Aufgabe (Kettenregel, 5 Punkte)
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Seien
f
,
g
:
C
→
C
{\displaystyle f,g\colon \mathbf {C} \to \mathbf {C} }
∂
∂
z
(
f
∘
g
)
=
∂
f
∂
z
∘
g
⋅
∂
g
∂
z
+
∂
f
∂
z
¯
∘
g
⋅
∂
g
¯
∂
z
{\displaystyle {\frac {\partial }{\partial z}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial z}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial z}}}
und
∂
∂
z
¯
(
f
∘
g
)
=
∂
f
∂
z
∘
g
⋅
∂
g
∂
z
¯
+
∂
f
∂
z
¯
∘
g
⋅
∂
g
¯
∂
z
¯
{\displaystyle {\frac {\partial }{\partial {\bar {z}}}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial {\bar {z}}}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}}
gelten.
Wir erinnern uns (Fischer/Lieb, Seite 21 unten): Für eine differenzierbare Funktion
f
:
U
→
C
{\displaystyle f\colon U\to \mathbf {C} }
z
{\displaystyle z}
z
¯
{\displaystyle {\bar {z}}}
A
1
,
A
2
:
U
→
C
{\displaystyle A_{1},A_{2}\colon U\to \mathbf {C} }
f
(
z
)
=
f
(
z
0
)
+
A
1
(
z
)
(
z
−
z
0
)
+
A
2
(
z
)
(
z
¯
−
z
¯
0
)
{\displaystyle f(z)=f(z_{0})+A_{1}(z)(z-z_{0})+A_{2}(z)({\bar {z}}-{\bar {z}}_{0})}
so ist
∂
z
f
(
z
0
)
=
A
1
(
z
0
)
{\displaystyle \partial _{z}f(z_{0})=A_{1}(z_{0})}
∂
z
¯
f
(
z
0
)
=
A
2
(
z
0
)
{\displaystyle \partial _{\bar {z}}f(z_{0})=A_{2}(z_{0})}
Diese Beschreibung der Wirtinger-Ableitungen wollen wir hier benutzen. Sei
z
0
∈
C
{\displaystyle z_{0}\in \mathbf {C} }
g
{\displaystyle g}
z
0
{\displaystyle z_{0}}
B
1
,
B
2
{\displaystyle B_{1},B_{2}}
g
(
z
)
=
g
(
z
0
)
+
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
{\displaystyle g(z)=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}
gilt. Damit ist
(
f
∘
g
)
(
z
)
=
f
(
g
(
z
)
)
=
f
(
g
(
z
0
)
+
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
)
{\displaystyle {\begin{array}{rl}(f\circ g)(z)&=f{\bigl (}g(z){\bigr )}\\&=f{\bigl (}g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\end{array}}}
Setze nun
w
:=
g
(
z
0
)
+
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
=
g
(
z
)
w
0
:=
g
(
z
0
)
{\displaystyle {\begin{array}{rl}w&:=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})=g(z)\\w_{0}&:=g(z_{0})\end{array}}}
Da
f
{\displaystyle f}
w
0
{\displaystyle w_{0}}
A
1
{\displaystyle A_{1}}
A
2
{\displaystyle A_{2}}
f
(
w
)
=
f
(
w
0
)
+
A
1
(
w
)
(
w
−
w
0
)
+
A
2
(
w
)
(
w
¯
−
w
¯
0
)
{\displaystyle f(w)=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})}
Setzen wir ein, ergibt das
(
f
∘
g
)
(
z
)
=
f
(
g
(
z
)
)
=
f
(
g
(
z
0
)
+
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
)
=
f
(
w
0
)
+
A
1
(
w
)
(
w
−
w
0
)
+
A
2
(
w
)
(
w
¯
−
w
¯
0
)
=
f
(
g
(
z
0
)
)
+
A
1
(
g
(
z
)
)
(
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
)
+
A
2
(
g
(
z
)
)
(
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
¯
)
=
(
f
∘
g
)
(
z
0
)
+
(
A
1
(
g
(
z
)
)
B
1
(
z
)
+
A
2
(
g
(
z
)
)
B
2
(
z
)
¯
)
⏟
C
1
(
z
)
:=
(
z
−
z
0
)
+
(
A
1
(
g
(
z
)
)
B
2
(
z
)
+
A
2
(
g
(
z
)
)
B
1
(
z
)
¯
)
⏟
C
2
(
z
)
:=
(
z
¯
−
z
¯
0
)
{\displaystyle {\begin{array}{rl}(f\circ g)(z)&=f{\bigl (}g(z){\bigr )}\\&=f{\bigl (}g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\\&=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})\\&=f{\bigl (}g(z_{0}){\bigr )}+A_{1}{\bigl (}g(z){\bigr )}{\bigl (}B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}+A_{2}{\bigl (}g(z){\bigr )}{\bigl (}{\overline {B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}}{\bigr )}\\&=(f\circ g)(z_{0})+\underbrace {{\Bigl (}A_{1}{\bigl (}g(z){\bigr )}B_{1}(z)+A_{2}{\bigl (}g(z){\bigr )}{\overline {B_{2}(z)}}{\Bigl )}} _{C_{1}(z):=}(z-z_{0})+\underbrace {{\Bigl (}A_{1}{\bigl (}g(z){\bigr )}B_{2}(z)+A_{2}{\bigl (}g(z){\bigr )}{\overline {B_{1}(z)}}{\Bigl )}} _{C_{2}(z):=}({\bar {z}}-{\bar {z}}_{0})\end{array}}}
Da
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
f
∘
g
{\displaystyle f\circ g}
∂
∂
z
(
f
∘
g
)
(
z
0
)
=
C
1
(
z
0
)
=
A
1
(
g
(
z
0
)
)
B
1
(
z
0
)
+
A
2
(
g
(
z
0
)
)
B
2
(
z
0
)
¯
=
∂
f
∂
z
(
w
0
)
∂
g
∂
z
(
z
0
)
+
∂
f
∂
z
¯
(
w
0
)
∂
g
∂
z
¯
(
z
0
)
¯
{\displaystyle {\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\end{array}}}
Den letzten Term schauen wir uns noch einmal an. Wenn wir
g
(
z
)
=
g
(
z
0
)
+
B
1
(
z
)
(
z
−
z
0
)
+
B
2
(
z
)
(
z
¯
−
z
¯
0
)
{\displaystyle g(z)=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}
konjugieren, erhalten wir
g
¯
(
z
)
=
g
¯
(
z
0
)
+
B
2
(
z
)
¯
(
z
−
z
0
)
+
B
1
(
z
)
¯
(
z
¯
−
z
¯
0
)
{\displaystyle {\bar {g}}(z)={\bar {g}}(z_{0})+{\overline {B_{2}(z)}}(z-z_{0})+{\overline {B_{1}(z)}}({\bar {z}}-{\bar {z}}_{0})}
da die Konjugation ein Körperautomorphismus ist. Wir lesen ab, dass
∂
g
¯
∂
z
(
z
0
)
=
∂
g
∂
z
¯
(
z
0
)
¯
{\displaystyle {\frac {\partial {\bar {g}}}{\partial z}}(z_{0})={\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}}
∂
g
¯
∂
z
¯
(
z
0
)
=
∂
g
∂
z
(
z
0
)
¯
{\displaystyle {\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})={\overline {{\frac {\partial g}{\partial z}}(z_{0})}}}
Setzen wir oben fort, folgt damit
∂
∂
z
(
f
∘
g
)
(
z
0
)
=
C
1
(
z
0
)
=
A
1
(
g
(
z
0
)
)
B
1
(
z
0
)
+
A
2
(
g
(
z
0
)
)
B
2
(
z
0
)
¯
=
∂
f
∂
z
(
w
0
)
∂
g
∂
z
(
z
0
)
+
∂
f
∂
z
¯
(
w
0
)
∂
g
∂
z
¯
(
z
0
)
¯
=
∂
f
∂
z
(
w
0
)
∂
g
∂
z
(
z
0
)
+
∂
f
∂
z
¯
(
w
0
)
∂
g
¯
∂
z
(
z
0
)
=
∂
f
∂
z
∘
g
(
z
0
)
∂
g
∂
z
(
z
0
)
+
∂
f
∂
z
¯
∘
g
(
z
0
)
∂
g
¯
∂
z
(
z
0
)
{\displaystyle {\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\\end{array}}}
wie behauptet. Analog folgt
∂
∂
z
¯
(
f
∘
g
)
(
z
0
)
=
C
2
(
z
0
)
=
A
1
(
g
(
z
0
)
)
B
2
(
z
0
)
+
A
1
(
g
(
z
0
)
)
B
1
(
z
0
)
¯
=
∂
f
∂
z
(
w
0
)
∂
g
∂
z
¯
(
z
0
)
+
∂
f
∂
z
¯
(
w
0
)
∂
g
∂
z
(
z
0
)
¯
=
∂
f
∂
z
(
w
0
)
∂
g
∂
z
¯
(
z
0
)
+
∂
f
∂
z
¯
(
w
0
)
∂
g
¯
∂
z
¯
(
z
0
)
=
∂
f
∂
z
∘
g
(
z
0
)
∂
g
∂
z
(
z
0
)
+
∂
f
∂
z
¯
∘
g
(
z
0
)
∂
g
¯
∂
z
¯
(
z
0
)
{\displaystyle {\begin{array}{rl}{\frac {\partial }{\partial {\bar {z}}}}(f\circ g)(z_{0})&=C_{2}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{2}(z_{0})+A_{1}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{1}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial z}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\\end{array}}}
