MAT3705 Assignment 5 Solutions 2026
UNISA
,Question 1
1(a) iv
1(b) iii
1(c) iv
1(d) ii
, QUESTION 2 — Poles
The assignment gives
5𝑧 3 + 2
𝑓(𝑧) =
(𝑧 − 1)3
and asks us to show, using two different methods, that 𝑧 = 1is a pole of order 3.
(a): Using the definition of a pole
Recall the definition
A point 𝑧 = 𝑧0 is a pole of order 𝑚of 𝑓if
lim (𝑧 − 𝑧0 )𝑚 𝑓(𝑧)
𝑧→𝑧0
exists and is finite and non-zero, while the corresponding limits for lower powers do not
give a finite non-zero value.
Here,
5𝑧 3 + 2
𝑓(𝑧) = .
(𝑧 − 1)3
We test 𝑚 = 3.
Multiply by (𝒛 − 𝟏)𝟑
5𝑧 3 + 2
(𝑧 − 1)3 𝑓(𝑧) = (𝑧 − 1)3 .
(𝑧 − 1)3
Cancel:
(𝑧 − 1)3 𝑓(𝑧) = 5𝑧 3 + 2.
Taking the limit as 𝑧 → 1,
lim(𝑧 − 1)3 𝑓(𝑧) = lim(5𝑧 3 + 2).
𝑧→1 𝑧→1
Substitute 𝑧 = 1:
= 5(1)3 + 2
UNISA
,Question 1
1(a) iv
1(b) iii
1(c) iv
1(d) ii
, QUESTION 2 — Poles
The assignment gives
5𝑧 3 + 2
𝑓(𝑧) =
(𝑧 − 1)3
and asks us to show, using two different methods, that 𝑧 = 1is a pole of order 3.
(a): Using the definition of a pole
Recall the definition
A point 𝑧 = 𝑧0 is a pole of order 𝑚of 𝑓if
lim (𝑧 − 𝑧0 )𝑚 𝑓(𝑧)
𝑧→𝑧0
exists and is finite and non-zero, while the corresponding limits for lower powers do not
give a finite non-zero value.
Here,
5𝑧 3 + 2
𝑓(𝑧) = .
(𝑧 − 1)3
We test 𝑚 = 3.
Multiply by (𝒛 − 𝟏)𝟑
5𝑧 3 + 2
(𝑧 − 1)3 𝑓(𝑧) = (𝑧 − 1)3 .
(𝑧 − 1)3
Cancel:
(𝑧 − 1)3 𝑓(𝑧) = 5𝑧 3 + 2.
Taking the limit as 𝑧 → 1,
lim(𝑧 − 1)3 𝑓(𝑧) = lim(5𝑧 3 + 2).
𝑧→1 𝑧→1
Substitute 𝑧 = 1:
= 5(1)3 + 2
