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.