TU Wien:Algebra und Diskrete Mathematik UE (diverse)/Übungen WS11/Beispiel 27

Man beweise $\overline{z_{1} z_{2}} = \overline{z_{1}} * \overline{z_{2}}$ und $\overline{z_{1} - z_{2}} = \overline{z_{1}} - \overline{z_{2}}$ .

Lösungsvorschlag

von --Christian.abila 14:11, 16. Jul. 2012 (CEST)

$z_{1}$ = $a_{1}$ + $b_{1}$ i, $\overline{z_{1}}$ = $a_{1}$ - $b_{1}$ i
$z_{2}$ = $a_{2}$ + $b_{2}$ i, $\overline{z_{2}}$ = $a_{2}$ - $b_{2}$ i

$z_{1} * z_{2} = (a_{1} + b_{1} i) * (a_{2} + b_{2} i) = a_{1} a_{2} + a_{1} b_{2} i + a_{2} b_{1} i - b_{1} b_{2} = a_{1} a_{2} - b_{1} b_{2} + i * (a_{1} b_{2} + a_{2} b_{1})$

$\overline{z_{1} z_{2}} = a_{1} a_{2} - b_{1} b_{2} - i * (a_{1} b_{2} + a_{2} b_{1})$

$\overline{z_{1}} * \overline{z_{2}} = (a_{1} - b_{1} i) * (a_{2} - b_{2} i) = a_{1} a_{2} - a_{1} b_{2} i - a_{2} b_{1} i - b_{1} b_{2} = a_{1} a_{2} - b_{1} b_{2} - i (a_{1} b_{2} + a_{2} b_{1})$

Ergo: $\overline{z_{1} z_{2}} = \overline{z_{1}} * \overline{z_{2}}$ ist wahr.

$\overline{z_{1} - z_{2}} = \overline{z_{1}} - \overline{z_{2}}$
$z_{1} - z_{2} = a_{1} + b_{1} i - a_{2} - b_{2} i = a_{1} - a_{2} + i (b_{1} - b_{2})$
$\overline{z_{1} - z_{2}} = (a_{1} - a_{2}) - i (b_{1} - b_{2})$

$\overline{z_{1}} - \overline{z_{2}} = a_{1} - b_{1} i - a_{2} + b_{2} i = (a_{1} - a_{2}) - i (b_{1} - b_{2})$

Ergo: $\overline{z_{1} - z_{2}} = \overline{z_{1}} - \overline{z_{2}}$ ist wahr.

TU Wien:Algebra und Diskrete Mathematik UE (diverse)/Übungen WS11/Beispiel 27

Lösungsvorschlag

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