Let
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
f
=
u
+
i
v
{\displaystyle f=u+iv}
z
=
x
+
i
y
∈
G
{\displaystyle z=x+iy\in G}
u
{\displaystyle u}
v
{\displaystyle v}
(
x
,
y
)
{\displaystyle \left(x,y\right)}
∂
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
{\displaystyle f}
z
{\displaystyle z}
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)}
Let
h
:=
k
+
i
0
(
k
∈
R
)
{\displaystyle h:=k+i0\left(k\in \mathbb {R} \right)}
f
′
(
z
)
=
lim
h
→
0
f
(
z
+
h
)
−
f
(
z
)
h
=
lim
k
→
0
u
(
x
+
k
,
y
)
+
i
v
(
x
+
k
,
y
)
−
u
(
x
,
y
)
−
i
v
(
x
,
y
)
k
=
lim
k
→
0
u
(
x
+
k
,
y
)
−
u
(
x
,
y
)
k
+
i
v
(
x
+
k
,
y
)
−
v
(
x
,
y
)
k
=
∂
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 _{k\to 0}{\dfrac {u\left(x+k,y\right)+iv\left(x+k,y\right)-u\left(x,y\right)-iv\left(x,y\right)}{k}}\\&=&\lim \limits _{k\to 0}{\dfrac {u\left(x+k,y\right)-u\left(x,y\right)}{k}}+i{\dfrac {v\left(x+k,y\right)-v\left(x,y\right)}{k}}\\&=&{\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)\end{array}}}
Let
h
:=
0
+
i
l
(
l
∈
R
)
{\displaystyle h:=0+il\left(l\in \mathbb {R} \right)}
f
′
(
z
)
=
lim
h
→
0
f
(
z
+
h
)
−
f
(
z
)
h
=
lim
l
→
0
u
(
x
,
y
+
l
)
+
i
v
(
x
,
y
+
l
)
−
u
(
x
,
y
)
−
i
v
(
x
,
y
)
i
l
=
lim
l
→
0
1
i
u
(
x
,
y
+
l
)
−
u
(
x
,
y
)
l
+
v
(
x
,
y
+
l
)
−
v
(
x
,
y
)
l
=
∂
v
∂
y
(
x
,
y
)
−
i
∂
u
∂
y
(
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 _{l\to 0}{\dfrac {u\left(x,y+l\right)+iv\left(x,y+l\right)-u\left(x,y\right)-iv\left(x,y\right)}{il}}\\&=&\lim \limits _{l\to 0}{\dfrac {1}{i}}{\dfrac {u\left(x,y+l\right)-u\left(x,y\right)}{l}}+{\dfrac {v\left(x,y+l\right)-v\left(x,y\right)}{l}}\\&=&{\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)\end{array}}}
Hence:
f
′
(
z
)
=
∂
u
∂
x
(
x
,
y
)
+
i
∂
v
∂
x
(
x
,
y
)
=
∂
v
∂
y
(
x
,
y
)
−
i
∂
u
∂
y
(
x
,
y
)
{\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)}
Equating the real and imaginary parts, we get the Cauchy-Riemann equations. The representation formula follows from the above line and the Cauchy-Riemann equations.
