Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 3.1:4d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
 
In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.
In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.
Divide both sides by <math>2+i</math>,
Divide both sides by <math>2+i</math>,
 +
{{Displayed math||<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}
-
<math>\bar{z}=\frac{1+i}{2+i}</math>
+
and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,
 +
{{Displayed math||<math>\begin{align}
 +
\bar{z}
 +
&= \frac{(1+i)(2-i)}{(2+i)(2-i)}
 +
= \frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\[5pt]
 +
&= \frac{2-i+2i+1}{4+1}
 +
= \frac{3+i}{5}
 +
= \frac{3}{5}+\frac{1}{5}\,i\,\textrm{.}
 +
\end{align}</math>}}
-
and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:
+
This means that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i\,</math>.
 +
We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,
-
<math>\begin{align}\bar{z}&=\frac{(1+i)(2-i)}{(2+i)(2-i)}=\frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\
+
{{Displayed math||<math>\begin{align}
-
&=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}</math>
+
\text{LHS}
-
 
+
&= (2+i)\bar{z}
-
 
+
= (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)}
-
This means that <math>z=\frac{3}{5}-\frac{1}{5}i.</math>
+
= (2+i)\Bigl(\frac{3}{5}+\frac{1}{5}\,i\Bigr)\\[5pt]
-
 
+
&= 2\cdot\frac{3}{5}+2\cdot\frac{1}{5}\,i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}\,i
-
We check that <math>z=\frac{3}{5}-\frac{1}{5}i</math> satisfies the original equation:
+
= \frac{6}{5}+\frac{2}{5}\,i+\frac{3}{5}\,i-\frac{1}{5}\\[5pt]
-
 
+
&=\frac{6-1}{5}+\frac{2+3}{5}\,i
-
 
+
= 1+i
-
<math>\begin{align}LHS &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\
+
= \text{RHS}\,\textrm{.}
-
&=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\
+
\end{align}</math>}}
-
&=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}</math>
+
-
{{NAVCONTENT_STOP}}
+

Version vom 08:09, 30. Okt. 2008

In the equation, z occurs only as z and, to begin with, we can therefore treat z as unknown.

Divide both sides by 2+i,

z=2+i1+i

and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,

z=(2+i)(2−i)(1+i)(2−i)=22−i212−1i+i2−ii=4+12−i+2i+1=53+i=53+51i.

This means that z=53−51i.

We check that z=53−51i satisfies the original equation,

LHS=(2+i)z=(2+i)53−51i=(2+i)53+51i=253+251i+i53+i51i=56+52i+53i−51=56−1+52+3i=1+i=RHS.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4d“