Identifikation der komplexen Zahlen IR2
Bearbeiten
Sei
R
:
C
→
R
2
,
x
+
i
y
↦
R
(
x
+
i
y
)
=
(
x
y
)
{\displaystyle R:\mathbb {C} \rightarrow \mathbb {R} ^{2},\ x+iy\mapsto R(x+iy)={\begin{pmatrix}x\\y\end{pmatrix}}}
R
{\displaystyle R}
R
−
1
:
R
2
→
C
,
(
x
y
)
↦
R
−
1
(
x
y
)
=
x
+
i
y
{\displaystyle R^{-1}:\mathbb {R} ^{2}\rightarrow \mathbb {C} ,\ {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto R^{-1}{\begin{pmatrix}x\\y\end{pmatrix}}=x+iy}
Vektoren aus dem
R
2
{\displaystyle \mathbb {R} ^{2}}
Zerlegt man nun eine Funktion
f
:
U
→
C
{\displaystyle f:U\rightarrow \mathbb {C} }
f
(
x
+
i
y
)
=
u
(
x
,
y
)
+
i
v
(
x
,
y
)
{\displaystyle f\left(x+iy\right)=u\left(x,y\right)+i\,v\left(x,y\right)}
u
:
U
R
→
R
{\displaystyle u:U_{R}\rightarrow \mathbb {R} }
v
:
U
R
→
R
{\displaystyle v:U_{R}\rightarrow \mathbb {R} }
U
R
⊂
R
2
{\displaystyle U_{R}\subset \mathbb {R} ^{2}}
U
=
{
x
+
i
y
∈
C
|
(
x
,
y
)
∈
U
R
}
{\displaystyle U=\{x+iy\in \mathbb {C} \ |\ (x,y)\in U_{R}\}}
f
R
:
U
R
→
R
2
,
(
x
,
y
)
↦
(
u
(
x
,
y
)
v
(
x
,
y
)
)
{\displaystyle f_{R}:U_{R}\rightarrow \mathbb {R} ^{2},(x,y)\mapsto {\begin{pmatrix}u\left(x,y\right)\\v\left(x,y\right)\end{pmatrix}}}
Jacobi-Matrix
(
∂
u
∂
x
∂
u
∂
y
∂
v
∂
x
∂
v
∂
y
)
.
{\displaystyle {\begin{pmatrix}{\frac {\partial u}{\partial x}}&{\frac {\partial u}{\partial y}}\\{\frac {\partial v}{\partial x}}&{\frac {\partial v}{\partial y}}\end{pmatrix}}.}
Geben Sie für die komplexwertige Funktion
f
:
C
→
C
,
z
↦
f
(
z
)
=
z
3
{\displaystyle f:\mathbb {C} \rightarrow \mathbb {C} ,\ z\mapsto f(z)=z^{3}}
u
,
v
{\displaystyle u,v}
f
(
x
+
i
y
)
=
u
(
x
,
y
)
+
i
v
(
x
,
y
)
{\displaystyle f\left(x+iy\right)=u\left(x,y\right)+i\,v\left(x,y\right)}
Auswertung der Jacobimatrix in einem Punkt
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Dabei liefert die Auswertung der Jacobimatrix in einem Punkt
(
x
o
,
y
o
)
∈
R
2
{\displaystyle (x_{o},y_{o})\in \mathbb {R} ^{2}}
x
o
+
i
y
o
∈
C
{\displaystyle x_{o}+iy_{o}\in \mathbb {C} }
(
∂
u
∂
x
(
x
o
,
y
o
)
∂
u
∂
y
(
x
o
,
y
o
)
∂
v
∂
x
(
x
o
,
y
o
)
∂
v
∂
y
(
x
o
,
y
o
)
)
{\displaystyle {\begin{pmatrix}{\frac {\partial u}{\partial x}}(x_{o},y_{o})&{\frac {\partial u}{\partial y}}(x_{o},y_{o})\\{\frac {\partial v}{\partial x}}(x_{o},y_{o})&{\frac {\partial v}{\partial y}}(x_{o},y_{o})\end{pmatrix}}}
Eine Funktion
f
{\displaystyle f}
z
o
:=
x
o
+
i
y
o
{\displaystyle z_{o}:=x_{o}+iy_{o}}
u
,
v
{\displaystyle u,v}
u
:
U
R
→
R
{\displaystyle u:U_{R}\rightarrow \mathbb {R} }
v
:
U
R
→
R
{\displaystyle v:U_{R}\rightarrow \mathbb {R} }
U
R
⊂
R
2
{\displaystyle U_{R}\subset \mathbb {R} ^{2}}
∂
u
∂
x
(
x
o
,
y
o
)
=
∂
v
∂
y
(
x
o
,
y
o
)
{\displaystyle {\frac {\partial u}{\partial x}}(x_{o},y_{o})={\frac {\partial v}{\partial y}}(x_{o},y_{o})}
∂
u
∂
y
(
x
o
,
y
o
)
=
−
∂
v
∂
x
(
x
o
,
y
o
)
{\displaystyle \displaystyle {\frac {\partial u}{\partial y}}(x_{o},y_{o})=-{\frac {\partial v}{\partial x}}(x_{o},y_{o})}
erfüllt sind.
Zusammenhang zwischen den partiellen Ableitungen
Bearbeiten
In dem folgenden Erläuterungen wird die Definition der Differenzierbarkeit in
C
{\displaystyle \mathbb {C} }
Wenn der folgende Limes für
f
:
G
→
C
{\displaystyle f:G\rightarrow \mathbb {C} }
z
o
∈
G
{\displaystyle z_{o}\in G}
G
⊂
C
{\displaystyle G\subset \mathbb {C} }
f
′
(
z
o
)
=
lim
z
→
z
o
f
(
z
)
−
f
(
z
o
)
z
−
z
o
{\displaystyle f'(z_{o})=\lim _{z\rightarrow z_{o}}{\frac {f(z)-f(z_{o})}{z-z_{o}}}}
bedeutet
lim
z
→
z
o
.
.
.
{\displaystyle \lim _{z\rightarrow z_{o}}...}
(
z
n
)
n
∈
N
{\displaystyle (z_{n})_{n\in \mathbb {N} }}
G
⊂
C
{\displaystyle G\subset \mathbb {C} }
lim
n
→
∞
z
n
=
z
o
{\displaystyle \lim _{n\rightarrow \infty }z_{n}=z_{o}}
f
′
(
z
o
)
=
lim
n
→
∞
f
(
z
n
)
−
f
(
z
o
)
z
n
−
z
o
{\displaystyle f'(z_{o})=\lim _{n\rightarrow \infty }{\frac {f(z_{n})-f(z_{o})}{z_{n}-z_{o}}}}
erfüllt ist.
Man betrachtet nun von diesen beliebigen Folgen nur die Folgen für die beiden folgenden Grenzwertprozessen mit
h
∈
R
{\displaystyle h\in \mathbb {R} }
f
′
(
z
o
)
=
lim
h
→
0
f
(
z
o
+
h
)
−
f
(
z
o
)
(
z
o
+
h
)
−
z
o
=
lim
h
→
0
f
(
z
o
+
h
)
−
f
(
z
o
)
h
{\displaystyle f'(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{(z_{o}+h)-z_{o}}}=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{h}}}
f
′
(
z
o
)
=
lim
i
h
→
0
f
(
z
o
+
i
h
)
−
f
(
z
o
)
(
z
o
+
i
h
)
−
z
o
=
lim
i
h
→
0
f
(
z
o
+
i
h
)
−
f
(
z
o
)
i
h
{\displaystyle f'(z_{o})=\lim _{ih\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{(z_{o}+ih)-z_{o}}}=\lim _{ih\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{ih}}}
Durch Einsetzen der Komponentenfunktionen für den Realteil und Imaginärteil
u
,
v
{\displaystyle u,v}
h
∈
R
{\displaystyle h\in \mathbb {R} }
f
′
(
z
o
)
=
lim
h
→
0
f
(
z
o
+
h
)
−
f
(
z
o
)
h
=
{\displaystyle f'(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{h}}=}
=
lim
h
→
0
u
(
x
o
+
h
,
y
o
)
−
u
(
x
o
,
y
o
)
h
+
i
lim
h
→
0
v
(
x
o
+
h
,
y
o
)
−
v
(
x
o
,
y
o
)
h
{\displaystyle =\lim _{h\rightarrow 0}{\frac {u(x_{o}+h,y_{o})-u(x_{o},y_{o})}{h}}+i\lim _{h\rightarrow 0}{\frac {v(x_{o}+h,y_{o})-v(x_{o},y_{o})}{h}}}
=
∂
u
∂
x
(
x
o
,
y
o
)
+
i
∂
v
∂
x
(
x
o
,
y
o
)
{\displaystyle ={\frac {\partial u}{\partial x}}(x_{o},y_{o})+i{\frac {\partial v}{\partial x}}(x_{o},y_{o})}
Bei der Anwendung auf die zweite Gleichung erhält man mit
h
∈
R
{\displaystyle h\in \mathbb {R} }
f
′
(
z
o
)
=
lim
i
h
→
0
f
(
z
o
+
i
h
)
−
f
(
z
o
)
i
h
{\displaystyle f'(z_{o})=\lim _{ih\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{ih}}}
=
lim
h
→
0
u
(
x
o
,
y
o
+
h
)
−
u
(
x
o
,
y
o
)
i
h
+
i
lim
h
→
0
v
(
x
o
,
y
o
+
h
)
−
v
(
x
o
,
y
o
)
i
h
{\displaystyle =\lim _{h\rightarrow 0}{\frac {u(x_{o},y_{o}+h)-u(x_{o},y_{o})}{ih}}+i\lim _{h\rightarrow 0}{\frac {v(x_{o},y_{o}+h)-v(x_{o},y_{o})}{ih}}}
=
−
i
lim
h
→
0
u
(
x
o
,
y
o
+
h
)
−
u
(
x
o
,
y
o
)
h
+
lim
h
→
0
v
(
x
o
,
y
o
+
h
)
−
v
(
x
o
,
y
o
)
h
{\displaystyle =-i\lim _{h\rightarrow 0}{\frac {u(x_{o},y_{o}+h)-u(x_{o},y_{o})}{h}}+\lim _{h\rightarrow 0}{\frac {v(x_{o},y_{o}+h)-v(x_{o},y_{o})}{h}}}
=
−
i
∂
u
∂
y
(
x
o
,
y
o
)
+
∂
v
∂
y
(
x
o
,
y
o
)
{\displaystyle =-i{\frac {\partial u}{\partial y}}(x_{o},y_{o})+{\frac {\partial v}{\partial y}}(x_{o},y_{o})}
Bearbeiten
Durch Gleichsetzung der Terme von (3) und (4) und Vergleich von Realteil und Imaginärteil erhält man die Cauchy-Riemann-Differentialgleichungen.
Realteil:
∂
u
∂
x
(
x
o
,
y
o
)
=
∂
v
∂
y
(
x
o
,
y
o
)
{\displaystyle {\frac {\partial u}{\partial x}}(x_{o},y_{o})={\frac {\partial v}{\partial y}}(x_{o},y_{o})}
Imaginärteil:
∂
u
∂
y
(
x
o
,
y
o
)
=
−
∂
v
∂
x
(
x
o
,
y
o
)
{\displaystyle \displaystyle {\frac {\partial u}{\partial y}}(x_{o},y_{o})=-{\frac {\partial v}{\partial x}}(x_{o},y_{o})}
Teil 6: Partielle Ableitung in Richtung Realteil
Bearbeiten
Die partiellen Ableitungen in
R
2
{\displaystyle \mathbb {R} ^{2}}
C
{\displaystyle \mathbb {C} }
f
:=
R
e
(
f
)
+
i
I
m
(
f
)
{\displaystyle f:={\mathfrak {Re}}(f)+i{\mathfrak {Im}}(f)}
R
e
(
f
)
:
C
→
R
{\displaystyle {\mathfrak {Re}}(f):\mathbb {C} \rightarrow \mathbb {R} }
I
m
(
f
)
:
C
→
R
{\displaystyle {\mathfrak {Im}}(f):\mathbb {C} \rightarrow \mathbb {R} }
h
∈
R
{\displaystyle h\in \mathbb {R} }
∂
f
∂
x
(
z
o
)
=
lim
h
→
0
f
(
z
o
+
h
)
−
f
(
z
o
)
h
∈
C
{\displaystyle {\frac {\partial f}{\partial x}}(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{h}}\in \mathbb {C} }
∂
R
e
(
f
)
∂
x
(
z
o
)
=
lim
h
→
0
R
e
(
f
)
(
z
o
+
h
)
−
R
e
(
f
)
(
z
o
)
h
∈
R
{\displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial x}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Re}}(f)(z_{o}+h)-{\mathfrak {Re}}(f)(z_{o})}{h}}\in \mathbb {R} }
∂
I
m
(
f
)
∂
x
(
z
o
)
=
lim
h
→
0
I
m
(
f
)
(
z
o
+
h
)
−
I
m
(
f
)
(
z
o
)
h
∈
R
{\displaystyle {\frac {\partial {\mathfrak {Im}}(f)}{\partial x}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Im}}(f)(z_{o}+h)-{\mathfrak {Im}}(f)(z_{o})}{h}}\in \mathbb {R} }
Bearbeiten
Die partiellen Ableitungen in
R
2
{\displaystyle \mathbb {R} ^{2}}
C
{\displaystyle \mathbb {C} }
f
:=
R
e
(
f
)
+
i
I
m
(
f
)
{\displaystyle f:={\mathfrak {Re}}(f)+i{\mathfrak {Im}}(f)}
R
e
(
f
)
:
C
→
R
{\displaystyle {\mathfrak {Re}}(f):\mathbb {C} \rightarrow \mathbb {R} }
I
m
(
f
)
:
C
→
R
{\displaystyle {\mathfrak {Im}}(f):\mathbb {C} \rightarrow \mathbb {R} }
h
∈
R
{\displaystyle h\in \mathbb {R} }
∂
f
∂
y
(
z
o
)
=
lim
h
→
0
f
(
z
o
+
i
h
)
−
f
(
z
o
)
h
∈
C
{\displaystyle {\frac {\partial f}{\partial y}}(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{h}}\in \mathbb {C} }
∂
R
e
(
f
)
∂
y
(
z
o
)
=
lim
h
→
0
R
e
(
f
)
(
z
o
+
i
h
)
−
R
e
(
f
)
(
z
o
)
h
∈
R
{\displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial y}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Re}}(f)(z_{o}+ih)-{\mathfrak {Re}}(f)(z_{o})}{h}}\in \mathbb {R} }
∂
I
m
(
f
)
∂
y
(
z
o
)
=
lim
h
→
0
I
m
(
f
)
(
z
o
+
i
h
)
−
I
m
(
f
)
(
z
o
)
h
∈
R
{\displaystyle {\frac {\partial {\mathfrak {Im}}(f)}{\partial y}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Im}}(f)(z_{o}+ih)-{\mathfrak {Im}}(f)(z_{o})}{h}}\in \mathbb {R} }
C
{\displaystyle \mathbb {C} }
Bearbeiten
Die partiellen Ableitungen der Cauchy-Riemann-Differentialgleichungen können auch in
C
{\displaystyle \mathbb {C} }
f
:=
f
x
+
i
f
y
{\displaystyle f:=f_{x}+if_{y}}
f
x
:=
R
e
(
f
)
{\displaystyle f_{x}:={\mathfrak {Re}}(f)}
f
y
:=
I
m
(
f
)
{\displaystyle f_{y}:={\mathfrak {Im}}(f)}
Realteil:
∂
R
e
(
f
)
∂
x
(
z
o
)
=
∂
I
m
(
f
)
∂
y
(
z
o
)
{\displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial x}}(z_{o})={\frac {\partial {\mathfrak {Im}}(f)}{\partial y}}(z_{o})}
Imaginärteil:
∂
R
e
(
f
)
∂
y
(
z
o
)
=
−
∂
I
m
(
f
)
∂
x
(
z
o
)
{\displaystyle \displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial y}}(z_{o})=-{\frac {\partial {\mathfrak {Im}}(f)}{\partial x}}(z_{o})}
