Cauchy Integral Theorem and Formula: Solutions to Assignment Questions, Assignments of Complex analysis

The solutions to assignment questions based on the cauchy integral theorem and formula. The questions involve integrating various functions over different contours, and the solutions are given in terms of the cauchy integral formula. Useful for students studying complex analysis, specifically those focusing on contour integration.

Typology: Assignments

2019/2020

Uploaded on 07/06/2020

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Assignment#2 (Solve the following questions with the help of Cauchy integral
theorem or Cauchy Integral Formula)
Questions
Answers
(i) 𝑒𝑧
𝑧2+𝜋2𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧|
𝐶= 4
0
(ii) sin 3𝑧
(𝑧+𝜋)
𝑐𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧|= 5
0
(iii) sin 𝜋𝑧2+cos 𝜋𝑧 2
(𝑧2−3𝑧+2) 𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧|= 3
𝐶
4𝜋𝑖
(iv) 𝑒3𝑖𝑧
(𝑧+𝜋)3𝑑𝑧
𝑐, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧 𝜋|= 3.2
0
(v) 𝑧3+𝑧+1
𝑧2−3𝑧+2 𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑙𝑙𝑖𝑝𝑠𝑒 4𝑥2+ 9𝑦2= 1
𝐶
0
(vi) 2𝑧2+𝑧
𝑧2−1 𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧 1|= 1
𝐶
3𝜋𝑖
(vii) 𝑒𝑖𝑧
5
𝐶𝑑𝑧, 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠 1 + 𝑖, −1 + 𝑖, −1 𝑖
0
(viii) 𝑧−1
(𝑧+1)2(𝑧−2)𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑙𝑙𝑖𝑝𝑠𝑒 𝑥2
9+(𝑦−1)2
16 = 1
𝐶
2𝜋𝑖
9
(ix) 𝑧2−2𝑧
(𝑧+1)2(𝑧2+4) 𝑑𝑧, 𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧|= 1
𝐶
8𝜋𝑖
25
(x) 𝑧
(𝑧+3)2(𝑧+1)𝑑𝑧,
𝐶
𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 𝑤𝑖𝑡ℎ 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠
1 𝑖, −3 + 𝑖, 1 + 𝑖, −3 𝑖
0
(xi) 2𝑧−1
𝑧3−𝑖𝑧2−𝑧2𝑑𝑧,
𝐶
𝑊ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑤𝑖𝑡ℎ 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠.
−𝑖, 3 𝑖, 2𝑖, 3 + 2𝑖
0
(xii) 𝑒2𝑧+𝑧2
(𝑧−1)5
𝐶𝑑𝑧 , 𝑤ℎ𝑒𝑟𝑒 𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 |𝑧|= 2
4𝜋𝑒2𝑖
3
Questions
Answers
(xiii) 1−cos(2(𝑧−3))
(𝑍−3)3
𝐶𝑑𝑧 , 𝑊ℎ𝑒𝑟𝑒 𝐶: |𝑧 3|= 1
4𝜋𝑖
pf2

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Assignment#2 (Solve the following questions with the help of Cauchy integral

theorem or Cauchy Integral Formula )

Questions Answers

(i) ∫

𝑒

𝑧

𝑧

2

+𝜋

2

𝐶

(ii) ∫

sin 3 𝑧

(𝑧+𝜋)

𝑐

(iii) ∫

sin 𝜋𝑧

2

+cos 𝜋𝑧

2

(𝑧

2

− 3 𝑧+ 2 )

𝐶

(iv) ∫

𝑒

3 𝑖𝑧

( 𝑧+𝜋

)

3

𝑐

(v) ∫

𝑧

3

+𝑧+ 1

𝑧

2

− 3 𝑧+ 2

2

2

𝐶

(vi) ∫

2 𝑧

2

+𝑧

𝑧

2

− 1

𝐶

(vii) ∫

𝑒

𝑖𝑧

5

𝐶

(viii) ∫

𝑧− 1

(𝑧+ 1 )

2

(𝑧− 2 )

𝑥

2

9

( 𝑦− 1

)

2

16

𝐶

(ix) ∫

𝑧

2

− 2 𝑧

( 𝑧+ 1

)

2

(𝑧

2

  • 4 )

𝐶

(x) ∫

𝑧

( 𝑧+ 3

)

2

( 𝑧+ 1

)

𝐶

(xi) ∫

2 𝑧− 1

𝑧

3

−𝑖𝑧

2

−𝑧

2

𝐶

(xii) ∫

𝑒

2 𝑧

+𝑧

2

( 𝑧− 1

)

5

𝐶

2

Questions Answers

(xiii) ∫

1 −cos( 2 (𝑧− 3 ))

(𝑍− 3 )

3

𝐶

(xiv) ∫

1 +𝑧

𝑧(𝑧− 2 )

𝐶