MAT3705 Assignment 3 2025 (Accurate Solutions) Due 24 July 2025

MAT3705
Assignment 3

Due 24 July 2025

, MAT3705 Assignment 3

Due Date: 24 July 2025




Question 1
Problem Statement
z+1
R
Let f (z) = z2
, and let C = {4eiθ : 0 ≤ θ ≤ π}. Compute C
f (z) dz using the definition
of the contour integral (Theorem 5.2.4).

Step 1: Parametrization
Define z(θ) = 4eiθ , 0 ≤ θ ≤ π. Then

d
dz = (4eiθ )dθ = 4ieiθ dθ.
dθ


Step 2: Express the integrand
Substitute z = 4eiθ into f (z):

z+1 4eiθ + 1
f (z) = = .
z2 16e2iθ

Thus,
4eiθ + 1 iθ i(4eiθ + 1)
f (z) dz = · 4ie dθ = dθ.
16e2iθ 4eiθ

Simplify:
i(4eiθ + 1) i
iθ
= i + e−iθ .
4e 4

Step 3: Evaluate the integral

Z Z π   Z π Z π
i −iθ i −iθ
f (z) dz = i+ e dθ = i dθ + e dθ.
C 0 4 0 0 4


First term: Z π
i dθ = iπ.
0




1

, Second term: Z π Z π
i −iθ i
e dθ = e−iθ dθ.
0 4 4 0


Compute:

π π
e−iθ

−1 −iπ −1
Z
−iθ 2
e dθ = = (e − 1) = (−2) = = −2i.
0 −i 0 i i i


Thus,
i −2i2 2 1
· (−2i) = = = .
4 4 4 2

Step 4: Final Result

Z
z+1 1
2
dz = iπ + .
C z 2




2