MAT3705
Assignment 4
Due 11 September 2025
, Student Name: MAT3705 Assignment 4
MAT3705 Assignment 4
Due date: 11 September 2025
Accurate Solutions
Question 1
Problem Statement: Let
1
f (z) = .
sin(1/z)
(a) What type of singularity is z = 0? Provide reasons.
1
(b) What type of singularity is z = 2π
? Provide reasons.
Step 1: Zeros of sin(1/z). The zeros of sin w occur at w = nπ for n ∈ Z. Substituting
w = 1/z gives singularities at
1
z= , n ∈ Z \ {0}.
nπ
These poles accumulate at z = 0.
Step 2: Analysis at z = 0. Since poles {1/(nπ)} cluster at z = 0, the point z = 0 is a
non-isolated singularity.
1 1
Step 3: Analysis at z = 2π
. At z0 = 2π
, sin(1/z0 ) = sin(2π) = 0 with multiplicity
1
one. Thus f (z) has a simple pole at z = 2π
.
Final Answer:
• (a) z = 0 is a non-isolated singularity.
• (b) z = 1
2π
is a simple pole.
—
Page 1
Assignment 4
Due 11 September 2025
, Student Name: MAT3705 Assignment 4
MAT3705 Assignment 4
Due date: 11 September 2025
Accurate Solutions
Question 1
Problem Statement: Let
1
f (z) = .
sin(1/z)
(a) What type of singularity is z = 0? Provide reasons.
1
(b) What type of singularity is z = 2π
? Provide reasons.
Step 1: Zeros of sin(1/z). The zeros of sin w occur at w = nπ for n ∈ Z. Substituting
w = 1/z gives singularities at
1
z= , n ∈ Z \ {0}.
nπ
These poles accumulate at z = 0.
Step 2: Analysis at z = 0. Since poles {1/(nπ)} cluster at z = 0, the point z = 0 is a
non-isolated singularity.
1 1
Step 3: Analysis at z = 2π
. At z0 = 2π
, sin(1/z0 ) = sin(2π) = 0 with multiplicity
1
one. Thus f (z) has a simple pole at z = 2π
.
Final Answer:
• (a) z = 0 is a non-isolated singularity.
• (b) z = 1
2π
is a simple pole.
—
Page 1
