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 3.1:4b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,
-
If we divide both sides by 2-i, we obtain z by itself on the left-hand side:
+
 +
{{Displayed math||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}
-
<math>z=\frac{3+2i}{2-i}.</math>
+
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
 +
{{Displayed math||<math>\begin{align}
 +
z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
 +
&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
 +
&= \frac{6+3i+4i-2}{4+1}\\[5pt]
 +
&= \frac{4+7i}{5}\\[5pt]
 +
&= \frac{4}{5}+\frac{7}{5}\,i\,\textrm{.}
 +
\end{align}</math>}}
-
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the numerator:
+
Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}z&= \frac{(3+2i)(2+i)}{(2-i)(2+i)}=\frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\
+
\text{LHS}
-
&=\frac{6+3i+4i-2}{4+1}=\frac{4+7i}{5}=\frac{4}{5}+\frac{7}{5}i\end{align}</math>
+
&= (2-i)z\\[5pt]
-
 
+
&= (2-i)\Bigl(\frac{4}{5}+\frac{7}{5}\,i\bigr)\\[5pt]
-
 
+
&= 2\cdot\frac{4}{5} + 2\cdot\frac{7}{5}\,i - i\cdot\frac{4}{5} - i\cdot\frac{7}{5}\,i\\[5pt]
-
Also, we substitute <math>z=\frac{4}{5}+\frac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly:
+
&= \frac{8}{5} + \frac{14}{5}\,i - \frac{4}{5}\,i + \frac{7}{5}\\[5pt]
-
 
+
&= \frac{8+7}{5} + \frac{14-4}{5}\,i\\[5pt]
-
 
+
&= \frac{15}{5} + \frac{10}{5}\,i\\[5pt]
-
<math>\begin{align}LHS &= (2-i)z=(2-i)(\frac{4}{5}+\frac{7}{5}i)\\
+
&= 3+2i\\[5pt]
-
&=2\cdot\frac{4}{5}-i\cdot\frac{4}{5}+2\cdot \frac{7}{5}i -i\cdot \frac{7}{5}i\\
+
&= \text{RHS.}\end{align}</math>}}
-
&=\frac{8}{5}-\frac{4}{5}i+\frac{14}{5}i+\frac{7}{5} = \frac{8+7}{5}+\frac{14-4}{5}i\\
+
-
&=\frac{15}{5}+\frac{10}{5}i=3+2i=RHS.\end{align}</math>
+
-
 
+
-
{{NAVCONTENT_STOP}}
+

Version vom 07:27, 30. Okt. 2008

If we divide both sides by 2−i, we obtain z by itself on the left-hand side,

z=2−i3+2i.

It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,

z=(2−i)(2+i)(3+2i)(2+i)=22−i232+3i+2i2+2ii=4+16+3i+4i−2=54+7i=54+57i.

Also, we substitute z=54+57i into the original equation to assure ourselves that we have calculated correctly,

LHS=(2−i)z=(2−i)54+57i=254+257i−i54−i57i=58+514i−54i+57=58+7+514−4i=515+510i=3+2i=RHS.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4b“