TU Wien:Algebra und Diskrete Mathematik UE (diverse)/Übungen WS11/Beispiel 28

Man beweise ${\overline {({\frac {z_{1}}{z_{2}}})}}={\frac {\overline {z_{1}}}{\overline {z_{2}}}}.$

Hilfreiches

Buch Mathematik für Informatik Seite 13:
"Die Division ${\frac {z_{1}}{z_{2}}}$ komplexer Zahlen kann nun so gesehen werden, dass der Quotient mit ${\overline {z_{2}}}$ erweitert wird, so dass der Nenner $z_{2}*{\overline {z_{2}}}$ insgesamt reell wird."

Lösungsvorschlag

von --Christian.abila 21:26, 16. Jul. 2012 (CEST)
$z_{1}=a_{1}+ib_{1},{\overline {z_{1}}}=a_{1}-ib_{1}$
$z_{2}=a_{2}+ib_{2},{\overline {z_{2}}}=a_{2}-ib_{2}$

${\frac {z_{1}}{z_{2}}}={\frac {a_{1}+ib_{1}}{a_{2}+ib_{2}}}={\frac {(a_{1}+ib_{1})(a_{2}-ib_{1})}{(a_{2}+ib_{2})(a_{2}-ib_{2})}}={\frac {a_{1}a_{2}-ia_{1}b_{2}+ia_{2}b_{1}+b_{1}b_{2}}{a_{2}^{2}-b_{2}^{2}}}={\frac {a_{1}a_{2}+b_{1}b_{2}+i(a_{2}b_{1}-a_{1}b_{2})}{a_{2}^{2}-b_{2}^{2}}}$

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

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