MAT3705 Assignment 3 Solutions 2026

MAT3705 Assignment 3 Solutions 2026
UNISA
All questions are fully answered, and multiple answers are solved for better
understanding.


DUE: 23 JULY 2026

,QUESTION 1: Multiple Choice

We are given

𝑒 3𝑧
𝑓(𝑧) =
(𝑧 2 + 4)(𝑧 − 2𝑖)


and

𝐶 = {𝑧 ∈ ℂ: ∣ 𝑧 − 𝑖 ∣= 2}.


The contour 𝐶is centred at 𝑖with radius 2.

The assignment states that only the completed answer table is marked, but it is
important to work through the problem first to select the correct options.

Factor the denominator

𝑧 2 + 4 = (𝑧 − 2𝑖)(𝑧 + 2𝑖).


Therefore,

𝑒 3𝑧
𝑓(𝑧) = .
(𝑧 − 2𝑖)2 (𝑧 + 2𝑖)


The singularities are therefore:

𝑧 = 2𝑖


and

𝑧 = −2𝑖.


Determine which singularities lie inside 𝑪

The contour is

∣ 𝑧 − 𝑖 ∣= 2.


For 𝑧 = 2𝑖:

∣ 2𝑖 − 𝑖 ∣=∣ 𝑖 ∣= 1 < 2.


Therefore,

, 𝑧 = 2𝑖


is inside 𝐶.

For 𝑧 = −2𝑖:

∣ −2𝑖 − 𝑖 ∣=∣ −3𝑖 ∣= 3 > 2.


Therefore,

𝑧 = −2𝑖


is outside 𝐶.

Hence, for Question 1(a):

iii. 𝑧 = 2𝑖 is interior to 𝐶 and 𝑧 = −2𝑖 is exterior to 𝐶.




1(b)

Since 𝑧 = 2𝑖is the only singularity inside 𝐶, rewrite 𝑓(𝑧)in the form required for the
Cauchy Integral Formula.

Because

𝑒 3𝑧
𝑓(𝑧) = ,
(𝑧 − 2𝑖)2 (𝑧 + 2𝑖)


we write
𝑔(𝑧)
𝑓(𝑧) = ,
(𝑧 − 2𝑖)2


where

𝑒 3𝑧
𝑔(𝑧) = .
𝑧 + 2𝑖


The function 𝑔(𝑧)is analytic inside and on 𝐶, because its only singularity is at

𝑧 = −2𝑖,