> and then multiply the magnitudes together,
		+
		+	{{Displayed math||<math>\begin{align}
		+	|(3-4i)(3+2i)|
		+	&= |3-4i|\cdot |3+2i|\\[5pt]
		+	&= \sqrt{3^2+(-4)^2}\cdot\sqrt{3^2+2^2}\\[5pt]
		+	&= \sqrt{9+16}\sqrt{9+4}\\[5pt]
		+	&= \sqrt{25}\sqrt{13}\\[5pt]
		+	&= 5\sqrt{13}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

One way to determine the magnitude is to calculate the product \displaystyle (3-4i)(3+2i) and then to take the magnitude of the result, but for products we have that

\displaystyle |zw| = |z|\cdot |w|

and we can take the magnitude of the factors \displaystyle 3-4i and \displaystyle 3+2i and then multiply the magnitudes together,

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