MAT3705 Assignment 4 Solutions Due 3 September 2026

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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MAT3705: Complex Analysis

Assignment 4 | 2026

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MAT3705
Module Code:
Complex Analysis
Module Name:
Residue Theory and Isolated Singularities
Assignment Topic:
Assignment 4
Assignment Number:
3 September 2026
Due Date:




Submitted in partial fulfilment of the requirements for MAT3705, UNISA 2026

, UNISA | MAT3705 Assignment 4 – Complex Analysis



Section A: Multiple Choice


Question 1


Question. 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}. Use residue theory to
∫︂ ∞
calculate P. V. f (x) dx.
−∞



(a) Singularities of f inside C(R) for R > 10


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


(x2 + 1)(x2 − 4x + 5) = 0.


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

Hence the four singularities are i, −i, 2 + i, 2 − i. Since C(R) is the upper semicircular contour
and R > 10, the singularities lying in the upper half-plane are z = i and z = 2 + i.



(a) iv


(b) Cauchy’s Residue Theorem applied to C(R)


By Cauchy’s Residue Theorem,

∫︂ 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




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