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Komplexe Zahlen/Reell differenzierbare Abbildung/dz konjugiert/Rückzug/Aufgabe/Lösung
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Komplexe Zahlen
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Reell differenzierbare Abbildung/dz konjugiert/Rückzug/Aufgabe
Es ist
d
z
¯
=
d
x
−
i
d
y
.
{\displaystyle {}d{\overline {z}}=dx-{\mathrm {i} }dy\,.}
Nach
Fakt
ist daher
φ
∗
(
d
z
¯
)
=
φ
∗
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d
x
−
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d
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=
∂
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1
∂
u
d
u
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∂
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∂
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d
v
−
i
(
∂
φ
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∂
u
d
u
+
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φ
2
∂
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d
v
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=
(
∂
φ
1
∂
u
−
i
∂
φ
2
∂
u
)
d
u
+
(
∂
φ
1
∂
v
−
i
∂
φ
2
∂
v
)
d
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=
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∂
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∂
u
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¯
d
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+
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∂
φ
∂
v
)
¯
d
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=
∂
φ
¯
∂
u
d
u
+
∂
φ
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∂
v
d
v
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(
1
2
⋅
∂
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∂
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∂
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∂
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1
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∂
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d
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∂
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∂
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1
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∂
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∂
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∂
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∂
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d
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=
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1
2
⋅
∂
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d
(
u
+
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+
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1
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∂
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d
(
u
−
i
v
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=
∂
φ
¯
∂
w
d
w
+
∂
φ
¯
∂
w
¯
d
w
¯
.
{\displaystyle {}{\begin{aligned}\varphi ^{*}(d{\overline {z}})&=\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\\&={\overline {\left({\frac {\partial \varphi }{\partial u}}\right)}}du+{\overline {\left({\frac {\partial \varphi }{\partial v}}\right)}}dv\\&={\frac {\partial {\overline {\varphi }}}{\partial u}}du+{\frac {\partial {\overline {\varphi }}}{\partial v}}dv\\&={\left({\frac {1}{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial u}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial v}}+{\frac {1}{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial u}}+{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial v}}\right)}du+{\left({\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \varphi }{\partial u}}+{\frac {1}{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial v}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial u}}+{\frac {1}{2}}\cdot {\frac {\partial \varphi }{\partial v}}\right)}dv\\&={\left({\frac {1}{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial u}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial v}}\right)}d{\left(u+{\mathrm {i} }v\right)}+{\left({\frac {1}{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial u}}+{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial {\overline {\varphi }}}{\partial v}}\right)}d{\left(u-{\mathrm {i} }v\right)}\\&={\frac {\partial {\overline {\varphi }}}{\partial w}}dw+{\frac {\partial {\overline {\varphi }}}{\partial {\overline {w}}}}d{\overline {w}}.\end{aligned}}}
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