Identification of the complex numbers IR 2
edit
Be
R
:
C
→
R
2
,
x
+
i
y
↦
R
(
x
+
i
y
)
=
(
x
y
)
{\textstyle R:\mathbb {C} \rightarrow \mathbb {R} ^{2},\ x+iy\mapsto R(x+iy)={\begin{pmatrix}x\\y\end{pmatrix}}}
R
{\textstyle R}
R
2
{\textstyle \mathbb {R} ^{2}}
Real part and imaginary part function
edit
The total function
f
:
U
→
C
{\textstyle f:U\rightarrow \mathbb {C} }
f
(
x
+
i
y
)
=
u
(
x
,
y
)
+
i
v
(
x
,
y
)
{\textstyle f\left(x+iy\right)=u\left(x,y\right)+i\,v\left(x,y\right)}
u
:
U
R
→
R
{\textstyle u:U_{R}\rightarrow \mathbb {R} }
(
∂
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}}.}
Specify the images
f
:
C
→
C
,
z
↦
f
(
z
)
=
z
3
{\textstyle f:\mathbb {C} \rightarrow \mathbb {C} ,\ z\mapsto f(z)=z^{3}}
u
,
v
{\textstyle u,v}
f
(
x
+
i
y
)
=
u
(
x
,
y
)
+
i
v
(
x
,
y
)
{\textstyle f\left(x+iy\right)=u\left(x,y\right)+i\,v\left(x,y\right)}
Evaluation of the Jacobimatrix in one point
edit
The evaluation of the Jacobi matrix in one point
(
x
o
,
y
o
)
∈
R
2
{\textstyle (x_{o},y_{o})\in \mathbb {R} ^{2}}
x
o
+
i
y
o
∈
C
{\textstyle 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}}}
Cauchy-Riemann differential equations
edit
A function
f
{\textstyle f}
z
o
:=
x
o
+
i
y
o
{\textstyle z_{o}:=x_{o}+iy_{o}}
u
,
v
{\textstyle u,v}
u
:
U
R
→
R
{\textstyle u:U_{R}\rightarrow \mathbb {R} }
∂
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})}
are fulfilled.
Relationship between the partial discharges
edit
In the following explanations, the definition of the differentiation in
C
{\textstyle \mathbb {C} }
If the following Limes exists for
f
:
G
→
C
{\textstyle f:G\rightarrow \mathbb {C} }
z
o
∈
G
{\textstyle z_{o}\in G}
G
⊂
C
{\textstyle 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}}}}
means
lim
z
→
z
o
.
.
.
{\textstyle \lim _{z\rightarrow z_{o}}...}
(
z
n
)
n
∈
N
{\textstyle (z_{n})_{n\in \mathbb {N} }}
G
⊂
C
{\textstyle G\subset \mathbb {C} }
lim
n
→
∞
z
n
=
z
o
{\textstyle \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}}}}
is fulfilled.
From these arbitrary consequences, one considers only the consequences for the two following limit processes with
h
∈
R
{\textstyle 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}}}
Part 3: Limitation process Real part
edit
By inserting the component functions for the real part and imaginary part
u
,
v
{\textstyle u,v}
h
∈
R
{\textstyle 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})}
Part 4: Limit for Imaginary part
edit
When applied to the second equation,
h
∈
R
{\textstyle 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})}
Part 5: Real part and imaginary part comparison
edit
The Cauchy Riemann differential equations are obtained by equation of the terms of (3) and (4) and comparison of the real part and the imaginary part.
Real part:
∂
u
∂
x
(
x
o
,
y
o
)
=
∂
v
∂
y
(
x
o
,
y
o
)
{\textstyle {\frac {\partial u}{\partial x}}(x_{o},y_{o})={\frac {\partial v}{\partial y}}(x_{o},y_{o})}
Imaginary part:
∂
u
∂
y
(
x
o
,
y
o
)
=
−
∂
v
∂
x
(
x
o
,
y
o
)
{\textstyle \displaystyle {\frac {\partial u}{\partial y}}(x_{o},y_{o})=-{\frac {\partial v}{\partial x}}(x_{o},y_{o})}
Part 6: Partial derivation towards real part
edit
The partial derivations in
R
2
{\textstyle \mathbb {R} ^{2}}
C
{\textstyle \mathbb {C} }
f
:=
R
e
(
f
)
+
i
I
m
(
f
)
{\textstyle f:={\mathfrak {Re}}(f)+i{\mathfrak {Im}}(f)}
R
e
(
f
)
:
C
→
R
{\textstyle {\mathfrak {Re}}(f):\mathbb {C} \rightarrow \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} }
Part 7: Partial derivation towards the imaginary part
edit
The partial discharges in
R
2
{\textstyle \mathbb {R} ^{2}}
C
{\textstyle \mathbb {C} }
f
:=
R
e
(
f
)
+
i
I
m
(
f
)
{\textstyle f:={\mathfrak {Re}}(f)+i{\mathfrak {Im}}(f)}
R
e
(
f
)
:
C
→
R
{\textstyle {\mathfrak {Re}}(f):\mathbb {C} \rightarrow \mathbb {R} }
I
m
(
f
)
:
C
→
R
{\textstyle {\mathfrak {Im}}(f):\mathbb {C} \rightarrow \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} }
