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Lösung 3.1:4d

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 3: Zeile 3:
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>}}
+
{{Abgesetzte Formel||<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,
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}
+
{{Abgesetzte Formel||<math>\begin{align}
\bar{z}
\bar{z}
&= \frac{(1+i)(2-i)}{(2+i)(2-i)}
&= \frac{(1+i)(2-i)}{(2+i)(2-i)}
Zeile 20: Zeile 20:
We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,
We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\text{LHS}
\text{LHS}
&= (2+i)\bar{z}
&= (2+i)\bar{z}

Version vom 13:07, 10. Mär. 2009

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“