MAT3705 Assignment 2 2026 | Due 4 June 2026 - Distinction Guaranteed

MAT3705
ASSIGNMENT 2 2026
DUE: 4 JUNE 2026 (MEMO)

, Questions 1:


Section A: Multiple choice



Please insert and complete the following table to answer the questions in Section A:



​




Section A: Multiple choice
​




Question 1(a) 1(b) 1(c) 1(d)
​




Answer iii iv iv ii



​ ​




i(x−iy) ix+y
= ∣z∣iz2 = x2 +y2 =
n f (z) ​




x2 +y 2 .
​




y x
s f (z) = x2 +y 2 + i x2 +y 2 . ​ ​




y x
So u(x, y) = x2 +y 2 , v(x, y) = ​




x2 +y 2
→ option iii for 1(a).

Partial derivatives:
x2 −y 2
ux = − (x22xy
​




+y 2 )2
, uy = ​ ​




(x2 +y 2 )2
,
y 2 −x2 x2 −y 2
vx =
​




(x2 +y 2 )2 =− (x2 +y 2 )2 , vy =
​ ​ − (x22xy
+y 2 )2 → option iv for 1(b).
​




Cauchy‑Riemann equations: ux = vy and uy = −vx . ​ ​ ​ ​




2xy
We have ux = − (x2 +y 2 )2 = vy (true everywhere except origin).
​ ​ ​




x2 −y 2 2 2 2 2
uy = ​ −vx = − (xy2 +y
(x2 +y 2 )2
,
−x x −y
2 )2 = (x2 +y 2 )2 also true everywhere except origin.
​ ​




Hence f is differentiable for all z = 0 (Theorem in §23, p. 66, Brown & Churchill). At z = 0 the function is not defined.
So differentiable at every point except z = 0 → option iv for 1(c).

Analyticity requires differentiability in a neighbourhood. The set of differentiable points is C ∖ {0} , which is open. But at any
point z0 = 0, there exists a neighbourhood contained in C ∖ {0} where f is differentiable, so f is analytic at every z = 0 . Hence f
​




is analytic on C ∖ {0}. However, the question asks “where the function is analytic” – the only correct choice among the given is
that it is analytic at every point except z = 0 → option iv for 1(d)? W
i. everywhere – false (not analytic at 0).
ii. nowhere – false.
iii. only at z = 0 – false (not defined at 0).
iv. at every point except z = 0 – true (analytic on C ∖ {0}).
​ ​




1(d) = iv.