MAT3705 Assignment 4 2026 Solutions Due 3 September 2026

UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology



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Complex Analysis
Assignment 4 — 2026


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Module Code: MAT3705

Module Name: Complex Analysis

Assignment No.: Assignment 4

Due Date: 3 September 2026

Semester: Semester 2, 2026




Submitted in partial fulfilment of the requirements for MAT3705
at the University of South Africa.

, UNISA | MAT3705 Complex Analysis — Assignment 4



Section A: Multiple Choice


Question 1


Let
x2
f (x) =
(x2 + 1)(x2 − 4x + 5)

and, for R > 0, let ∫︂C(R) = CR ∪ [−R, R], where CR = {Reit : 0 ≤ t ≤ π}. Residue theory is used
∞
to calculate P. V. f (x) dx.
−∞



Question 1(a)


The singularities of f are found by setting the denominator equal to zero:


(x2 + 1)(x2 − 4x + 5) = 0.


From x2 + 1 = 0, z = ±i. From x2 − 4x + 5 = 0, the quadratic formula gives
√
4± 16 − 20 4 ± 2i
z= = = 2 ± i.
2 2

The four singularities are i, −i, 2 + i, 2 − i. Since C(R) is the upper semicircular contour, only
the singularities in the upper half-plane lie inside C(R) for R > 10, namely z = i and z = 2 + i.

Answer: (a) iv


Question 1(b)


By Cauchy’s Residue Theorem, with z1 = i and z2 = 2 + i the singularities enclosed by C(R),

∫︂ 2
∑︂
f (z) dz = 2πi Resz=zk f (z).
C(R) k=1


Since C(R) = [−R, R] ∪ CR ,

∫︂ R ∫︂ 2
∑︂
f (x) dx + f (z) dz = 2πi Resz=zk f (z).
−R CR k=1



Answer: (b) iii




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