MAT3705
Assignment 4
DUE 11 September 2025
, MAT3705 Assignment 4 - Complete Worked Solutions
Learning Units 7 and 8 (Chapters 6 and 7)
Due date: 11 September 2025
Question 1
1
Let 𝑓(𝑧) = .
sin(1/𝑧)
(a) What type of singularity is 𝑧 = 0 ? Provide reasons for your answer.
The singularities of 𝑓(𝑧) occur where sin(1/𝑧) = 0 .
1
This happens when 1/𝑧 = 𝑛𝜋 for some integer 𝑛 ∈ ℤ , which means 𝑧 = . As 𝑛 → ∞ ,
𝑛𝜋
1
the singularities 𝑧 = approach 0.
𝑛𝜋
Since we have an infinite number of singularities in any neighborhood of 𝑧 = 0 , the point
𝑧 = 0 is a non-isolated essential singularity.
A singularity is non-isolated if every neighborhood of the singularity contains other
singularities of the function.
1
(b) What type of singularity is 𝑧 = ? Provide reasons for your answer.
2𝜋
1
The singularity at 𝑧 = is an isolated singularity because there exists a punctured disk
2𝜋
around this point that contains no other singularities of 𝑓(𝑧).
To classify it, we examine the limit of the function as 𝑧 approaches the singularity.
1
Let 𝑧0 = .
2𝜋
1 1
We can write 1/𝑧 = 1/(𝑧 0 + ℎ) = ≈ (1 − ℎ/𝑧0 ). As 𝑧 → 𝑧 0 , 1/𝑧 → 2𝜋 .
𝑧0 (1+ℎ/𝑧 0 ) 𝑧0
Assignment 4
DUE 11 September 2025
, MAT3705 Assignment 4 - Complete Worked Solutions
Learning Units 7 and 8 (Chapters 6 and 7)
Due date: 11 September 2025
Question 1
1
Let 𝑓(𝑧) = .
sin(1/𝑧)
(a) What type of singularity is 𝑧 = 0 ? Provide reasons for your answer.
The singularities of 𝑓(𝑧) occur where sin(1/𝑧) = 0 .
1
This happens when 1/𝑧 = 𝑛𝜋 for some integer 𝑛 ∈ ℤ , which means 𝑧 = . As 𝑛 → ∞ ,
𝑛𝜋
1
the singularities 𝑧 = approach 0.
𝑛𝜋
Since we have an infinite number of singularities in any neighborhood of 𝑧 = 0 , the point
𝑧 = 0 is a non-isolated essential singularity.
A singularity is non-isolated if every neighborhood of the singularity contains other
singularities of the function.
1
(b) What type of singularity is 𝑧 = ? Provide reasons for your answer.
2𝜋
1
The singularity at 𝑧 = is an isolated singularity because there exists a punctured disk
2𝜋
around this point that contains no other singularities of 𝑓(𝑧).
To classify it, we examine the limit of the function as 𝑧 approaches the singularity.
1
Let 𝑧0 = .
2𝜋
1 1
We can write 1/𝑧 = 1/(𝑧 0 + ℎ) = ≈ (1 − ℎ/𝑧0 ). As 𝑧 → 𝑧 0 , 1/𝑧 → 2𝜋 .
𝑧0 (1+ℎ/𝑧 0 ) 𝑧0