TU Wien:Algebra und Diskrete Mathematik UE (diverse)/Übungen WS11/Beispiel 32

Man zeige $|{\frac {z_{1}+z_{2}}{2}}|^{2}+|{\frac {z_{1}-z_{2}}{2}}|^{2}={\frac {1}{2}}(|z_{1}|^{2}+|z_{2}|^{2})$

Hilfreiches

$|z|={\sqrt {z*{\overline {z}}}}$

$|z|^{2}=z*{\overline {z}}$

${\overline {z_{1}+z_{2}}}={\overline {z_{1}}}+{\overline {z_{2}}}$

${\overline {z_{1}-z_{2}}}={\overline {z_{1}}}-{\overline {z_{2}}}$

Lösung(svorschlag)

von --Christian.abila 23:59, 16. Jul. 2012 (CEST)

${\frac {1}{4}}(|z_{1}+z_{2}|^{2}+|z_{1}-z_{2}|^{2})={\frac {1}{4}}((z_{1}+z_{2})*{\overline {(z_{1}+z_{2})}}+(z_{1}-z_{2})*{\overline {(z_{1}-z_{2})}})=$

${\frac {1}{4}}((z_{1}{\overline {z_{1}}}+z_{1}{\overline {z_{2}}}+z_{2}{\overline {z_{1}}}+z_{2}{\overline {z_{2}}})+(z_{1}{\overline {z_{1}}}-z_{1}{\overline {z_{2}}}-z_{2}{\overline {z_{1}}}+z_{2}{\overline {z_{2}}}))={\frac {1}{2}}(|z_{1}|^{2}+|z_{2}|^{2})$