PDE & Complex Variables
P13-1
Lesson 13 Laurent Series and Residue (EK 16)
Laurent Series (EK 16.1)
Motivation
Singularities are routine for complex functions, for bounded entire functions must be constant
(Liouville’s theorem, Lesson 10). We have to represent complex functions by Laurent series
if there is singularity in the region of interest.
Laurent theorem
If f(z) is analytic in an open annulus D: R2<|z−z0|<R1,⇒
∞
∞
bn
n
n =1 ( z − z0 )
f(z)= ∑ a n ( z − z 0 ) n + ∑
n =0
an=
f ( z ′)
1
dz ′ ,
∫
C
2πi ( z ′ − z 0 ) n +1
bn=
1
( z ′ − z0 ) n−1 f ( z ′)dz′
∫
C
2πi
(13.1)
for every point z∈D, where C⊂D is a simple closed path.
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Proof: Let C1, C2 ⊂D are two concentric circles of radii r1, r2. By Cauchy’s integral formula
for doubly connected domain [eq. (10.10)]: f(z)=
f (z* ) *
1
1
f ( z* ) *
dz
−
dz ≡
2πi ∫C2 z * − z
2πi ∫C1 z * − z
g(z)+h(z) = (analytic part)+(principal part), for any z∈D.
(1) Since z lies within C1, and f(z) is analytic on C1, by the proof of Taylor series in Lesson
12: g(z)=
1
f (z* )
f ( z* ) * ∞
1
n
dz
=
a
(
z
−
z
)
,
where
a
=
dz * .
n
∑
n
0
∫
C1 ( z * − z ) n +1
2πi ∫C1 z * − z
2
i
π
n =0
0
<Comment>
The proof of Taylor series consists of two steps: (i) f(z)=
is analytic everywhere inside C. (ii)
∫
C
f (z* ) *
dz =
z* − z
∞
f (z* ) *
1
dz , which requires f(z)
2πi ∫C z * − z

∑  ∫
n =0

C

f ( z ′)
dz ′ ( z − z 0 ) n , which
n +1
( z′ − z0 )

only requires f(z) is analytic on C. Here we simply use (ii).
1
(2) Since z lies outside C2, *
=
z −z
z * − z0
1
−p
, q=
, |q|<1.
=
, p≡
1− q
z − z0
 z * − z0  z − z 0

( z − z0 )1 −
z − z0 

−1
 z * − z0
 q n +1 
 z * − z0
1
n
 , ⇒ p= 1 +
+ ... + 
=(1+q+…+q )+ 
1− q
z − z0
 z − z0
1− q 

⇒ 2πi⋅h(z)=
∫
C2
 p
f ( z * )
 z − z0
 *
 1
dz = 
 z − z0


1
+…+
( z * − z0 ) n f ( z * )dz *  +
n +1 ∫C
2
( z − z0 )

⇒ h(z)=
n +1
bm
∑ (z − z )
m=1
0
m
∫
C2
f ( z * )dz * +

1

 ( z − z ) n +1
0

+ Rn* ( z ) , where bm=
n
 
( z * − z0 ) n+1
.
  +
*
n
  ( z − z )( z − z0 )
1
( z − z0 ) 2
∫
C2
( z * − z0 ) f ( z * )dz *
( z * − z 0 ) n +1 f ( z * ) * 
dz  .
∫C2
(z − z* )

1
( z * − z0 ) m−1 f ( z * )dz * , and Rn* ( z ) =
2πi ∫C2
1
( z − z0 ) n+1 f ( z * ) *
dz . (We still need to prove bm exists, and Rn* ( z ) →0.)
2πi ( z − z0 ) n+1 ∫C2
( z − z* )
*
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By the ML-inequality and the same procedure in proving Taylor theorem eq’s (12.1-2), we
∞
1
bn
, where bn=
( z * − z 0 ) n −1 f ( z * )dz * .
n
∫
C
2
2πi
n =1 ( z − z0 )
derive lim Rn* ( z ) =0, ⇒ h(z)= ∑
n →∞
(3) By Cauchy’s integral theorem 4 [eq. (10.8)], an=
bn=
1
1
[" ]dz * =
[" ]dz ′ ,
∫
2πi C1
2πi ∫C
1
1
[" ]dz * =
[" ]dz ′ , for any contour C⊂ D.
∫
2πi C2
2πi ∫C
Eq. (13.1) remains valid if we continuously move C1 outward and C2 inward until they reach
some singularity. ⇒ ROC of Laurent series is the open annulus D: R2<|z-z0|<R1.
<Comment>
1) A function may have different power series because of different ROCs or different centers.
E.g.
1
= (i)
1− z
∞
∑z
n
, if |z|<1; (ii)
n =0
∞
−1
− 1 ∞ −n
=
⋅
z
=
−
z −( n +1) , if |z|>1; (iii)
∑
∑
−1
z (1 − z ) z n =0
n =0
−1
, if 0<|z−1|<∞.
z −1
∞
2) A power series only has one closed function form. E.g. f(z)= − ∑ z −( n +1) . The series will
n =0
diverge if we substitute z=
∞
1
1
into it to evaluate f   . ⇒ Deriving the closed form of
2
2
f(z) first. f(z)= −(z-1) ∑ ( z −1 ) n =
n =0
− z −1
1
1
=
, ⇒ f   =2.
−1
1− z
1− z
2
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Singularities and Zeros (EK 16.2)
Definitions
1) A function f(z) has a singularity at z=z0 if f(z) is not analytic at z=z0.
2) z=z0 is an isolated singularity if it has a neighborhood without any other singularity.
E.g. tan z has isolated singularities at z= ±
( )
E.g. tan z −1
±
2n + 1
π.
2
2
1
2n + 1
2
π , ⇒ z= ±
=±
=± ,
z
2
(2n + 1)π
π
is not continuous/analytic for
2
, …. For an arbitrarily small disk |z|<ε, you can always find infinitely many
3π
singularities within it, ⇒ z=0 is a non-isolated singularity (other sigularities are isolated).
3) If the Laurent series of f(z) centered at z=z0 has nonzero coefficient(s) up to m-th order
(bm≠0), ⇒ z0 is a pole of order m of f(z), which can be removed by multiplying (z−z0)m. It
is called a simple pole if m=1.
E.g.
1
1 z 7z 3
1
=
+ +
+... has a simple
,
by
long
division,
=
sin z z − ( z 3 / 3!) + ( z 5 / 5!) − ...
z 6 360
pole at z=0.
E.g. To find the order of pole of tan z at z=
π
2
, we change the variable: u=z−
π
2
: f(z)=
π
π
 1 u u3


−
− ... , which has a simple pole at u=0, i.e. z= .
=
−tan
u=
tan u + 
 − −
2
2

 u 3 45

4) If the Laurent series of f(z) centered at z=z0 has infinitely many nonzero coefficients, z0 is
an essential singularity of f(z).
∞
1
has an isolated essential singularity at z=0.
n
n =0 n! z
E.g. e1/z = ∑
( )
E.g. tan z −1 =
sin(1 / z ) 1
1
= + 3 +... has a non-isolated essential singularity at z=0.
cos(1 / z ) z 3z
5) If an analytic function f(z) has a property of f(z0)= f'(z0)=…= f
(n-1)
(z0)=0 (i.e. the first n
Taylor coefficients a0=a1=…=an-1=0), ⇒ z=z0 is a zero of order n of f(z).
E.g. sin z has simple zeros (n=1) at z=0, ±π, ±2π, …; sin2 z has second-order zeros at
these points.
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PDE & Complex Variables
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Properties
1) If f(z) is analytic except for a pole at z=z0, ⇒ |f(z)|→∞ as z→z0 in any manner.
2) Picard’s theorem: If f(z) is analytic except for an essential singularity at z=z0, ⇒ f(z) can
be equal to any complex number in an arbitrarily small neighborhood of z0.
E.g. f(z)=e1/z: (i) if z=x=(+∞→0+), f(z)→∞; (ii) if z= −x=(−∞→0-), f(z)→0; (iii) For any
cos θ − i sin θ 
given complex number c=c0⋅eiα, we can find some z=r⋅eiθ, s.t. f(z)= exp 
 =c,
r

⇒

 cosr θ
sin θ
= c0 , −
= α + 2 nπ  ;
e
r


⇒
r = [(ln c0 ) 2 + (α + 2nπ ) 2 ]
−
1
2
,
tanθ =
− (α + 2nπ )
. The solution z can be made arbitrarily close to 0 (r→0) by increasing n.
ln c0
3) If f(z) is analytic and has an nth-order zero at z0, ⇒
1
has an nth-order pole at z=z0.
f ( z)
4) The properties of f(z) at large |z| can be investigated by: set z=1/w, investigate
g(w)=f(1/w)=f(z) in the neighborhood of w=0.
E.g. ez has an isolated essential singularity at ∞, for e1/w has that at w=0.
Residue Integration (EK 16.3)
Evaluate contour integral by residue
To evaluate I= ∫ f ( z )dz for some arbitrary contour C:
C
1) If f(z) is analytic for every point on and within C, ⇒ I=0 by Cauchy’s integral theorem 1.
2) If f(z) has only one singularity at z=z0 inside C, its “closest” Laurent series with ROC:
∞
∞
bn
,⇒
n
n =1 ( z − z0 )
{0<|z−z0|<R} is: f(z)= ∑ a n ( z − z 0 ) n + ∑
n =0
I=2πi⋅ Res f ( z )
z = z0
(13.2)
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PDE & Complex Variables
P13-6
where Res f ( z ) =b1 is called the residue of f(z) at z0. [eq. (10.9) is a special case.]
z = z0
Proof: by
2πi, if m = −1
m
−
=
,
z
z
dz
(
)

0
∫C
0, otherwise
E.g. I= ∫ f ( z )dz , where f(z)=
C
∫
C
f ( z )dz = b1 ∫
C
1
dz =2πib1.
z − z0
1
1
, C: |z|= , in counterclockwise sense.
z (1 − z )
2
Since only one singularity z=0 lies inside C, ⇒ find the Laurent series centered at z=0: (1)
1
1
1
f(z)= +1+z2+…., for 0<|z|<1, ⇒ b1=1. (2) f(z)= −  2 + 3 + ... , for |z|>1, ⇒ b1=0. ⇒
z
z
z

Choose series-1: b1=1, and I=2πi.
<Comment>
We always choose the closest Laurent series to evaluate residue, even when part of the
integral path C falls outside its ROC. E.g. Let f(z)=
1
1
, C: z=  + i  + re iθ ,
z (1 − z )
4 
θ∈[0,2π], which passes through ROCs of the two Laurent series and only encloses one
singularity z=0 if
17 4 <r< 5 4 . ⇒ I=2πi for arbitrary
17 4 <r< 5 4 .
3) If f(z) has finitely many isolated singularities {zi; i=1−N} inside C, ⇒
N
I= 2πi ⋅ ∑ Res[ f ( z )]
k =1
z = zk
(13.3)
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N
Proof: By Cauchy’s integral theorem 4 [eq. (10.6)], ⇒ ∫ f ( z )dz = ∑ ∫ f ( z )dz , where
C
k =1
Ck
Ck is a circle that only encloses one singularity zk, and is separated from all the other
circles. By eq. (13.2),
∫
Ck
f ( z )dz =2π i⋅ Res[ f ( z )] , ⇒ ….
z = zk
How to evaluate the residue of a singularity
1) If z=z0 is (i) a pole of unknown order, (ii) essential singularity, (iii) non-isolated
singularity of f(z), ⇒ try to derive the (partial) Laurent series.
2) If z=z0 is a simple pole of f(z), ⇒
Res f ( z ) = lim ( z − z 0 ) f ( z )
z = z0
Proof: f(z)=
(13.4)
z → z0
b1
+ a0 + a1 ( z − z 0 ) + a 2 ( z − z 0 ) 2 + ... , (z−z0)f(z)=b1+a0(z−z0)+a1(z−z0)2
z − z0
+…, = b1+ (z−z0)[ a0+a1(z−z0)+…].
4) If z=z0 is a simple pole of
p( z )
, where p(z0)≠0, q(z) has a simple zero at z0, ⇒
q( z )
Res
z = z0
p( z ) p( z 0 )
=
q( z ) q ′( z 0 )
(13.5)
q ′′( z 0 )
(z−z0)2 +…, by eq. (13.4):
2!
( z − z 0 ) p( z )
p( z 0 )
p( z )
p( z )
Res
= lim ( z − z 0 )
= lim
=
.
z → z0
z = z 0 ( z − z )[q ′( z ) + ( z − z ) q ′′( z ) / 2!+...]
z = z0 q ( z )
q( z )
q ′( z 0 )
0
0
0
0
Proof: If z=z0 is a simple zero of q(z), q(z)= q ′( z 0 ) (z−z0)+
5) If z=z0 is mth order pole of f(z), ⇒
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Res f ( z ) =
z = z0
1
d m −1
[( z − z 0 ) m f ( z )]z = z0
(m − 1)! dz m −1
 bm
b 
Proof: f(z)= 
+ ... + 1  +
m
z − z0 
 ( z − z0 )
∞
∑a
n =0
n
(13.6)
( z − z 0 ) n , g(z)≡(z−z0)m f(z)= bm+ bm-1(z−z0)
∞
+ … +b1(z−z0)m-1 + ∑ a n ( z − z 0 ) n + m ; ⇒ b1 is the Taylor coefficient of the power (z−z0)m-1
n =0
of g(z), ⇒ b1=
( m −1)
g
( z0 )
.
(m − 1)!
E.g. Find the residues of f(z)=
lim zf ( z ) = lim( z + 2) −3 =
z→0
z →0
1
at all poles. (1) z=0 is a simple pole. Res f ( z ) =
z= 0
z ( z + 2) 3
1
. (2) z=−2 is a 3rd order pole, by eq. (13.6), Res f ( z ) =
z = −2
8
1 d2
[( z + 2) 3 f ( z )]z =−2 = − 1 .
2
8
2! dz
Note: Partial fractions reveal all the residues: f(z)=
− 1/ 2
− 1/ 8
− 1/ 4
1/ 8
+
.
+
+
2
z
z + 2 ( z + 2)
( z + 2) 3
(*) E.g. Find the residue of f(z)=(cot z) at z=π.
Ans: Find the order first. Let u=z−π, cot z =cot u, ⇒ {order of pole z=π of cot z}={order of
pole u=0 of cot u}. cot u=
cos u 1 − (u 2 / 2!) + (u 4 / 4!) − ... 1 u
= − +…, ⇒ u=0 is a simple
=
sin u u − (u 3 / 3!) + (u 5 / 5!) − ... u 3
pole, ⇒ z=π is a simple pole of cot z. By eq. (13.4), Res
z =π
cos( z ) cos(π )
=
=1.
sin( z ) sin ′(π )
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Appendix 13A − Residue at Infinity
If the contour C of ∫ f ( z )dz encloses a non-isolated singularity or a large number of
C
isolated singularities of f(z), evaluating the “interior” residues becomes inefficient. Instead,
we can derive
∫
C
f ( z )dz by evaluating “exterior” residues of isolated singularities z1, z2, , zN
outside C, and perhaps a residue at infinity:
Creating an infinite contour C∞. By eq. (13.3):
∫
C
f ( z )dz = ∫
∫
N
C∞
N
C∞
f ( z )dz − ∫ f ( z )dz = 2πi ⋅ ∑ Res[ f ( z )] , ⇒
f ( z )dz − 2πi ⋅ ∑ Res[ f ( z )] . To evaluate
k =1
z = zk
C
∫
C∞
k =1
z = zk
f ( z )dz , let C∞: z=reiθ, r→∞,
θ =[0,2π] (counterclockwise). By change of variable w=1/z, C∞ in the z-plane becomes C'∞ in
the w-plane: w≡ρeiφ=(1/r)e-iθ, ρ→0, φ=[0,-2π] (clockwise), f(z)=f(1/w), ⇒
∫
C∞′
f (1 / w)
− dw
=
w2
∫
-C∞′
∫
C∞
f ( z )dz =
f (1 / w)
dw , since − C ∞′ only encloses one point w=0 in a
w2
 f (1 / w) 
counterclockwise sense, =2πi⋅ Res 
,⇒
2
w=0
 w 
∫
C


 f (1 / w) 
[
]
f ( z )dz = 2πi Res 
Re
s
f
(
z
)
−

∑
2
w= 0
 w  zk outside C


(13A.1)
<Comment>
Res
w= 0
f (1 / w)
= Res f ( z ) . Proof: Let Res f ( z ) =b1, ⇒ the Laurent series of f(z) centered at
z =∞
z =∞
w2
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b1
1
+ a 0 + a1 ( z − ∞) + ... } ⇒. By change of variable: w= , f(z)= f(1/w)=
z−∞
z
a
b
a
a1
f (1 / w) 



= ... + 1 + 02 + 13 + ... , where b1 is the residue at
... + b1 w + a 0 + + ... , ⇒
2
w
w w
w
w




z=∞ as: f(z)={…+
w=0.
E.g. Evaluate
z 99e1 / z
∫ C z100 + 1 dz , C: |z|=3.
Ans: There are 101 singularities within C (z=0, and 100 points fall on the unit circle), and
w−99e w
ew
z 99e1 / z
none outside C, ⇒ use residue at infinity. Res 100
= Res
= Res
,
z =∞ z
+ 1 w=0 w2 ( w−100 + 1) w=0 w(1 + w100 )
ew
=1. ⇒
w→0 1 + w100
since w=0 is a simple pole, = lim
∫
C
f ( z )dz =2πi⋅ Res f ( z ) =2πi.
z =∞
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