Lösung 3.2:4c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
One way to determine the magnitude is to calculate the product
+
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
-
<math>\left( \text{3}-\text{4}i \right)\left( \text{3}+\text{2}i \right)</math>
+
-
and then to take the magnitude of the result, but for products
+
 +
{{Displayed math||<math>|zw| = |z|\cdot |w|</math>}}
-
<math>\left| zw \right|=\left| z \right|\centerdot \left| w \right|</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}
-
and we can take the magnitude of the factors
+
|(3-4i)(3+2i)|
-
<math>\text{3}-\text{4}i</math>
+
&= |3-4i|\cdot |3+2i|\\[5pt]
-
and
+
&= \sqrt{3^2+(-4)^2}\cdot\sqrt{3^2+2^2}\\[5pt]
-
<math>\text{3}+\text{2}i</math>
+
&= \sqrt{9+16}\sqrt{9+4}\\[5pt]
-
and then multiply the magnitudes together:
+
&= \sqrt{25}\sqrt{13}\\[5pt]
-
 
+
&= 5\sqrt{13}\,\textrm{.}
-
 
+
\end{align}</math>}}
-
<math>\begin{align}
+
-
& \left| \left( \text{3}-\text{4}i \right)\left( \text{3}+\text{2}i \right) \right|=\left| \left( \text{3}-\text{4}i \right) \right|\centerdot \left| \left( \text{3}+\text{2}i \right) \right| \\
+
-
& =\sqrt{3^{2}+\left( -4 \right)^{2}}\centerdot \sqrt{3^{2}+2^{2}} \\
+
-
& =\sqrt{9+16}\centerdot \sqrt{9+4} \\
+
-
& =\sqrt{25}\centerdot \sqrt{13}=5\sqrt{13} \\
+
-
\end{align}</math>
+

Version vom 12:05, 29. Okt. 2008

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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:4c“