Komplexe Zahlen/Reell differenzierbare Abbildung/dz/Rückzug/Aufgabe/Lösung

Es ist

${}dz=dx+{\mathrm {i} }dy\,.$

Nach Fakt ist daher

${}{\begin{aligned}\varphi ^{*}(dz)&=\varphi ^{*}(dx+{\mathrm {i} }dy)\\&={\frac {\partial \varphi _{1}}{\partial u}}du+{\frac {\partial \varphi _{1}}{\partial v}}dv+{\mathrm {i} }{\left({\frac {\partial \varphi _{2}}{\partial u}}du+{\frac {\partial \varphi _{2}}{\partial v}}dv\right)}\\&={\left({\frac {\partial \varphi _{1}}{\partial u}}+{\mathrm {i} }{\frac {\partial \varphi _{2}}{\partial u}}\right)}du+{\left({\frac {\partial \varphi _{1}}{\partial v}}+{\mathrm {i} }{\frac {\partial \varphi _{2}}{\partial v}}\right)}dv\\&={\frac {\partial \varphi }{\partial u}}du+{\frac {\partial \varphi }{\partial v}}dv\\&={\left({\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial u}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial v}}+{\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial u}}+{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial v}}\right)}du+{\left({\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial u}}+{\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial v}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial u}}+{\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial v}}\right)}dv\\&={\left({\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial u}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial v}}\right)}d{\left(u+{\mathrm {i} }v\right)}+{\left({\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial u}}+{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial v}}\right)}d{\left(u-{\mathrm {i} }v\right)}\\&={\frac {\partial \varphi }{\partial w}}dw+{\frac {\partial \varphi }{\partial {\overline {w}}}}d{\overline {w}}.\end{aligned}}$

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