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MAT3705 Assignment 3 (Accurate Solutions) Due 24 July 2025

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Uploaded on
July 22, 2025
Number of pages
18
Written in
2024/2025
Type
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MAT3705
Assignment 3

Due 24 July 2025

,MAT3705 Assignment 3

Due Date: 24 July 2025

Problem 1


Problem Statement:
z+1
Let f (z) = z2
and let C = {4eiθ : 0 ≤ θ ≤ π}. Calculate the contour integral
Z
z+1
dz
C z2

using Theorem 5.2.4, which refers to the definition of a contour integral.

Step 1: Parametrize the contour


We are given a semicircular contour: z(θ) = 4eiθ , for 0 ≤ θ ≤ π.
Then,
d
dz = (4eiθ ) dθ = 4ieiθ dθ


Step 2: Write f (z) in terms of θ


Substitute z = 4eiθ into f (z):

4eiθ + 1 4eiθ + 1
f (z(θ)) = =
(4eiθ )2 16e2iθ

So the integral becomes:
π
4eiθ + 1
Z
2iθ
· 4ieiθ dθ
0 16e

Step 3: Simplify the integrand


Simplify:
(4eiθ + 1)(4ieiθ ) 16ie2iθ + 4ieiθ i
2iθ
= 2iθ
= i + iθ
16e 16e 4e




1

, Step 4: Integrate term by term



Z π   Z π Z π
i i
i + iθ dθ = i dθ + iθ

0 4e 0 0 4e

First integral: Z π
i dθ = iπ
0

Second integral:

π π  −iθ π
i e−iπ − 1 i −2
Z Z
i i −iθ i e i 2 1

dθ = e dθ = · = · = · = · =
0 4e 4 0 4 −i 0 4 −i 4 −i 4 i 2


Final Answer:

Z
z+1 1
2
dz = iπ +
C z 2




2

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