21MAB102T-ADVANCED
CALCULUS AND COMPLEX
ANALYSIS
COMPLEX INTEGRATION
Dr. V. Poongothai
Assistant Professor, Department of Mathematics
SRM Institute of Science and Technology, Kattankulathur- 603 203.
1
INTRODUCTION
Let f (z) be a continuous function of the complex variable 𝑧 = z + iy defined at
every point of a curve ′𝐶′ whose end points are 𝐴 and 𝐵. Let us divide the curve 𝐶
into ′𝑛′ parts by points 𝐴 = 𝑝0 𝑧0 , 𝑝1 𝑧1 , 𝑝2 𝑧2 , … 𝑝𝑖 𝑧𝑖 = 𝐵. Let 𝛿𝑧𝑖 = 𝑧𝑖−1 −
𝑧𝑖 and let 𝜀𝑖 be a point on the arc 𝑝𝑖−1 − 𝑝𝑖 . Then the limit of the sum
σ𝑛𝑖=1 𝑓 𝜀𝑖 𝛿𝑧𝑖 𝑎𝑠 𝑛 → ∞ in such a way that each 𝛿𝑧𝑖 → 0, if it exits is called the
line integral of 𝒇(𝒛) along 𝑪 and is denoted by ‫𝑧𝑑 𝑧 𝑓 𝐶ׯ‬.
2
If 𝐶 is a closed curve, i.e if 𝑝0 and 𝑝𝑛 coincide the integral is called the
contour integral and is denoted by ‫𝑧𝑑)𝑧(𝑓 𝑐ׯ‬.
3
Complex Integration
The integral of a complex function is defined as the limit of a certain sum in the
same manner as integral of a real function along the 𝑥-axis is defined.
Multiple point
If a curve intersects itself at a point then the point is said to be a multiple point of
the curve.
Simple curve
A continuous curve which does not have a point of self-intersection is called a
simple curve. Simple curves are called Jordan curves.
4
Positively oriented simple closed curve
A simple closed curve 𝐶 encloses a region, if the region lies to the left of a person
when he travels along ′𝐶 ′ , then the curve ′𝐶 ′ is called positively oriented simple
closed curve.
5
Content Integral
An integral along a simple closed curve is called a contour integral.
Simply - connected region
A region 𝐷 is said to he Simply – Connected if, for every closed curve ′𝐶′ in 𝐷, 𝐶𝑖 is
wholly connected in D. Simply – Connectedness of a region is equivalent to the
absence of holes in it or to the situation in which every closed curve in 𝐷 can shrink
to a point, all the while being in 𝐷 itself.
Multiply - Connected region
A region which is not simply - connected is called multiply - connected region.
6
Evaluation of line integral
The evaluation of a line integral is reduced to the evaluation of two real line
integrals as follows:
Since 𝑧 = 𝑥 + 𝑖𝑦, 𝑑𝑧 = 𝑑𝑥 + 𝑖𝑑𝑦, 𝑓(𝑧) = 𝑢 + 𝑖𝑣
‫׬ = 𝑧𝑑)𝑧(𝑓 𝑐׬‬c (𝑢 + 𝑖𝑣)(𝑑𝑥 + 𝑖𝑑𝑦) = ‫ 𝑥𝑑𝑢( 𝑐׬‬− 𝑣𝑑𝑦) + 𝑖 ‫ 𝑥𝑑𝑣( 𝑐׬‬+ 𝑢𝑑𝑦)
7
Example 1
𝑑𝑧
Evaluate ‫ 𝐶׬‬′
where 𝑛 is an integer and 𝐶 is a circle
𝑧 −𝑧0 𝑛+1
z − 𝑧0 = 𝑟.
Solution:
Let 𝑧 − 𝑧0 = 𝑟𝑒 𝑖θ , so that 𝜃 varies from 0 to 2𝜋 as 𝑧 describes the circle 𝐶.
2𝜋 𝑟𝑒 𝑖θ 𝑖
𝑖 2𝜋 −𝑖𝑛𝜃
𝑙 = ‫׬‬0
𝑑𝜃 = 𝑛 ‫׬‬0 𝑒
𝑑𝜃
(𝑟𝑒 𝑖θ )𝑛+1
𝑟
2π
Case (i): when 𝑛 = 0, 𝑙 = 𝑖‫׬‬0 𝑑𝜃 = 2𝜋𝑖
Case (ii): when 𝑛 ≠ 0,
1 2𝜋
1
𝑙 = 𝑛 ‫׬‬0 (cos 𝑛𝜃 − 𝑖sin 𝑛𝜃)𝑑𝜃 = 𝑛 (sin 𝑛𝜃 + 𝑖cosn 𝜃)2𝜋
0 = 0.
𝑟
𝑛𝑟
8
Example 2
1+𝑖
Evaluate ‫׬‬0
i.
𝑥 − 𝑦 + 𝑖𝑥 2 𝑑𝑧
along the straight line between the limits (0,0) and 1,1
ii. over the path along the lines 𝑦 = 0, 𝑥 = 1
iii. over the path along the lines 𝑥 = 0 and 𝑦 = 1
iv. along the parabola 𝑦 2 = 𝑥.
Solution:
(i) Along the path 𝑂𝑃, 𝑦 = 𝑥, 𝑑𝑦 = 𝑑𝑥
1
∴ 𝐼 = ‫׬‬0
1
𝑥 − 𝑦 + 𝑖𝑥 2 (𝑑𝑥 + 𝑖𝑑𝑦) = ‫׬‬0
𝑥 − 𝑥 + 𝑖𝑥 2 (1 + 𝑖)𝑑𝑥
9
1 2
1
= (𝑖 − 1) ‫׬‬0 𝑥 𝑑𝑥 = (𝑖 − 1)
3
(ii) Along the path 𝑂𝐴𝑃
𝐼 = ‫ 𝑥 𝐴𝑂׬‬− 𝑦 + 𝑖𝑥 2 𝑑𝑧 + ‫ 𝑥 𝑃𝐴׬‬− 𝑦 + 𝑖𝑥 2 𝑑𝑧 = 𝑙1 + 𝐼2
(1)
Along the line 𝑂𝐴, 𝑦 = 0, 𝑑𝑦 = 0
10
1
T 2
3+2𝑖
𝐼1 = ‫׬‬0 𝑥𝑑𝑥 + 𝑖 ‫׬‬0 𝑥 𝑑𝑥 =
6
Along the line 𝐴𝑃, 𝑥 = 1, 𝑑𝑥 = 0
1
1
−2+𝑖
𝐼2 = ‫׬‬0 − 𝑦𝑑𝑦 + 𝑖 ‫׬‬0 (1 − 𝑦)𝑑𝑦 =
2
Equation (1) becomes 𝐼 =
3+2𝑖
𝑖−2
−3+5𝑖
+
=
6
2
6
(iii) Along the path 𝑂𝐵𝑃
𝐼 = along the line 𝑂𝐵 + along the line 𝐵𝑃 = 𝐼1 + 𝐼2
(2)
Along the line 𝑂𝐵, 𝑥 = 1, 𝑑𝑥 = 0
11
1
𝑖
𝐼2 = 𝑖 ‫׬‬0 − ydy = −
2
Along the line 𝐵𝑃, 𝑦 = 1, 𝑑𝑦 = 0
12
1
1 2
−1
𝑖
𝐼2 = ‫׬‬0 (𝑥 − 1)𝑑𝑥 + 𝑖 ‫׬‬0 𝑥 𝑑𝑥 = +
2
3
𝑖
Equation (2) becomes 𝐼 = − +
2
1
1
− +
2
3
−3−𝑖
=
6
(iv) Along the parabola 𝑦 2 = 𝑥 ⇒ 2𝑦𝑑𝑦 = 𝑑𝑥
1
∴ 𝐼 = ‫׬‬0
=
1
𝑥 − 𝑦 + 𝑖𝑥 2 𝑑𝑧 = ‫׬‬0
𝑦 2 − 𝑦 + 𝑖𝑦 + (2𝑦 + 𝑖)𝑑𝑦
1
𝑦4
2𝑦 3
𝑦6
𝑦3
𝑦2
𝑦5
−11
𝑖
−
+𝑖 +𝑖 −𝑖 −
=
+ .
2
3
3
3
2
5 0
30
6
13
Example 3
Evaluate ‫𝑧𝑑𝑧 𝑐׬‬
᪄ from 𝐴(0,0) to 𝐵(4,2) along the curve 𝐶 and 𝑧 = 𝑡 2 + 𝑖𝑡
Solution:
Let 𝑧᪄ = 𝑥 − 𝑖𝑦, 𝑧 = 𝑥 + 𝑖𝑦 = 𝑡 2 + 𝑖𝑡 ⇒ 𝑥 = t 2 , 𝑦 = t so that 𝑑𝑥 = 2𝑡𝑑𝑡, 𝑑𝑦 = 𝑑𝑡
and 𝑑𝑧 = 𝑑𝑥 + 𝑖𝑑𝑦 = 2𝑡𝑑𝑡 + 𝑖𝑑𝑡 = (2𝑡 + 𝑖)𝑑𝑡. Also 𝑥 = 0, 4 ⇒ 𝑡 = 0, 2 and 𝑦 =
0, 2 ⇒ 𝑡 = 0, 2
𝑡
Hence 𝐼 = ‫𝑧𝑑𝑧 𝐶׬‬
᪄ = ‫׬‬0
.
8
2
𝑡 − 𝑖𝑡 (2𝑡 + 𝑖)𝑑𝑡 = 10 − 𝑖
3
𝜕𝑢
𝜕𝑣
Note. The integral ‫ 𝑥𝑑𝑢 ׬‬+ 𝑣𝑑𝑦 is independent of the path if ƶ = .
𝜕𝑦
𝜕𝑥
14
Example 4
Find the value of the integral ‫ 𝑐׬‬− (𝑥 + 𝑦)𝑑𝑥 + 𝑥 2 𝑦𝑑𝑦,
(i) along 𝑦 = 𝑥 2 having (0,0), (3,9) as end points,
(ii) along 𝑦 = 3𝑥 between the same points. Do the values depend upon the path?
Solution:
Let 𝐼 = ‫ 𝑥( ׬‬+ 𝑦)𝑑𝑥 + 𝑥 2 𝑑𝑦 = ‫ 𝑥𝑑𝑢 ׬‬+ 𝑣𝑑𝑦
Here 𝑢 = 𝑥 + 𝑦 and 𝑣 = 𝑥 2 𝑦 and we have
(1)
𝜕𝑢
𝜕𝑣
𝜕
𝜕𝑣
= 1,
= 𝑥 2 so that 𝑢 ≠ .
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦
Hence the integral is dependent on path.
15
(i) Along 𝑦 = 𝑥 2 , 𝑑𝑦 = 2𝑥𝑑𝑥 and 𝑥 varies from 0 to 3.
Now equation (1) becomes
3
𝐼 = ‫׬‬0
=
2
2 2
𝑥 + 𝑥 𝑑𝑥 + 𝑥 𝑥
3
2𝑥𝑑𝑥 = ‫׬‬0
𝑥 + 𝑥 2 + 2𝑥 5 𝑑𝑥
3
𝑥2
𝑥3
𝑥6
9
27
729
+ +
= + +
= 256.5
2
3
3 0
2
3
3
(ii) Along 𝑦 = 3𝑥, 𝑑y = 3𝑑𝑥 and 𝑥 varies from 0 to 3.
Now equation (1) becomes
3
𝐼 = ‫׬‬0 (𝑥 + 3𝑥)𝑑𝑥 + 𝑥 2 (3𝑥)3𝑑𝑥 =
4 3
9𝑥
2𝑥 2 +
= 200.25
4 0
16
Example 5
Evaluate ‫ 𝐶׬‬3𝑦 2 𝑑𝑥 + 2𝑦𝑑𝑦 where 𝐶 is a circle 𝑥 2 + 𝑦 2 = 1 counter clockwise from
(1,0) to (0,1).
Solution:
Let 𝐼 = ‫ 𝐶׬‬3𝑦 2 𝑑𝑥 + 2𝑦𝑑𝑦.
Given 𝑥 2 + 𝑦 2 = 1.
Differentiate w. r. to 𝑥 and 𝑦,
we get 2𝑥𝑑𝑥 + 2𝑦𝑑𝑦 = 0, ⇒ 𝑦𝑑𝑦 + 𝑥𝑑𝑥 = 0 ⇒ 𝑦𝑑𝑦 = −𝑥𝑑𝑥.
0
∴ 𝐼 = ‫׬‬1 3 1 − 𝑥 2 𝑑𝑥 − 2𝑥𝑑𝑥 = 3𝑥 − 𝑥 3 − 𝑥 2 10 = −1.
17
Cauchy's theorem (or) Cauchy's Integral theorem
Statement:
If 𝑓(𝑧) is analytic and 𝑓 ′ (𝑧) is continuous at all points inside and on a simple closed
curve 𝐶, then ‫ 𝑓 ׯ‬′ (𝑧)𝑑𝑧 = 0.
Proof:
Let 𝑓 𝑧 = 𝑢 + 𝑖𝑣, 𝑧 = 𝑥 + 𝑖𝑦 and 𝑑𝑧 = 𝑑𝑥 + 𝑖𝑑y.
Then ‫ 𝑢 𝑐ׯ = 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬+ iv 𝑑𝑥 + 𝑖𝑑y = ‫ 𝑥𝑑𝑢(𝑐ׯ‬+ 𝑖𝑢𝑑𝑦 + 𝑖𝑣𝑑𝑥 − 𝑣𝑑𝑦)
= ‫ 𝑥𝑑𝑢 𝑐ׯ‬− 𝑣dy + 𝑖(𝑢𝑑𝑦 + 𝑣𝑑𝑥)
We know, by Green's theorem in a plane, ‫ 𝑥𝑑𝑢 𝑐ׯ‬+ 𝑣𝑑y = ‫׭‬
𝜕𝑢
𝜕𝑣
− +
𝜕𝑦
𝜕𝑥
(1)
𝑑𝑥𝑑𝑦
18
Equation (1) becomes
𝜕𝑢
𝜕𝑣
− 𝑑𝑥𝑑𝑦 − 𝑑𝑥𝑑𝑦
𝜕𝑦
𝜕𝑥
+𝑖
=‫׭‬
𝜕𝑢
𝜕𝑣
− −
𝜕𝑦
𝜕𝑥
𝑑𝑥𝑑𝑦 + 𝑖 ‫׭‬
𝜕𝑢
𝜕𝑣
−
𝜕𝑥
𝜕𝑦
𝑑𝑥𝑑𝑦
=‫׭‬
𝜕𝑢
𝜕𝑢
− +
𝜕𝑦
𝜕𝑦
𝑑𝑥𝑑𝑦 + 𝑖 ‫׭‬
𝜕𝑢
𝜕𝑢
−
𝜕𝑥
𝜕𝑥
𝑑𝑥𝑑𝑦 = 0,
‫׭ = 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬
𝜕𝑢
𝜕𝑣
𝑑𝑥𝑑𝑦 − 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑢
𝜕𝑣 𝜕𝑢
𝜕𝑣
by C-R equations = ,
=− .
𝜕𝑥
𝜕𝑦 𝜕𝑦
𝜕𝑥
That is ‫ = 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬0.
19
Cauchy's theorem for multiply connected region
Statement:
If 𝑓(𝑧) is analytic and 𝑓 ′ (𝑧) is continuous at all points in the region bounded by the
simple closed curves 𝐶𝑙 and 𝐶2 , then‫𝑧𝑑)𝑧(𝑓 𝑐ׯ = 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬.
1
2
Corollary:
If 𝑓(𝑧) is analytic and 𝑓 ′ (𝑧) is continuous at all points in the region between the
curves 𝐶𝑙 , 𝐶2 , 𝐶3 … . 𝐶𝑛 Which lies entirely within a simple closed curve 𝐶, then
‫ 𝑧𝑑)𝑧(𝑓 𝑐ׯ = 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬+ ‫ 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬+ ‫ 𝑧𝑑)𝑧(𝑓 𝑐ׯ‬+ ⋯ . + ‫𝑧𝑑)𝑧(𝑓 𝑐ׯ‬
1
2
3
𝑛
20
Cauchy's Fundamental Formula (OR) Cauchy's Integral Formula
Statement:
If 𝒇(𝑧) is analytic and 𝑓 ′ (𝑧) is continuous and if ′𝑎′ is any point inside ′𝐶 ′ then
𝑓(𝑎) =
1
𝑓(𝑧)
𝑑𝑧.
‫ׯ‬
𝑐
2𝜋𝑖
𝑧−𝑎
Proof:
The function 𝐹(𝑧) =
𝑓(𝑧)
𝑓(𝑧)
is analytic at all points except at 𝑧 = 𝑎. Since
is
𝑧−𝑎
𝑧−𝑎
analytic with ' 𝑎 ' as center and radius ' 𝑟 ' draw a small circle 𝐶𝑙 which lies entirely
within 𝐶. By Cauchy's theorem,
𝑓(𝑧)
𝑓(𝑧)
‫𝑧 𝑐ׯ‬−𝑎 𝑑𝑧 = ‫𝑧 𝑐ׯ‬−𝑎 𝑑𝑧
1
21
Let 𝑧 − 𝑎 = 𝑟𝑒 𝑖𝜃 ⇒ 𝑧 = 𝑎 + 𝑟𝑒 𝑖𝜃 ⇒ 𝑑𝑧 = 0 + 𝑟𝑖𝑒 𝑖𝜃 𝑑𝜃 = 𝑟𝑖𝑒 𝑖𝜃 𝑑𝜃.
Equation (1) becomes
𝑓 𝑎+𝑟𝑒 𝑖𝜃 𝑟𝑖𝑒 𝑖𝜃
𝑓(𝑧)
𝑖𝜃
𝑑𝜃
=
i
𝑓
𝑎
+
𝑟𝑒
‫𝑧 𝑐ׯ‬−𝑎 𝑑𝑧 = ‫𝑐ׯ‬
‫ׯ‬
𝑐
𝑟𝑒 𝑖𝜃
1
𝑑𝜃
As 𝑟 → 0, the circle 𝐶𝑙 shrinks to the point ' 𝑎 ', but 𝜃 = 0 to 2𝜋.
Equation (2) becomes
2𝜋
𝑓(𝑧)
‫𝑧 𝑐ׯ‬−𝑎 𝑑𝑧 = 𝑖 ‫׬‬0 𝑓(𝑎)𝑑𝜃 = 𝑖𝑓(𝑎)[𝜃]2𝜋
0 = 𝑖 𝑓(𝑎)2𝜋 = 2𝜋𝑖𝑓(𝑎)
⇒ 𝑓 𝑎 =
1
𝑓 𝑧
𝑑𝑧.
‫ׯ‬
2𝜋𝑖 𝑐 𝑧−𝑎
22
Corollary:
Differentiate (3) w. r. to 𝑎, we get
𝑓 ′ (𝑎) =
1
(−1)
(−1)
‫ׯ‬
2
𝐶
2𝜋𝑖
(𝑧−𝑎)
𝑓(𝑧)𝑑𝑧 ⇒ 𝑓 ′ (𝑎) =
1
𝑓(𝑧)𝑑𝑧
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)2
Differentiate again w. r. to 𝑎, we get
𝑓 ′′ (𝑎) =
1
(−2)
(−1)
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)3
𝑓(𝑧)𝑑𝑧 ⇒ 𝑓 ′′ (𝑎) =
2!
𝑓(𝑧)𝑑𝑧
.
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)3
Differentiate again w. r. to 𝑎, we get
𝑓 ′′′ 𝑎
2!
=
‫ׯ‬
2𝜋𝑖 𝐶
−3
(𝑧−𝑎)4
−1
2!×3
𝑓 𝑧 𝑑𝑧
𝑓 𝑧 𝑑𝑧 =
.
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)4
3!
𝑓(𝑧)𝑑𝑧
etc.
‫ׯ‬
4
𝐶
2𝜋𝑖
(𝑧−𝑎)
𝑛!
𝑓 𝑧 𝑑𝑧
=
.
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)𝑛+1
That is 𝑓 ′′′ (𝑎) =
In general 𝑓 𝑛 𝑎
23
Note.
We recall that a point 𝑧0 at which a function 𝑓(𝑧) is not analytic is known as a
singular point or a singularity of 𝑓(𝑧). To find the singularity of 𝑓(𝑧), equate the
denominator of 𝑓(𝑧) to zero and solve it for 𝑧. For example the singularity of
𝑧+3
𝑓(𝑧) =
is obtained as (𝑧 − 1)(𝑧 − 2) = 0
(𝑧−1)(𝑧−2)
⇒ 𝑧 − 1 = 0, 𝑧 − 2 = 0 ⇒ 𝑧 = 1, z = 2.
24
Example 1
𝑒 −𝑧
1
Evaluate ‫𝐶ׯ‬
𝑑𝑧 where 𝐶 is a circle (𝑖)|𝑧| = 2, (𝑖𝑖)|𝑧| = .
𝑧+1
2
Solution:
Singular points are obtained by putting 𝑧 + 1 = 0 ⇒ 𝑧 = −1.
(i) The singular point 𝑧 = −1 lies within the circle |𝑧| = 2. Then by Cauchy's
Integral formula
𝑓 𝑎
1
𝑓 𝑧
𝑓 𝑧
=
𝑑𝑧 ⇒ ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖 𝑓(𝑎)
‫ׯ‬
2𝜋𝑖 𝐶 𝑧−𝑎
𝑧−𝑎
𝑒 −𝑧
𝑒 −𝑧
we have ‫𝐶ׯ‬
𝑑𝑧 = ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖 𝑓(−1).
(𝑧+1)
[𝑧−(−1)]
25
Here 𝑓(𝑧) = 𝑒 −𝑧 , 𝑎 = −1 and 𝑓(−1) = 𝑒.
𝑒 −𝑧
Hence ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖𝑒.
(𝑧+1)
1
(ii) The singular point 𝑧 = −1 lies outside |𝑧| = .
2
𝑓(𝑧)
Then ‫𝐶ׯ‬
𝑑𝑧 = 0 by Cauchy's theorem.
𝑧+1
26
Example 2
cos 𝜋𝑧 2
Evaluate ‫𝐶ׯ‬
𝑑𝑧 where 𝐶 is a circle |𝑧| = 3.
(𝑧−1)(𝑧−2)
Solution:
Singular points are obtained by putting 𝑧 − 1 𝑧 − 2 = 0 ⇒ 𝑧 = 1, 2,
Both the singular points lies within |𝑧| = 3.
Now
1
𝐴
𝐵
=
+
(𝑧−1)(𝑧−2)
𝑧−1
𝑧−2
(1)
Multiply both sides of (1) by (𝑧 − 1)(𝑧 − 2), we get 1 = 𝐴(𝑧 − 2) + B(𝑧 − 1).
When 𝑧 = 2, 1 = 0 + 𝐵(2 − 1) ⇒ 𝐵 = 1.
When 𝑧 = 1, 1 = 𝐴(1 − 2) = −𝐴 ⇒ 𝐴 = −1.
27
Hence equation (1) becomes
1
−1
1
=
+
.
(𝑧−1)(𝑧−2)
(𝑧−1)
(𝑧−2)
cos 𝜋𝑧 2
1
Now ‫𝐶ׯ‬
𝑑𝑧 = ‫𝐶ׯ‬
(𝑧−1)(𝑧−2)
(𝑧−1)(𝑧−2)
cos 𝜋𝑧 2 𝑑𝑧
cos 𝜋𝑧 2
cos 𝜋𝑧 2
= − ‫𝐶ׯ‬
𝑑𝑧 + ‫𝐶ׯ‬
𝑑𝑧
(𝑧−1)
(𝑧−2)
= ‫𝐶ׯ‬
−1
1
+
(𝑧−1)
(𝑧−2)
= −𝐼1 + 𝐼2
𝑐𝑜𝑠 𝜋𝑧 2
Now 𝐼1 = ‫𝐶ׯ‬
𝑑𝑧
(𝑧−1)
cos 𝜋𝑧 2 𝑑𝑧
(2)
(3)
28
By Cauchy's Integral formula
𝑓 𝑎 =
1
𝑓 𝑧
𝑓 𝑧
𝑑𝑧
=
𝑑𝑧 = 2𝜋𝑖 𝑓(𝑎)
‫ׯ‬
‫ׯ‬
𝐶 𝑧−𝑎
2𝜋𝑖 𝐶 𝑧−𝑎
(4)
Comparing (3) and (4), we get 𝑎 = 𝐼,
𝑓(𝑧) = cos 𝜋𝑧 2 ⇒ 𝑓(1) = cos 𝜋 = −1.
Equation (3) implies that
cos 𝜋𝑧 2
𝐼1 = ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖 𝑓(1) = 2𝜋𝑖(−1) = −2𝜋𝑖
(𝑧−1)
29
cos 𝜋𝑧 2
Also 𝐼2 = ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖 𝑓 2
(𝑧−2)
(5)
We have 𝑓(𝑧) = cos 𝜋𝑧 2 ⇒ 𝑓(2) = cos 4𝜋 = 1.
Equation (5) becomes
cos 𝜋𝑧 2
𝐼2 = ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖 𝑓(2) = 2𝜋𝑖(1) = 2𝜋𝑖.
(𝑧−2)
cos 𝜋𝑧 2
Therefore ‫𝐶ׯ‬
𝑑𝑧 = −𝐼1 + 𝐼2 = −(−2𝜋𝑖) + 2𝜋𝑖 = 4𝜋𝑖.
(𝑧−1)(𝑧−2)
30
Example 3
𝑒 2𝑧
Using Cauchy's Integral formula evaluate ‫𝐶ׯ‬
𝑑𝑧 where 𝐶 is a circle |𝑧| = 2.
(𝑧+1)4
Solution:
Singular points are obtained by putting (𝑧 + 1)4 = 0 ⇒ 𝑧 = −1 which lies inside
the circle C. By Cauchy's Integral formula
3!
𝑓(𝑧)
𝑓(𝑧)
2𝜋𝑖 ′′′
′′′
𝑓 (𝑎) =
𝑑𝑧 ⇒ ‫𝐶ׯ‬
𝑑𝑧 =
𝑓 (𝑎).
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)4
(𝑧−𝑎)4
3!
𝑒 2𝑧
𝑓(𝑧)
2𝜋𝑖 ′′′
Hence ‫𝐶ׯ‬
𝑑𝑧 = ‫𝐶ׯ‬
𝑑𝑧 =
𝑓 (−1)
(𝑧+1)4
[𝑧−(−1)]4
3!
(1)
31
where 𝑓(𝑧) = 𝑒 2𝑧 , 𝑓 ′ (𝑧) = 2𝑒 2𝑧 , 𝑓 ′′ (𝑧) = 4𝑒 2𝑧 , 𝑓 ′′′ (𝑧) = 8𝑒 2𝑧
and 𝑓 ′′′ (−1) = 8𝑒 −2 =
8
. Now equation (1) becomes
𝑒2
𝑒 2𝑧
2𝜋𝑖 ′′′
2𝜋𝑖 8
𝑑𝑧
=
𝑓
(−1)
=
‫𝑧( ׬‬+1)4
3!
3! 𝑒 2
16𝜋𝑖𝑒 −2
8𝜋𝑖𝑒 −2
=
=
.
6
3
Example 4
3𝑧 2 +𝑧
Evaluate ‫ 𝐶ׯ‬2 𝑑𝑧 where 𝐶 is a circle |𝑧 − 1| = 1.
𝑧 −1
Solution:
Singular points are obtained by putting 𝑧 2 − 1 = 0 ⇒ 𝑧 2 = 1 ⇒ 𝑧 = 1, −1. The
singular point 𝑧 = 1 lies within the circle |𝑧 − 1| = 1 but 𝑧 = −1 lies outside the
circle |𝑧 − 1| = 1.
32
1
1
𝐴
𝐵
Now 2 =
=
+
𝑧 −1
(𝑧+1)(𝑧−1)
𝑧+1
𝑧−1
(1)
Multiply both sides of (1) by (𝑧 − 1)(𝑧 + 1), we get 1 = 𝐴 𝑧 − 1 + 𝐵(𝑧 + 1).
When 𝑧 = 1,
When 𝑧 = −1,
1 = 0 + 2𝐵 ⇒ 𝐵 = 1/2
1 = 𝐴(−2) ⇒ 𝐴 = −1/2.
Hence equation (1) becomes
1
𝑧 2 −1
1
1
1
=
=−
+
.
(𝑧+1)(𝑧−1)
2(𝑧+1)
2(𝑧−1)
3𝑧 2 +𝑧
1
‫ 𝑧 𝐶ׯ‬2 −1 𝑑𝑧 = ‫ 𝐶ׯ‬− 2(𝑧+1)
3𝑧 2 + 𝑧 𝑑𝑧 +
1
‫ 𝐶ׯ‬2(𝑧−1)
3𝑧 2 + 𝑧 𝑑𝑧
33
3𝑧 2 +𝑧
3𝑧 2 +𝑧
1
1
= − ‫𝐶ׯ‬
𝑑𝑧 + ‫𝐶ׯ‬
𝑑𝑧
2
(𝑧+1)
2
(𝑧−1)
1
= − 2𝜋𝑖 𝑓 −1
2
1
+ 2𝜋𝑖 𝑓(1), where 𝑓(𝑧) =
2
3𝑧 2 + 𝑧
= −𝜋𝑖 × 0 + 𝜋𝑖 × 4 = 4𝜋𝑖. where 𝑓(1) = 4.
Hence ‫𝐶ׯ‬
3𝑧 2 +𝑧
𝑧 2 −1
𝑑𝑧 = 4𝜋𝑖.
34
Example 5
𝑧𝑒 2𝑧
Evaluate ‫𝐶ׯ‬
dz where 𝐶 is a circle |𝑧 + 𝑖| = 2.
(𝑧−1)3
Solution:
Singular points are obtained by putting (𝑧 − 1)3 = 0 ⇒ 𝑧 = 1 is a singular point lies
inside the circle |𝑧 + 𝑖| = 2.
35
By Cauchy's Integral formula 𝑓 ′′′ (𝑎) =
2!
𝑓(𝑧)𝑑𝑧
.
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)3
𝑧𝑒 2𝑧
𝑓(𝑧)
2𝜋𝑖 ′′
‫𝑧( 𝐶ׯ‬−1)3 𝑑𝑧 = ‫𝑧( 𝐶ׯ‬−1)3 𝑑𝑧 = 2! 𝑓 (1), where 𝑓(𝑧) = 𝑧𝑒 2𝑧
Let 𝑓 𝑧 = 𝑧𝑒 2𝑧 , 𝑓 ′ 𝑧 = 2𝑧𝑒 2𝑧 + 𝑒 2𝑧 ,
𝑓 ′′ (𝑧) = 2 2𝑧𝑒 2𝑧 + 𝑒 2𝑧 + 2𝑒 2𝑧
and 𝑓 ′′ (−1) = 2 2𝑒 2 + 𝑒 2 + 2𝑒 2 = 8𝑒 2 .
𝑧𝑒 2𝑧
2𝜋𝑖 ′′
Equation (1) becomes ‫𝐶ׯ‬
=
𝑓 1
(𝑧−1)3
2!
2𝜋𝑖
=
2
8𝑒 2 = 8𝜋𝑖𝑒 2 .
36
Example 6
𝑑𝑧
over the circle |𝑧 − 2| = 1.
𝑧 −2𝑧
Evaluate ‫ 𝐶ׯ‬2
Solution:
Singular points are obtained by putting 𝑧 2 − 2𝑧 = 0 ⇒ 𝑧 𝑧 − 2 = 0 ⇒ 𝑧 = 0, 2
and |𝑧 − 2| = 1 is a circle with center (2,0) and radius 1. Only the singular point
𝑧 = 2 lies within the circle |𝑧 − 2| = 1. Hence by Cauchy's integral formula
𝑑𝑧
𝐼 = ‫ 𝐶ׯ‬2
𝑧 −2𝑧
1
𝑑𝑧
𝑑𝑧
𝑓(𝑧)𝑑𝑧
𝑧
= ‫𝐶ׯ‬
= ‫𝐶ׯ‬
= ‫𝐶ׯ‬
= 2𝜋𝑖 𝑓(2),
𝑧(𝑧−2)
(𝑧−2)
(𝑧−2)
1
1
where 𝑓(𝑧) = , 𝑓(2) =
𝑧
2
= 2𝜋𝑖
1
2
= 𝜋𝑖.
37
Example 7
Evaluate the integral ‫𝐶ׯ‬
cos 𝑧𝑑𝑧
where 𝐶 is an ellipse 9𝑥 2 + 4𝑦 2 = 1.
𝑧
Solution:
𝑥2
𝑦2
Here 𝑧 = 0 is a singular point which lies within an ellipse
+
= 1.
1/9
1/4
Hence by Cauchy's integral formula
cos 𝑧
𝑓(𝑧)
𝐼 = ‫𝐶ׯ‬
𝑑𝑧 = ‫𝐶ׯ‬
𝑑𝑧 = 2𝜋𝑖 𝑓(0) = 2𝜋𝑖. 1 = 2𝜋𝑖, where 𝑓(𝑧) = cos z.
𝑧
𝑧
38
Example 8
𝑧 2 +1
Evaluate ‫ 𝐶ׯ‬2 𝑑𝑧 where 𝐶 is a circle with unit radius and centre 1.
𝑧 −1
Solution:
Singular points are 𝑧 2 − 1 = 0 ⇒ 𝑧 = ±1, hence only 𝑧 = 1 lies within the circle.
Hence by Cauchy's integral formula
𝑧2 +1
𝑧+1
𝑧 2 +1
𝑧 2 +1
𝑓(𝑧)
𝐼 = ‫ 𝐶ׯ‬2 𝑑𝑧 = ‫𝐶ׯ‬
𝑑𝑧 = ‫𝐶ׯ‬
𝑑𝑧 = ‫𝐶ׯ‬
𝑑𝑧
𝑧 −1
(𝑧−1)(𝑧+1)
(𝑧−1)
(𝑧−1)
𝑧 2 +1
= 2𝜋𝑖 𝑓(1) = 2𝜋𝑖, where 𝑓(𝑧) =
and 𝑓(1) = 1.
𝑧+1
39
Example 9
𝑑𝑧
Evaluate ‫ 𝐶ׯ‬3
where 𝐶 is a circle |𝑧| = 2.
𝑧 (𝑧+4)
Solution:
Singular points are obtained by putting 𝑧 3 (𝑧 + 4) = 0 ⇒ 𝑧 = 0 and 𝑧 = −4 of
which only 𝑧 = 0 lies inside the given circle. By Cauchy's integral formula
1
(𝑧+4)
𝐶 𝑧3
𝑑𝑧
=‫ׯ‬
𝑧 (𝑧+4)
I = ‫ 𝐶ׯ‬3
𝑑𝑧 = ‫𝐶ׯ‬
𝑓(𝑧)
1
𝑑𝑧,
where
𝑓(𝑧)
=
(𝑧−0)3
𝑧+4
2𝜋𝑖 ′′
1
2
′
′′
=
𝑓 (0), where 𝑓 (𝑧) = −
, 𝑓 (𝑧) =
2!
(𝑧+4)2
(𝑧+4)3
= 𝜋𝑖
1
32
, where 𝑓 ′′ 0
2
2
1
𝜋𝑖
=
= = = .
(0+4)3
64
32
32
40
TAYLOR SERIES AND LAURENT'S SERIES
Now, we develop Taylor's series and Laurent's series to expand an analytic function
𝑓(𝑧) in powers in 𝑧.
Taylor's theorem (Taylor's series)
A function 𝑓(𝑧) is analytic at all points inside a circle 𝐶, with its center at the point
'𝑎' and radius 𝑅, we can expand
(𝑧 − 𝑎) ′
(𝑧 − 𝑎)2 ′′
(𝑧 − 𝑎)3 ′′′
𝑓 𝑧 = 𝑓(𝑎) +
𝑓 (𝑎) +
𝑓 (𝑎) +
𝑓 (𝑎)
1!
2!
3!
(𝑧 − 𝑎)𝑛 𝑛
+⋯+
𝑓 (𝑎) + ⋯ ∞
𝑛!
(𝑧−𝑎)
= σ∞
𝑛=1
𝑛!
𝑛
𝑓 𝑛 (𝑎), where 𝑓 𝑛 (𝑎) denotes the nth derivative.
41
Corollary 1
Putting 𝑎 = 0, we get
𝑧 ′
𝑧 2 ′′
𝑧 3 ′′′
𝑧𝑛 𝑛
𝑓(𝑧) = 𝑓(𝑎) + 𝑓 (𝑎) + 𝑓 (𝑎) + 𝑓 (𝑎) + ⋯ + 𝑓 (𝑎) + ⋯ ∞.
1!
2!
3!
𝑛!
Corollary 2
Putting 𝑧 − 𝑎 = ℎ ⇒ 𝑧 = 𝑎 + ℎ, we get
ℎ ′
ℎ2 ′′
ℎ3 ′′′
ℎ𝑛 𝑛
𝑓(𝑎 + ℎ) = 𝑓(𝑎) + 𝑓 (𝑎) + 𝑓 (𝑎) + 𝑓 (𝑎) + ⋯ + 𝑓 (𝑎) + ⋯ ∞
1!
2!
3!
𝑛!
42
Laurent's theorem (Laurent's series)
If 𝑓(𝑧) is analytic on two concentric circles 𝐶1 and 𝐶2 of radii 𝑟1 and 𝑟2 𝑟1 > 𝑟2
with center at ′𝑎′ and also on the annular region 𝑅 bounded by 𝐶1 and 𝐶2 then for all
𝑧 in 𝑅.
∞
𝑏𝑛
𝑛 = 𝐼1 + 𝐼2
(𝑧−𝑎)
𝑛=1
𝑛
𝑓(𝑧) = σ∞
𝑧=0 𝑎𝑛 (𝑧 − 𝑎) + ෍
𝐼1 called the regular part (contain positive powers) and 𝐼2 is called the principal part
(contain
negative
powers)
of
𝑓(𝑧)
and
𝑎𝑛 =
1
𝑓(𝑧)
𝑑𝑧, 𝑏𝑛 =
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)𝑛+1
1
𝑓(𝑧)
𝑑𝑧 both the integrals being taken anticlockwise direction.
‫ׯ‬
2𝜋𝑖 𝐶 (𝑧−𝑎)1−𝑛
43
Note. Note the two formulae:
(1 + 𝑥)−1 = 1 − 𝑥 + 𝑥 2 − 𝑥 3 + ⋯
(1 − 𝑥)−1 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯
(1 + 𝑥)−1 = 1 − 2𝑥 + 3𝑥 2 …
Example 1
Obtain the expansion of 𝑓(𝑧) =
𝑧−1
in Taylor's series in powers of (𝑧 − 1) and give
𝑧2
the region of validity and Laurent's series for 𝑧 − 1 > 1.
Solution:
(i) Let 𝑓(𝑧) =
𝑧−1
1
1
=
−
. Now we want 𝑓(𝑧) as a series in powers of 𝑧 − 1. The
𝑧2
𝑧
𝑧2
function is not analytic at 𝑧 = 0.
44
The
region
should
Let 𝑢 = 𝑧 − 1 ⇒ 𝑧 = 𝑢 + 1
𝑓(𝑧) =
not
contain
𝑧=0
in
|𝑧 − 1| > 1
.
∴ |𝑢| < 1
𝑧−1
𝑢
−2
=
=
𝑢(1
+
𝑢)
𝑧2
(𝑢+1)2
= 𝑢 1 − 2𝑢 + 3𝑢2 − 4𝑢3 + ⋯ = 𝑢 − 2𝑢2 + 3𝑢3 − ⋯
= (𝑧 − 1) − 2(𝑧 − 1)2 + 3(𝑧 − 1)3 − ⋯ is a Taylor series valid for |𝑧 − 1| > 1.
(ii) Let 𝑢 = 𝑧 − 1
∴ 𝑧 = 𝑢 + 1,
𝑢 > 1 𝑖. 𝑒 1 < 𝑢 ,
𝑧−1
𝑢
𝑢
1
Now 𝑓(𝑧) = 2 =
=
=
1 2
𝑧
(𝑢+1)2
𝑢
2
𝑢 1+
1
<1
|𝑢|
1 −2
1
2
3
1+
= − 2+ 3−⋯
𝑢
𝑢
𝑢
𝑢
𝑢
1
2
3
=
−
+
− ⋯ valid expansion for |𝑧 − 1| < 1.
𝑧−1
(𝑧−1)2
(𝑧−1)3
45
Example 2
2
Expand 𝑧𝑒 −𝑧 in Laurent's series about 𝑧 = 0.
Solution:
2
−𝑧
Let 𝑧𝑒
=𝑧
𝑧2
𝑧4
𝑧6
1− + − +⋯
1!
2!
3!
𝑧3
𝑧5
𝑧7
=𝑧− + − +⋯
1!
2!
3!
Here this Laurent series does not contain negative powers of 𝑧.
46
Example 3
Expand
1−cos 𝑧
in Laurent's series about 𝑧 = 0.
𝑧
Solution:
1−cos 𝑧
1
Let
=
𝑧
𝑧
1−
𝑧2
𝑧4
1− + −⋯
2!
4!
𝑧
𝑧3
𝑧5
= − + −⋯
2!
4!
6!
though 𝑧 = 0 appears to be a singularity if (1 − cos 𝑧)/𝑧, Laurent series of (1 −
cos 𝑧)/𝑧 at 𝑧 = 0 does not contain -ve powers of 𝑧.
47
Example 4
Find the residue at 𝑧 = 0 of (i) 𝑓(𝑧) = 𝑒1/2 (ii) 𝑓(𝑧) = sin 𝑧/𝑧 4 .
Solution:
1
1
(i) Here 𝑓(𝑧) = 1 + + 2 + ⋯
𝑧
𝑧 /2!
Residue of 𝑓(𝑧) at 𝑧 = 0 = coefficient of 1/𝑧 in the Laurent's expansion = 1.
1
(ii) Let 𝑓(𝑧) = 4
𝑧
𝑧3
𝑧5
𝑧− + −⋯
3!
5!
1
𝑧
= 3−
1
𝑧
+ −⋯
3!𝑧
5!
Residue of 𝑓(𝑧) at 𝑧 = 0 = coefficient of 1/𝑧 = −1/6.
48
Example 5
sin 𝑧
Expand
about 𝑧 = 𝜋.
𝑧−𝜋
Solution:
Put 𝑧 − 𝜋 = 𝑡. Then
sin 𝑧
sin(𝜋+𝑡)
sin 𝑡
=
=−
𝑧−𝜋
𝑡
𝑡
−1
=
𝑡
𝑡3
𝑡5
1− + −⋯
3!
5!
(𝑧−𝜋)2
(𝑧−𝜋)4
= −1 +
−
+⋯
3!
5!
49
Example 6
Expand 𝑓(𝑧) =
𝑧
about 𝑧 = −2.
(𝑧+1)(𝑧+2)
Solution:
Put 𝑧 + 2 = 𝑡 ⇒ 𝑧 = 𝑡 − 2.
Then 𝑓(𝑧) =
𝑡−2
(2−𝑡)
=
(1 − 𝑡)−1
𝑡(𝑡−1)
𝑡
1
=
𝑡
2 + 𝑡 + 𝑡2 + ⋯
2
𝑡
= + 1 + 𝑡 + 𝑡2 + ⋯
=
2
+ 1 + (𝑧 + 2) + (𝑧 + 2)2 + ⋯.
𝑧+2
50
Example 7
1+𝑧
Expand log
at 𝑧 = 0 using Taylor's series.
1−𝑧
Solution:
Let 𝑓(𝑧) = log
Then 𝑓 ′ (𝑧) =
1+𝑧
1−𝑧
= log(1 + 𝑧) − log(1 − 𝑧) ⇒ 𝑓(0) = 0.
1
1
−1
1
′′ (0) = 0 etc.
+
⇒ 𝑓 ′ (0) = 2, 𝑓 ′′ (𝑧) =
+
⇒
𝑓
1+𝑧
1−𝑧
(1+𝑧)2
(1−𝑧)2
The Taylor series of 𝑓(𝑧) about the point 𝑧 = 𝑎 is
(𝑧−𝑎)
′
𝑓(𝑧) = 𝑓(𝑎) + (𝑧 − 𝑎)𝑓 (𝑎) +
2!
2
𝑓 ′′ (𝑎) + ⋯
The Taylor series of 𝑓(𝑧) about the point 𝑧 = 0 is
𝑓 𝑧 =𝑓 0
+ 𝑧𝑓 ′ 0
𝑧 2 ′′
+ 𝑓 0
2!
+ ⋯ = 2𝑧 +
4 3
𝑧 +⋯
3!
51
Example 8
1
Expand 𝑓(𝑧) =
as Laurent's series in powers valid in |𝑧| <∣ and |𝑧| > 1.
𝑧(𝑧−1)
Solution:
1
1
1
1
=− −
= − − (1 − 𝑧)−1
𝑧(𝑧−1)
𝑧
1−𝑧
𝑧
1 𝑛
(i) Let 𝑓(𝑧) = −
− σ∞
𝑧 𝑛 is a valid expansion for |𝑧| < 1.
0
𝑧
1
1
1
1
(ii) Let |𝑧| > 1 ⇒ < 1. Then
=− +
1
|𝑧|
𝑧(𝑧−1)
𝑧
𝑧 1−
Let
𝑧
1 −1
1−
𝑧
=
1
−
𝑧
1
+
𝑧
=
1
−
𝑧
1
+ σ∞
𝑧 0
1 𝑛
is a valid expansion for |𝑧| > 1.
𝑧
52
Example 9
1
If 𝑓(𝑧) =
about 𝑧 = 0 valid in (i) |𝑧| < 1, (ii) 1 < |𝑧 ∣< 2, and (iii) |𝑧| > 2.
(𝑧−1)(𝑧−2)
Solution:
Let 𝑓(𝑧) =
(i) In
1
1
1
=−
+
.
(𝑧−1)(𝑧−2)
𝑧−1
𝑧−2
1
1
|𝑧| < 1, 𝑓(𝑧) =
−
𝑧
1−𝑧
2 1−
=
2
1 + 𝑧 + 𝑧2 + ⋯
1 ∞
− σ0
2
𝑧 𝑛
2
is a valid
expansion for |𝑧| < 1.
1
|𝑧|
(ii) Let 1 < |𝑧 ∣< 2, ⇒ < 1,
< 1.
𝑧
2
1
1
Then 𝑓(𝑧) = −
+
.
𝑧−1
𝑧−2
53
𝑓 𝑧
1
1
1
=−
=−
1 +
𝑧
𝑧
𝑧 1−𝑧
2 2−1
1 −1
1
1−
−
𝑧
2
𝑧 −1
1−
2
1 ∞
1 𝑛
1 ∞
𝑧 𝑛
= − σ𝑛=0
− σ𝑛=0
is a valid expansion for 1 <
𝑧
𝑧
2
2
𝑧 < 2.
2
(iii) Let |𝑧| > 2 ⇒ < 1.
|𝑧|
1
1
1
Then 𝑓(𝑧) = −
+
=−
1
𝑧−1
𝑧−2
𝑧 1−
𝑧
1
=−
𝑧
1 −1
1
1−
+
𝑧
𝑧
1 ∞
= − σ𝑛=0
𝑧
+
1
2
𝑧
𝑧 1−
2 −1
1−
𝑧
1 𝑛
1 ∞
+ σ𝑛=0
𝑧
𝑧
2 𝑛
is a valid expansion for |𝑧| > 2.
𝑧
54
Example 10
1
Expand 𝑓(𝑧) =
in the region 0 < |𝑧 − 1| < 1.
(𝑧−1)(𝑧−2)
Solution:
Let 𝑓(𝑧) =
1
1
1
=
−
(𝑧−1)(𝑧−2)
𝑧−2
𝑧−1
1
1
=
−
[(𝑧−1)−1]
𝑧−1
= −[1 − (𝑧 − 1)]−1 − (𝑧 − 1)−1
= −(𝑧 − 1)−1 + 1 + (𝑧 − 1) + (𝑧 − 1)2 + ⋯
55
Example 11
4𝑧+3
Represent the function
in Laurent series, (i) within |𝑧| = 2, (ii) in the
𝑧(𝑧−3)(𝑧+2)
annular region between |𝑧| = 2 and |𝑧| = 3 and (iii) exterior to |𝑧| = 3.
Solution:
4𝑧+3
𝐴
𝐵
𝐶
Let
= +
+
.
𝑧(𝑧−3)(𝑧+2)
𝑧
𝑧−3
𝑧+2
⇒ 4𝑧 + 3 = 𝐴(𝑧 − 3)(𝑧 + 2) + 𝐵𝑧(𝑧 + 2) + 𝐶𝑧(𝑧 − 3)
(1)
Put 𝑧 = 0 in (1), we get 3 = 𝐴(6) ⇒ 𝐴 = 1/2.
Put 𝑧 = −2 in (1), we get −5 = 𝐶(−2)(−5) ⇒ 𝐶 = −1/2 and put 𝑧 = 3 in (1), we
get 15 = 𝐵(3) ⇒ 𝐵 = 5.
56
4𝑧+3
1
5
1
Hence
= +
−
.
𝑧(𝑧−3)(𝑧+2)
2𝑧
𝑧−3
2(𝑧+2)
|𝑧|
|𝑧|
(i) With in |𝑧| = 2 ⇒ |𝑧| < 2 ⇒ < 1 and < 1
2
3
4𝑧+3
Now
𝑧 𝑧−3 𝑧+2
1
5
1
= +
−
2𝑧
𝑧−3
2 𝑧+2
𝑧 −1
5
=
+
𝑧
2
−3 1−
3
𝑧 −1
5
=
−
2
3
−
1
4 1+
𝑧
2
𝑧 −1
1
1−
−
3
4
𝑧 −1
1+
2
𝑧 −1
5
𝑧 𝑛
1
𝑧 𝑛
𝑛
=
− σ
− σ(−1)
2
3
3
4
2
57
(ii) In the annular region between |𝑧| = 2 and |𝑧| = 3
⇒2< 𝑧
2
|𝑧|
and |𝑧| < 3 ⇒ < 1 and < 1.
|𝑧|
3
4𝑧+3
Now
𝑧 𝑧−3 𝑧+2
1
5
= +
3
2𝑧
𝑧 1−
𝑧
𝑧 −1
5
=
+
2
𝑧
−
1
𝑧
4 1+2
3 −1
1
1−
−
𝑧
4
𝑧 −1
1+
2
𝑧 −1
5
𝑧 𝑛
1
𝑧 𝑛
𝑛
=
− σ
− σ(−1)
2
3
3
4
2
58
3
(iii) Exterior to |𝑧| = 3 ⇒ |𝑧| > 3 ⇒ < 1.
|𝑧|
4𝑧+3
1
5
1
Now
= +
−
𝑧(𝑧−3)(𝑧+2)
2𝑧
𝑧−3
2(𝑧+2)
𝑧 −1
5
=
−
3
2
𝑧 1−
𝑧
𝑧 −1
5
=
−
2
𝑧
−
1
2
2𝑧 1+𝑧
3 −1
1
1−
−
𝑧
2𝑧
𝑧 −1
5
=
− σ
2
𝑧
2 −1
1+
𝑧
3 𝑛
1
2 𝑛
𝑛
− σ (−1)
.
𝑧
2𝑧
𝑧
59