Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 3.1:1d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
We expand the expression by multiplying each term in the first bracket with every term in the second bracket,
-
We expand the expression by multiplying each term in the first bracket with every term in the second bracket:
+
-
<math>\begin{align}(3-2i)(7+5i)&=3\cdot 7 + 3 \cdot 5i + \cdots\\
+
{{Displayed math||<math>\begin{align}
-
&=3\cdot 7 + 3 \cdot 5i-2i\cdot 7 -2i \cdot 5i.\end{align}</math>
+
(3-2i)(7+5i)
 +
&= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt]
 +
&= 3\cdot 7 + 3 \cdot 5i - 2i\cdot 7 - 2i \cdot 5i\,\textrm{.}
 +
\end{align}</math>}}
-
Then, we use that <math>i^2=-1</math> and write the real and imaginary parts together:
+
Then, we use that <math>i^2=-1</math> and write the real and imaginary parts together,
-
<math>\begin{align}(3-2i)(7+5i)&=21+15i-14i-10i^2\\
+
{{Displayed math||<math>\begin{align}
-
&=21+15i-14i-10\cdot(-1)\\
+
(3-2i)(7+5i)
-
&=(21+10)+(15i-14i)\\
+
&=21+15i-14i-10i^2\\[5pt]
-
&=31+1i\\
+
&=21+15i-14i-10\cdot(-1)\\[5pt]
-
&=31+i\end{align}</math>
+
&=(21+10)+(15i-14i)\\[5pt]
-
 
+
&=31+(15-14)i\\[5pt]
-
{{NAVCONTENT_STOP}}
+
&=31+i\,\textrm{.}
 +
\end{align}</math>}}

Version vom 14:54, 29. Okt. 2008

We expand the expression by multiplying each term in the first bracket with every term in the second bracket,

(3−2i)(7+5i)=37+35i+=37+35i−2i7−2i5i.

Then, we use that i2=−1 and write the real and imaginary parts together,

(3−2i)(7+5i)=21+15i−14i−10i2=21+15i−14i−10(−1)=(21+10)+(15i−14i)=31+(15−14)i=31+i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:1d“