Man beweise (z1z2)¯=z1¯z2¯,(z1,z2∈C){\displaystyle {\overline {\begin{pmatrix}{\frac {z_{1}}{z_{2}}}\end{pmatrix}}}={\frac {\overline {z_{1}}}{\overline {z_{2}}}},(z_{1},z_{2}\in \mathbb {C} )}.
(z1z2)¯=[r1,φ1][r2,φ2]¯=[r1r2,φ1−φ2]¯=[r1r2,−φ1+φ2]=[r1,−φ1][r2,−φ2]=z1¯z2¯{\displaystyle {\overline {\left({\frac {z_{1}}{z_{2}}}\right)}}={\overline {\frac {[r_{1},\varphi _{1}]}{[r_{2},\varphi _{2}]}}}={\overline {\left[{\frac {r_{1}}{r_{2}}},{\varphi _{1}-\varphi _{2}}\right]}}=\left[{\frac {r_{1}}{r_{2}}},{-\varphi _{1}+\varphi _{2}}\right]={\frac {[r_{1},-\varphi _{1}]}{[r_{2},-\varphi _{2}]}}={\frac {\overline {z_{1}}}{\overline {z_{2}}}}}
Das Stricherl über dem z{\displaystyle z} heißt, dass das z¯{\displaystyle {\bar {z}}} die konjugiert Komplexe Zahl von z{\displaystyle z} ist.
für Leute, die lieber im Tabloid-Style rechnen :-),
Rechenregeln:
Konjunktion: z¯=[r,φ]¯=[r,−φ]{\displaystyle {\overline {z}}={\overline {[r,\varphi ]}}=[r,-\varphi ]}
Division: [r1,φ1][r2,φ2]=[r1/r2,φ1−φ2]{\displaystyle {\frac {[r_{1},\varphi _{1}]}{[r_{2},\varphi _{2}]}}=[r_{1}/r_{2},\varphi _{1}-\varphi _{2}]}
Also:
(z1z2)¯=z1¯z2¯{\displaystyle {\overline {\left({\frac {z_{1}}{z_{2}}}\right)}}={\frac {\overline {z_{1}}}{\overline {z_{2}}}}}
([r1,φ1][r2,φ2])¯=[r1,−φ1][r2,−φ2]{\displaystyle {\overline {\left({\frac {[r_{1},\varphi _{1}]}{[r_{2},\varphi _{2}]}}\right)}}={\frac {[r_{1},-\varphi _{1}]}{[r_{2},-\varphi _{2}]}}}
[r1/r2,φ1−φ2]¯=[r1/r2,−φ1+φ2]{\displaystyle {\overline {[r_{1}/r_{2},\varphi _{1}-\varphi _{2}]}}=[r_{1}/r_{2},-\varphi _{1}+\varphi _{2}]}
[r1/r2,−φ1+φ2]=[r1/r2,−φ1+φ2]{\displaystyle [r_{1}/r_{2},-\varphi _{1}+\varphi _{2}]=[r_{1}/r_{2},-\varphi _{1}+\varphi _{2}]}
--Baccus 05:29, 26. Nov 2006 (CET)
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![{\displaystyle {\overline {\left({\frac {z_{1}}{z_{2}}}\right)}}={\overline {\frac {[r_{1},\varphi _{1}]}{[r_{2},\varphi _{2}]}}}={\overline {\left[{\frac {r_{1}}{r_{2}}},{\varphi _{1}-\varphi _{2}}\right]}}=\left[{\frac {r_{1}}{r_{2}}},{-\varphi _{1}+\varphi _{2}}\right]={\frac {[r_{1},-\varphi _{1}]}{[r_{2},-\varphi _{2}]}}={\frac {\overline {z_{1}}}{\overline {z_{2}}}}}](/index.php?title=Spezial:MathShowImage&hash=e43d64a2f0444e9628d716e441d7b011&mode=mathml)
heißt, dass das 
die konjugiert Komplexe Zahl von ist.
![{\displaystyle {\overline {z}}={\overline {[r,\varphi ]}}=[r,-\varphi ]}](/index.php?title=Spezial:MathShowImage&hash=1e11e1d026403c312ca9fb6331467171&mode=mathml)
![{\displaystyle {\frac {[r_{1},\varphi _{1}]}{[r_{2},\varphi _{2}]}}=[r_{1}/r_{2},\varphi _{1}-\varphi _{2}]}](/index.php?title=Spezial:MathShowImage&hash=8bc3580308f4c89193f43677c0a4d7cb&mode=mathml)
![{\displaystyle {\overline {\left({\frac {[r_{1},\varphi _{1}]}{[r_{2},\varphi _{2}]}}\right)}}={\frac {[r_{1},-\varphi _{1}]}{[r_{2},-\varphi _{2}]}}}](/index.php?title=Spezial:MathShowImage&hash=ecb3449cdc110d98be9bfe2660beb72f&mode=mathml)
![{\displaystyle {\overline {[r_{1}/r_{2},\varphi _{1}-\varphi _{2}]}}=[r_{1}/r_{2},-\varphi _{1}+\varphi _{2}]}](/index.php?title=Spezial:MathShowImage&hash=d436f8bd80c4d80bbfaa3d038752eb29&mode=mathml)
![{\displaystyle [r_{1}/r_{2},-\varphi _{1}+\varphi _{2}]=[r_{1}/r_{2},-\varphi _{1}+\varphi _{2}]}](/index.php?title=Spezial:MathShowImage&hash=f675f1710c4797f757e88c6b756b413d&mode=mathml)