SERIES REPRESENTATIONS IN COMPLEX ANALYSIS -
TAYLOR AND LAURENT SERIES
1 INTRODUCTION TO SERIES REPRESENTATIONS
In complex analysis, series representations play a crucial role in understanding and
manipulating complex functions. We’ll explore two fundamental types of series: Taylor series
and Laurent series. Through exercises and discussions, we’ll delve into their properties,
applications, and relationships.
2 EXERCISE 1: TAYLOR SERIES BASICS
Consider the function 𝑓(𝑧)=𝑒𝑧.
1. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
2. What is the radius of convergence of this series?
3. Explain why this series converges for all complex numbers.
Solution:
1. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
2. The radius of convergence is infinite (𝑅=∞).
3. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
3 EXERCISE 2: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
1. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
2. Use this approximation to estimate sin(0.1).
3. Compare your approximation with the actual value of sin(0.1).
Solution:
1. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
2. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
3. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
4 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
1. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
2. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
3. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
1. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
2. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
3. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
5 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
6 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
1. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
2. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
3. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
1. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
2. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
3. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
4. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
7 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
1. Use the Taylor series method to find the solution 𝑦(𝑥).
2. Identify the resulting function and its Taylor series.
3. How does this relate to the uniqueness theorem for differential equations?
Solution:
1. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
2. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
3. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
8 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
1. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
2. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
3. Verify your result using the residue theorem.
Solution:
1. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
2. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
3. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
4. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
9 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
1. Every convergent power series represents an analytic function within its disk of
convergence.
2. The derivatives of an analytic function can be easily computed using its power series
representation.
3. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
4. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
5. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
6. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
7. What is the radius of convergence of this series?
8. Explain why this series converges for all complex numbers.
Solution:
9. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
10. The radius of convergence is infinite (𝑅=∞).
11. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
10 EXERCISE 9: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
12. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
13. Use this approximation to estimate sin(0.1).
14. Compare your approximation with the actual value of sin(0.1).
Solution:
15. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
16. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
17. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
11 EXERCISE 10: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
18. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
19. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
20. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
21. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
22. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
23. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
12 EXERCISE 11: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
13 EXERCISE 12: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
24. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
25. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
26. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
27. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
28. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
29. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
30. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
14 EXERCISE 13: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
31. Use the Taylor series method to find the solution 𝑦(𝑥).
32. Identify the resulting function and its Taylor series.
33. How does this relate to the uniqueness theorem for differential equations?
Solution:
34. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
35. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
36. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
15 EXERCISE 14: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
37. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
38. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
39. Verify your result using the residue theorem.
Solution:
40. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
41. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
42. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
43. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
16 EXERCISE 15: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
44. Every convergent power series represents an analytic function within its disk of
convergence.
45. The derivatives of an analytic function can be easily computed using its power series
representation.
46. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
47. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
48. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
49. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
50. What is the radius of convergence of this series?
51. Explain why this series converges for all complex numbers.
Solution:
52. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
53. The radius of convergence is infinite (𝑅=∞).
54. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
17 EXERCISE 16: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
55. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
56. Use this approximation to estimate sin(0.1).
57. Compare your approximation with the actual value of sin(0.1).
Solution:
58. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
59. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
60. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
18 EXERCISE 17: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
61. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
62. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
63. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
64. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
65. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
66. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
19 EXERCISE 18: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
20 EXERCISE 19: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
67. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
68. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
69. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
70. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
71. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
72. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
73. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
21 EXERCISE 20: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
74. Use the Taylor series method to find the solution 𝑦(𝑥).
75. Identify the resulting function and its Taylor series.
76. How does this relate to the uniqueness theorem for differential equations?
Solution:
77. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
78. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
79. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
22 EXERCISE 21: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
80. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
81. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
82. Verify your result using the residue theorem.
Solution:
83. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
84. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
85. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
86. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
23 EXERCISE 22: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
87. Every convergent power series represents an analytic function within its disk of
convergence.
88. The derivatives of an analytic function can be easily computed using its power series
representation.
89. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
90. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
91. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
92. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
93. What is the radius of convergence of this series?
94. Explain why this series converges for all complex numbers.
Solution:
95. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
96. The radius of convergence is infinite (𝑅=∞).
97. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
24 EXERCISE 23: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
98. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
99. Use this approximation to estimate sin(0.1).
100. Compare your approximation with the actual value of sin(0.1).
Solution:
101. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
102. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
103. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
25 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
104. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
105. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
106. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
107. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
108. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
109. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
26 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
27 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
110. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
111. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
112. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
113. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
114. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
115. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
116. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
28 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
117. Use the Taylor series method to find the solution 𝑦(𝑥).
118. Identify the resulting function and its Taylor series.
119. How does this relate to the uniqueness theorem for differential equations?
Solution:
120. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
121. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
122. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
29 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
123. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
124. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
125. Verify your result using the residue theorem.
Solution:
126. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
127. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
128. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
129. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
30 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
130. Every convergent power series represents an analytic function within its disk of
convergence.
131. The derivatives of an analytic function can be easily computed using its power series
representation.
132. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
133. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
134. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
135. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
136. What is the radius of convergence of this series?
137. Explain why this series converges for all complex numbers.
Solution:
138. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
139. The radius of convergence is infinite (𝑅=∞).
140. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
31 EXERCISE 2: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
141. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
142. Use this approximation to estimate sin(0.1).
143. Compare your approximation with the actual value of sin(0.1).
Solution:
144. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
145. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
146. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
32 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
147. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
148. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
149. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
150. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
151. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
152. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
33 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
34 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
153. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
154. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
155. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
156. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
157. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
158. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
159. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
35 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
160. Use the Taylor series method to find the solution 𝑦(𝑥).
161. Identify the resulting function and its Taylor series.
162. How does this relate to the uniqueness theorem for differential equations?
Solution:
163. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
164. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
165. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
36 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
166. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
167. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
168. Verify your result using the residue theorem.
Solution:
169. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
170. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
171. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
172. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
37 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
173. Every convergent power series represents an analytic function within its disk of
convergence.
174. The derivatives of an analytic function can be easily computed using its power series
representation.
175. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
176. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
177. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
178. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
179. What is the radius of convergence of this series?
180. Explain why this series converges for all complex numbers.
Solution:
181. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
182. The radius of convergence is infinite (𝑅=∞).
183. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
38 EXERCISE 2: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
184. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
185. Use this approximation to estimate sin(0.1).
186. Compare your approximation with the actual value of sin(0.1).
Solution:
187. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
188. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
189. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
39 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
190. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
191. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
192. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
193. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
194. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
195. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
40 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
41 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
196. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
197. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
198. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
199. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
200. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
201. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
202. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
42 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
203. Use the Taylor series method to find the solution 𝑦(𝑥).
204. Identify the resulting function and its Taylor series.
205. How does this relate to the uniqueness theorem for differential equations?
Solution:
206. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
207. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
208. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
43 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
209. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
210. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
211. Verify your result using the residue theorem.
Solution:
212. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
213. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
214. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
215. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
44 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
216. Every convergent power series represents an analytic function within its disk of
convergence.
217. The derivatives of an analytic function can be easily computed using its power series
representation.
218. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
219. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
220. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
221. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
222. What is the radius of convergence of this series?
223. Explain why this series converges for all complex numbers.
Solution:
224. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
225. The radius of convergence is infinite (𝑅=∞).
226. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
45 EXERCISE 2: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
227. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
228. Use this approximation to estimate sin(0.1).
229. Compare your approximation with the actual value of sin(0.1).
Solution:
230. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
231. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
232. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
46 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
233. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
234. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
235. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
236. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
237. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
238. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
47 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
48 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
239. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
240. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
241. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
242. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
243. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
244. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
245. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
49 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
246. Use the Taylor series method to find the solution 𝑦(𝑥).
247. Identify the resulting function and its Taylor series.
248. How does this relate to the uniqueness theorem for differential equations?
Solution:
249. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
250. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
251. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
50 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
252. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
253. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
254. Verify your result using the residue theorem.
Solution:
255. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
256. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
257. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
258. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
51 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
259. Every convergent power series represents an analytic function within its disk of
convergence.
260. The derivatives of an analytic function can be easily computed using its power series
representation.
261. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
262. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
263. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
264. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
265. What is the radius of convergence of this series?
266. Explain why this series converges for all complex numbers.
Solution:
267. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
268. The radius of convergence is infinite (𝑅=∞).
269. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
52 EXERCISE 2: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
270. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
271. Use this approximation to estimate sin(0.1).
272. Compare your approximation with the actual value of sin(0.1).
Solution:
273. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
274. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
275. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
53 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
276. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
277. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
278. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
279. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
280. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
281. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
54 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
55 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
282. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
283. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
284. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
285. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
286. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
287. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
288. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
56 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
289. Use the Taylor series method to find the solution 𝑦(𝑥).
290. Identify the resulting function and its Taylor series.
291. How does this relate to the uniqueness theorem for differential equations?
Solution:
292. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
293. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
294. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
57 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
295. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
296. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
297. Verify your result using the residue theorem.
Solution:
298. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
299. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
300. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
301. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
58 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
302. Every convergent power series represents an analytic function within its disk of
convergence.
303. The derivatives of an analytic function can be easily computed using its power series
representation.
304. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
305. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
306. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
Consider the function 𝑓(𝑧)=𝑒𝑧.
307. Write the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
308. What is the radius of convergence of this series?
309. Explain why this series converges for all complex numbers.
Solution:
310. The Taylor series for 𝑒𝑧 centered at 𝑧0=0 is:
𝑒𝑧= ∑𝑧𝑛
𝑛!
∞
𝑛=0 =1+𝑧+𝑧2
2!+𝑧3
3!+⋯
311. The radius of convergence is infinite (𝑅=∞).
312. This series converges for all complex numbers because:
– The ratio test gives lim𝑛→∞|𝑧𝑛+1/(𝑛+1)!
𝑧𝑛/𝑛! |=lim𝑛→∞|𝑧
𝑛+1|=0 for any fixed 𝑧.
– Since this limit is always less than 1, the series converges absolutely for all 𝑧.
59 EXERCISE 2: TAYLOR SERIES APPROXIMATION
Let 𝑓(𝑧)=sin(𝑧).
313. Write the first four non-zero terms of the Taylor series for 𝑓(𝑧) centered at 𝑧0=0.
314. Use this approximation to estimate sin(0.1).
315. Compare your approximation with the actual value of sin(0.1).
Solution:
316. The Taylor series for sin(𝑧) centered at 𝑧0=0 is:
sin(𝑧)=𝑧−𝑧3
3!+𝑧5
5!−𝑧7
7!+⋯
317. Using the first four terms to approximate sin(0.1):
sin(0.1)≈0.1−(0.1)3
6+(0.1)5
120 −(0.1)7
5040
≈0.1−0.000166667+0.000000083−0.000000000
≈0.099833416
318. The actual value of sin(0.1) to 9 decimal places is 0.099833417. Our approximation is
accurate to 7 decimal places, demonstrating the power of Taylor series approximations.
60 EXERCISE 3: LAURENT SERIES
Consider the function 𝑓(𝑧)=1
𝑧(𝑧−1).
319. Find the Laurent series for 𝑓(𝑧) in the annulus 0<|𝑧|<1.
320. Find the Laurent series for 𝑓(𝑧) in the region |𝑧|>1.
321. Identify the singularities of 𝑓(𝑧) and classify them.
Solution:
322. In 0<|𝑧|<1, we can write:
1
𝑧(𝑧−1)=−1
𝑧⋅1
1−𝑧=−1
𝑧(1+𝑧+𝑧2+𝑧3+⋯)
=−1
𝑧−1−𝑧−𝑧2−⋯
323. In |𝑧|>1, we can write:
1
𝑧(𝑧−1)=1
𝑧⋅1
1−1
𝑧=1
𝑧(1+1
𝑧+1
𝑧2+1
𝑧3+⋯)
=1
𝑧+1
𝑧2+1
𝑧3+⋯
324. Singularities:
– 𝑧=0: This is a simple pole (order 1).
– 𝑧=1: This is also a simple pole (order 1).
61 EXERCISE 4: RESIDUES AND LAURENT SERIES
For the function 𝑓(𝑧)=𝑒𝑧
𝑧2, find the residue at 𝑧=0 using the Laurent series expansion.
Solution: To find the residue, we need the coefficient of 1
𝑧 in the Laurent series expansion
around 𝑧=0.
Expanding 𝑒𝑧 as a Taylor series:
𝑒𝑧
𝑧2=1
𝑧2(1+𝑧+𝑧2
2!+𝑧3
3!+⋯)
=1
𝑧2+1
𝑧+1
2!+𝑧
3!+⋯
The coefficient of 1
𝑧 is 1, so the residue of 𝑓(𝑧) at 𝑧=0 is 1.
62 EXERCISE 5: ANALYTIC CONTINUATION
Consider the function 𝑓(𝑧)=1
1−𝑧.
325. Find the Taylor series of 𝑓(𝑧) centered at 𝑧=0. What is its radius of convergence?
326. Use the concept of analytic continuation to extend the domain of 𝑓(𝑧) beyond this
radius of convergence.
327. How does this relate to the Laurent series of 𝑓(𝑧)?
Solution:
328. The Taylor series of 𝑓(𝑧)=1
1−𝑧 centered at 𝑧=0 is:
𝑓(𝑧)=1+𝑧+𝑧2+𝑧3+⋯= ∑𝑧𝑛
∞
𝑛=0
329. The radius of convergence is 1, as this geometric series converges for |𝑧|<1.
330. We can analytically continue 𝑓(𝑧) to the entire complex plane except for 𝑧=1:
– For |𝑧|<1, we use the Taylor series ∑𝑧𝑛
∞
𝑛=0 .
– For |𝑧|>1, we can use the Laurent series −∑1
𝑧𝑛
∞
𝑛=1 .
331. The Laurent series representation allows us to define 𝑓(𝑧) in different regions:
– In |𝑧|<1, it matches the Taylor series.
– In |𝑧|>1, it provides a different series representation.
– The point 𝑧=1 remains a singularity (simple pole) of the function.
63 EXERCISE 6: TAYLOR SERIES AND DIFFERENTIAL EQUATIONS
Consider the differential equation 𝑦′=𝑦 with initial condition 𝑦(0)=1.
332. Use the Taylor series method to find the solution 𝑦(𝑥).
333. Identify the resulting function and its Taylor series.
334. How does this relate to the uniqueness theorem for differential equations?
Solution:
335. Let 𝑦(𝑥)=∑𝑎𝑛
∞
𝑛=0 𝑥𝑛. Then 𝑦′=∑𝑛
∞
𝑛=1 𝑎𝑛𝑥𝑛−1. From 𝑦′=𝑦 and 𝑦(0)=1, we get:
– 𝑎0=1
– 𝑛𝑎𝑛=𝑎𝑛−1 for 𝑛≥1
This gives 𝑎𝑛=1
𝑛! for all 𝑛≥0.
336. The resulting function is 𝑦(𝑥)=𝑒𝑥, and its Taylor series is:
𝑒𝑥= ∑𝑥𝑛
𝑛!
∞
𝑛=0
337. This result aligns with the uniqueness theorem for differential equations:
– The theorem states that for a first-order ODE with given initial conditions, there
exists a unique solution.
– Our Taylor series method found this unique solution, which is the exponential
function.
– The fact that we derived the Taylor series for 𝑒𝑥 confirms that this is indeed the
unique solution satisfying the given differential equation and initial condition.
64 EXERCISE 7: LAURENT SERIES AND COMPLEX INTEGRATION
Let 𝑓(𝑧)=1
(𝑧−1)(𝑧−2).
338. Find the Laurent series of 𝑓(𝑧) in the annulus 1<|𝑧|<2.
339. Use this series to evaluate the integral ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧.
340. Verify your result using the residue theorem.
Solution:
341. In 1<|𝑧|<2, we can write:
𝑓(𝑧)=1
𝑧−1−1
𝑧−2=1
(𝑧−1)+1
2−𝑧
=1
(𝑧−1)+1
2⋅1
1−𝑧
2
=1
𝑧−1+1
2(1+𝑧
2+(𝑧
2)2+⋯)
=1
𝑧−1+1
2+𝑧
4+𝑧2
8+⋯
342. In the Laurent series, only the term 1
𝑧−1 contributes to the integral:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)=2𝜋𝑖
343. Using the residue theorem:
∮ 𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅Res(𝑓,1)
344. The residue at 𝑧=1 is lim𝑧→1(𝑧−1)𝑓(𝑧)=1
1−2 =−1 So, ∮𝑓
|𝑧|=3
2(𝑧)𝑑𝑧=2𝜋𝑖⋅(−1)=
−2𝜋𝑖 This confirms our result from part (b).
65 EXERCISE 8: POWER SERIES AND ANALYTIC FUNCTIONS
Theorem 1.
If
𝑓(𝑧)=∑𝑎𝑛
∞
𝑛=0 𝑧𝑛
converges in a disk
|𝑧|<𝑅
, then
𝑓(𝑧)
is analytic in this disk
and its derivatives can be obtained by term-by-term differentiation of the series.
Discuss the implications of this theorem for Taylor series representations of analytic functions.
Discussion: This theorem has several important implications:
345. Every convergent power series represents an analytic function within its disk of
convergence.
346. The derivatives of an analytic function can be easily computed using its power series
representation.
347. The Taylor series of an analytic function 𝑓(𝑧) at 𝑧0 converges to 𝑓(𝑧) in some
neighborhood of 𝑧0.
348. The coefficients of the Taylor series are uniquely determined by the function and its
derivatives at the center point.
349. The radius of convergence of a Taylor series is the distance from the center to the
nearest singularity of the function.
These implications highlight the deep connection between analytic functions and their Taylor
series representations, forming a cornerstone of complex analysis.
