A slightly more general form of the expansion above is the following: Let
0
≤
r
1
<
r
2
{\displaystyle 0\leq r_{1}<r_{2}}
r
1
=
0
{\displaystyle r_{1}=0}
A
r
1
,
r
2
:=
z
∈
C
:
r
1
<
|
z
−
z
0
|
<
r
2
{\displaystyle A_{r_{1},r_{2}}:={z\in \mathbb {C} :r_{1}<|z-z_{0}|<r_{2}}}
z
0
{\displaystyle z_{0}}
f
:
A
r
1
,
r
2
→
C
{\displaystyle f\colon A_{r_{1},r_{2}}\to \mathbb {C} }
Laurent Series
f
(
z
)
=
∑
n
=
−
∞
∞
a
n
(
z
−
z
0
)
n
{\displaystyle f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}}
with
a
n
∈
C
{\displaystyle a_{n}\in \mathbb {C} }
f
{\displaystyle f}
A
r
1
,
r
2
{\displaystyle A_{r_{1},r_{2}}}
z
∈
A
r
1
,
r
2
{\displaystyle z\in A_{r_{1},r_{2}}}
Every holomorphic function on
A
r
1
,
r
2
{\displaystyle A_{r_{1},r_{2}}}
z
0
{\displaystyle z_{0}}
a
n
{\displaystyle a_{n}}
a
n
=
1
2
π
i
∫
|
z
−
z
0
|
=
r
f
(
z
)
(
z
−
z
0
)
n
+
1
d
z
{\displaystyle a_{n}={\frac {1}{2\pi i}}\int _{|z-z_{0}|=r}{\frac {f(z)}{(z-z_{0})^{n+1}}}\,dz}
for a radius
r
{\displaystyle r}
r
1
<
r
<
r
2
{\displaystyle r_{1}<r<r_{2}}
The coefficients are uniquely determined by:
a
n
=
1
2
π
i
∫
|
z
−
z
0
|
=
r
f
(
z
)
(
z
−
z
0
)
n
+
1
d
z
{\displaystyle a_{n}={\frac {1}{2\pi i}}\int _{|z-z_{0}|=r}{\frac {f(z)}{(z-z_{0})^{n+1}}}\,dz}
Proof of Existence and Uniqueness of the Laurent Representation
edit
Uniqueness follows from the Identity Theorem for Laurent Series . To prove existence, choose a radius
r
{\displaystyle r}
r
1
<
r
<
r
2
{\displaystyle r_{1}<r<r_{2}}
R
1
,
R
2
{\displaystyle R_{1},R_{2}}
r
1
<
R
1
<
r
<
R
2
<
r
2
{\displaystyle r_{1}<R_{1}<r<R_{2}<r_{2}}
z
∈
A
R
1
,
R
2
{\displaystyle z\in A_{R_{1},R_{2}}}
A
R
1
,
R
2
{\displaystyle A_{R_{1},R_{2}}}
D
1
{\displaystyle D_{1}}
D
2
{\displaystyle D_{2}}
∂
K
R
2
−
∂
K
R
1
{\displaystyle \partial K_{R_{2}}-\partial K_{R_{1}}}
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
A
{\displaystyle A}
D
1
{\displaystyle D_{1}}
D
2
{\displaystyle D_{2}}
z
{\displaystyle z}
C
1
{\displaystyle C_{1}}
Cauchy Integral Theorem , we have:
f
(
z
)
=
1
2
π
i
∫
C
1
f
(
w
)
w
−
z
d
w
{\displaystyle f(z)={\frac {1}{2\pi i}}\int _{C_{1}}{\frac {f(w)}{w-z}}\,dw}
and
0
=
1
2
π
i
∫
C
2
f
(
w
)
w
−
z
d
w
{\displaystyle 0={\frac {1}{2\pi i}}\int _{C_{2}}{\frac {f(w)}{w-z}}\,dw}
since
C
2
{\displaystyle C_{2}}
z
{\displaystyle z}
C
1
+
C
2
=
∂
K
R
2
−
∂
K
R
1
{\displaystyle C_{1}+C_{2}=\partial K_{R_{2}}-\partial K_{R_{1}}}
f
(
z
)
=
1
2
π
i
∫
|
w
−
z
0
|
=
R
2
f
(
w
)
w
−
z
d
w
−
1
2
π
i
∫
|
w
−
z
0
|
=
R
1
f
(
w
)
w
−
z
d
w
{\displaystyle f(z)={\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{w-z}}\,dw-{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{w-z}}\,dw}
For
|
w
−
z
0
|
=
R
2
{\displaystyle |w-z_{0}|=R_{2}}
1
w
−
z
=
1
(
w
−
z
0
)
−
(
z
−
z
0
)
=
1
w
−
z
0
⋅
1
1
−
z
−
z
0
w
−
z
0
=
1
w
−
z
0
∑
n
=
0
∞
(
z
−
z
0
)
n
(
w
−
z
0
)
n
{\displaystyle {\begin{array}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\cdot {\frac {1}{1-{\frac {z-z_{0}}{w-z_{0}}}}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\sum _{n=0}^{\infty }{\frac {(z-z_{0})^{n}}{(w-z_{0})^{n}}}\end{array}}}
The series converges absolutely because
|
z
−
z
0
|
<
|
w
−
z
0
|
{\displaystyle |z-z_{0}|<|w-z_{0}|}
1
2
π
i
∫
|
w
−
z
0
|
=
R
2
f
(
w
)
w
−
z
d
w
=
1
2
π
i
∫
|
w
−
z
0
|
=
R
2
1
w
−
z
0
⋅
f
(
w
)
(
z
−
z
0
)
n
(
w
−
z
0
)
n
d
w
=
1
2
π
i
∑
n
=
0
∞
∫
|
w
−
z
0
|
=
R
2
f
(
w
)
(
w
−
z
0
)
n
+
1
d
w
⋅
(
z
−
z
0
)
n
=
1
2
π
i
∑
n
=
0
∞
∫
|
w
−
z
0
|
=
r
f
(
w
)
(
w
−
z
0
)
n
+
1
d
w
⋅
(
z
−
z
0
)
n
{\displaystyle {\begin{array}{rl}{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{w-z}}\,dw&={\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {1}{w-z_{0}}}\cdot {\frac {f(w)(z-z_{0})^{n}}{(w-z_{0})^{n}}}\,dw\\&={\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\\&={\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{|w-z_{0}|=r}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\end{array}}}
Now, consider the integral over the inner circle, which is analogous to the above for
|
w
−
z
0
|
=
R
1
{\displaystyle |w-z_{0}|=R_{1}}
1
w
−
z
=
1
(
w
−
z
0
)
−
(
z
−
z
0
)
=
−
1
z
−
z
0
⋅
1
1
−
w
−
z
0
z
−
z
0
=
−
1
z
−
z
0
∑
n
=
0
∞
(
w
−
z
0
)
n
(
z
−
z
0
)
n
{\displaystyle {\begin{array}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {-1}{z-z_{0}}}\cdot {\frac {1}{1-{\frac {w-z_{0}}{z-z_{0}}}}}\\&=\displaystyle {\frac {-1}{z-z_{0}}}\sum _{n=0}^{\infty }{\frac {(w-z_{0})^{n}}{(z-z_{0})^{n}}}\end{array}}}
Thus, due to
R
1
=
|
w
−
z
0
|
<
|
z
−
z
0
|
{\displaystyle R_{1}=|w-z_{0}|<|z-z_{0}|}
−
1
2
π
i
∫
|
w
−
z
0
|
=
R
1
f
(
w
)
w
−
z
d
w
=
1
2
π
i
∫
|
w
−
z
0
|
=
R
1
−
1
z
−
z
0
⋅
f
(
w
)
(
w
−
z
0
)
n
(
z
−
z
0
)
n
d
w
=
1
2
π
i
∑
n
=
0
∞
∫
|
w
−
z
0
|
=
R
1
f
(
w
)
(
w
−
z
0
)
−
n
d
w
⋅
(
z
−
z
0
)
−
n
−
1
=
1
2
π
i
∑
n
=
0
∞
∫
|
w
−
z
0
|
=
r
f
(
w
)
(
w
−
z
0
)
−
n
d
w
⋅
(
z
−
z
0
)
−
n
−
1
{\displaystyle {\begin{array}{rl}-{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{w-z}}\,dw&={\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\frac {-1}{z-z_{0}}}\cdot {\frac {f(w)(w-z_{0})^{n}}{(z-z_{0})^{n}}}\,dw\\&={\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{(w-z_{0})^{-n}}}\,dw\cdot (z-z_{0})^{-n-1}\\&={\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{|w-z_{0}|=r}{\frac {f(w)}{(w-z_{0})^{-n}}}\,dw\cdot (z-z_{0})^{-n-1}\end{array}}}
Thus, it follows that for
z
∈
A
R
1
,
R
2
{\displaystyle z\in A_{R_{1},R_{2}}}
f
(
z
)
=
1
2
π
i
∑
n
=
−
∞
∞
∫
|
w
−
z
0
|
=
r
f
(
w
)
(
w
−
z
0
)
n
+
1
d
w
⋅
(
z
−
z
0
)
n
{\displaystyle f(z)={\frac {1}{2\pi i}}\sum _{n=-\infty }^{\infty }\int _{|w-z_{0}|=r}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}}
which proves the existence of the claimed Laurent expansion.
