Question 6:
--- 6(a) ---
6(a) | Defines and in terms of two
variables ∗
for exam𝑧𝑧ple a𝑧𝑧nd | 1.1a | M1 | Let then
∗
𝑧𝑧 = 𝑥𝑥+i𝑦𝑦 𝑧𝑧 = 𝑥𝑥−i𝑦𝑦
∗
2𝑧𝑧−𝑧𝑧 = 𝑥𝑥+3i𝑦𝑦
∗ ∗
(2𝑧𝑧−𝑧𝑧 ) = 𝑥𝑥−3i𝑦𝑦
Re: 2 2 2 .........①
𝑧𝑧 = 𝑥𝑥 −𝑦𝑦 +2i𝑥𝑥𝑦𝑦
2 2
Im: .........②
𝑥𝑥 = 𝑥𝑥 −𝑦𝑦
② or
−3𝑦𝑦 = 2𝑥𝑥𝑦𝑦
3
⇒ 𝑦𝑦 = 0 𝑥𝑥 = −2
If then ① or 1
𝑦𝑦 = 0 ⇒ 𝑥𝑥 = 0
If then ①
3 3 9 2
an𝑥𝑥d = −2 ⇒ −2 = 4−𝑦𝑦
√15
𝑦𝑦 = ± 2
So the only solutions are
and
𝑧𝑧 = 0, 𝑧𝑧 = 1,
3 √15 3 √15
𝑧𝑧 = −2+ 2 i 𝑧𝑧 = −2− 2 i
Hence these are the only solutions
there are exactly four solutions
𝑥𝑥 𝑦𝑦
Obtains correct expressions for
and
∗ ∗ 2 | 1.1b | A1
( 2𝑧𝑧−𝑧𝑧 ) 𝑧𝑧
Uses their expressions for
and to form a pair
of simu∗lta∗neous 2equations
(2𝑧𝑧−𝑧𝑧 ) 𝑧𝑧 | 3.1a | M1
Deduces that the second
equation implies the result
“ or ”
3 | 2.2a | A1
𝑦𝑦 = 0 𝑥𝑥 = −2
Deduces that
Implies the result “ or 1”
PI and 𝑦𝑦 = 0 | 2.2a | A1
𝑥𝑥 = 0
Ob𝑧𝑧ta=ins0 any t𝑧𝑧w=o c1orrect
solutions in the form | 1.1b | A1
𝑧𝑧 = ⋯
Produces a clear argument to
conclude that there are exactly
four solutions stating them in the
form | 2.1 | R1
𝑧𝑧 = ⋯ Total | 7
Q | Marking Instructions | AO | Marks | Typical solution
--- 6(b)(i) ---
6(b)(i) | Shows their four points correctly
on Argand diagram
Condone no labelling | 1.1b | B1F
Connects their four points to
obtain shape with correct
symmetry | 1.1b | B1F
Total | 2
Q | Marking Instructions | AO | Marks | Typical solution
--- 6(b)(ii) ---
6(b)(ii) | Produces a clear argument to
show the required result | 2.1 | R1 | Area of upper triangle = Area of lower
triangle
1 √15 √15
= ×1× =
2 2 4
Total area
√15 √15
= 2× 4 = 2
as required
Total | 1
Question total | 10
Q | Marking Instructions | AO | Marks | Typical solution