In the following lesson, we first make an identification of the complex numbers
C
{\displaystyle \mathbb {C} }
R
{\displaystyle \mathbb {R} }
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
[ edit | edit source ] Let
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}
R
2
{\displaystyle \mathbb {R} ^{2}}
Now, if we decompose a function
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}}}
Jacobian matrix as its representation:
(
∂
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}}.}
For the complex-valued function
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)}
Task
The evaluation of the Jacobian matrix at a point
(
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}}}
Evaluation of partial derivatives at a point
A function
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})}
In the following explanations, the definition of differentiability in
C
{\displaystyle \mathbb {C} }
If the following limit exists for
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}}}}
(
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}}}}
Now consider only the sequences for the two following limit processes with
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}}}
By inserting the component functions for the real and imaginary parts
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})}
Applying this to the second equation, we get with
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})}
Limit Process in the Direction of the Imaginary Part
In the first summand, the fraction is extended by
i
{\displaystyle i}
i
{\displaystyle i}
h
{\displaystyle h}
By equating the terms from (3) and (4) and comparing the real and imaginary parts, we obtain the Cauchy-Riemann differential equations.
Real part:
∂
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})}
Imaginary part:
∂
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})}
Comparison of Real and Imaginary Parts of the Derivatives
Part 6: Partial Derivative in the Direction of the Real Part [ edit | edit source ] The partial derivatives 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} }
Part 7: Partial Derivative in the Direction of the Imaginary Part [ edit | edit source ] The partial derivatives 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} }
[ edit | edit source ] The partial derivatives of the Cauchy-Riemann differential equations can also be expressed 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)}
∂
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})}
∂
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})}
Let
G
⊆
C
{\textstyle G\subseteq \mathbb {C} }
f
=
u
+
i
v
{\textstyle f=u+iv}
z
=
x
+
i
y
∈
G
{\textstyle z=x+iy\in G}
u
{\textstyle u}
v
{\textstyle v}
(
x
,
y
)
∈
R
2
{\textstyle \left(x,y\right)\in \mathbb {R} ^{2}}
∂
u
∂
x
(
x
,
y
)
=
∂
v
∂
y
(
x
,
y
)
{\displaystyle {\dfrac {\partial u}{\partial x}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)}
∂
u
∂
y
(
x
,
y
)
=
−
∂
v
∂
x
(
x
,
y
)
{\displaystyle {\dfrac {\partial u}{\partial y}}\left(x,y\right)=-{\dfrac {\partial v}{\partial x}}\left(x,y\right)}
In this case, the derivative of
f
{\textstyle f}
z
∈
C
{\textstyle z\in \mathbb {C} }
u
{\textstyle u}
v
{\textstyle v}
f
′
(
z
)
=
∂
u
∂
x
(
x
,
y
)
−
i
∂
u
∂
y
(
x
,
y
)
=
∂
v
∂
y
(
x
,
y
)
+
i
∂
v
∂
x
(
x
,
y
)
{\displaystyle f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)}
The proof considers two directional derivatives:
(DG1) the derivative in the direction of the real part and (DG2) the derivative in the direction of the imaginary part. Since these coincide for complex differentiability, the Cauchy-Riemann differential equations are obtained by setting them equal and comparing the real and imaginary parts.
In the first step, let
h
∈
C
{\displaystyle h\in \mathbb {C} }
h
:=
h
1
+
i
⋅
0
{\textstyle h:=h_{1}+i\cdot 0}
h
1
∈
R
{\textstyle h_{1}\in \mathbb {R} }
f
=
u
+
i
v
{\textstyle f=u+iv}
u
{\displaystyle u}
v
{\displaystyle v}
f
′
(
z
)
=
lim
h
→
0
f
(
z
+
h
)
−
f
(
z
)
h
=
lim
h
1
→
0
u
(
x
+
h
1
,
y
)
+
i
v
(
x
+
h
1
,
y
)
−
u
(
x
,
y
)
−
i
v
(
x
,
y
)
h
1
=
lim
h
1
→
0
u
(
x
+
h
1
,
y
)
−
u
(
x
,
y
)
h
1
+
i
v
(
x
+
h
1
,
y
)
−
v
(
x
,
y
)
h
1
=
∂
u
∂
x
(
x
,
y
)
+
i
∂
v
∂
x
(
x
,
y
)
{\displaystyle {\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\&=&\lim \limits _{h_{1}\to 0}{\dfrac {u\left(x+h_{1},y\right)+iv\left(x+h_{1},y\right)-u\left(x,y\right)-iv\left(x,y\right)}{h_{1}}}\\&=&\lim \limits _{h_{1}\to 0}{\dfrac {u\left(x+h_{1},y\right)-u\left(x,y\right)}{h_{1}}}+i{\dfrac {v\left(x+h_{1},y\right)-v\left(x,y\right)}{h_{1}}}\\&=&{\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)\end{array}}}
Step 3 - Derivative in the Direction of the Imaginary Part [ edit | edit source ] Similarly, the partial derivative for the imaginary part can be considered with
h
:=
0
+
i
⋅
h
2
{\textstyle h:=0+i\cdot h_{2}}
h
2
∈
R
{\textstyle h_{2}\in \mathbb {R} }
Step 4 - Calculation of the Derivative - Imaginary Part [ edit | edit source ]
f
′
(
z
)
=
lim
h
→
0
f
(
z
+
h
)
−
f
(
z
)
h
=
lim
h
2
→
0
u
(
x
,
y
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{\textstyle {\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\&=&\lim \limits _{h_{2}\to 0}{\dfrac {u\left(x,y+h_{2}\right)+iv\left(x,y+h_{2}\right)-u\left(x,y\right)-iv\left(x,y\right)}{i\cdot h_{2}}}\\&=&\lim \limits _{l\to 0}{\dfrac {1}{i}}{\dfrac {u\left(x,y+h_{2}\right)-u\left(x,y\right)}{h_{2}}}+{\dfrac {v\left(x,y+h_{2}\right)-v\left(x,y\right)}{h_{2}}}\\&=&{\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)\end{array}}}
By equating the two derivatives, one can compare the real and imaginary parts of the two derivatives (DG1) and (DG2):
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{\displaystyle f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)}
Two complex numbers are equal if and only if their real and imaginary parts are equal. This results in the Cauchy-Riemann differential equations. The two representation formulas follow from the above equation and the Cauchy-Riemann equations.
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