OCR Further Pure Core AS 2020 November — Question 1 6 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypePure square root finding
DifficultyStandard +0.8 This is a standard Further Maths complex numbers question requiring the algebraic method (setting z² = -77 - 36i, equating real and imaginary parts, solving simultaneous equations). While it involves multiple steps and careful algebra, it's a well-practiced technique in Further Pure with no novel insight required. The 6 marks reflect computational work rather than conceptual difficulty, placing it moderately above average but well within typical Further Pure territory.
Spec4.02h Square roots: of complex numbers

In this question you must show detailed reasoning. Use an algebraic method to find the square roots of \(-77 - 36\text{i}\). [6]

Question 1:
AnswerMarks
1DR
(a + bi)2 = a2 – b2 + 2abi
a2 – b2 = –77 and 2ab = –36 (where a and b are
real)
2
18  18
b=− ⇒a2 −−  =−77
a  a 
⇒a4 +77a2 −324=0
(a2 – 4)(a2 + 81) = 0
a2 = 4 or b2 = 81 only
AnswerMarks
2 – 9i and –2 + 9iB1
M1
M1
B1
A1ft
A1
AnswerMarks
[6]1.1
1.1
1.1
2.1
2.3
AnswerMarks
1.1Seen or implied in solution
Comparing real and imaginary parts (no
i unless later recovered) from a 3 term
expansion
Eliminating b or a to obtain 3 term
quadratic in a2 or b2. Unknowns must
not be in denominator and non-zero
terms on same side. = 0 seen or implied
by solution.
Rogue solutions; a2 = –81, b2 = – 4 .
Any rogue solutions must be discarded
before A1 awarded.
Both roots. Can use ± but not ±2 ± 9i
and not ± 2 – 9i. ±(2 – 9i), ±(–2 + 9i)
AnswerMarks
or ±2 m 9i are all acceptable.Allow equating real and imaginary
considering (a+ib)(c+id)
(b4 – 77b2 – 324 = 0)
Factorised forms:
(b2 – 81)(b2 + 4)
DR requires evidence of solving
quadratic in a2 (or b2). Can be
implied by sight of all 4 solutions.
For follow through need to discard
rogue solutions
2 – 9i and – 2 + 9i without working
from quartic could score
B1M1M1B0A0A1
Question 1:
1 | DR
(a + bi)2 = a2 – b2 + 2abi
a2 – b2 = –77 and 2ab = –36 (where a and b are
real)
2
18  18
b=− ⇒a2 −−  =−77
a  a 
⇒a4 +77a2 −324=0
(a2 – 4)(a2 + 81) = 0
a2 = 4 or b2 = 81 only
2 – 9i and –2 + 9i | B1
M1
M1
B1
A1ft
A1
[6] | 1.1
1.1
1.1
2.1
2.3
1.1 | Seen or implied in solution
Comparing real and imaginary parts (no
i unless later recovered) from a 3 term
expansion
Eliminating b or a to obtain 3 term
quadratic in a2 or b2. Unknowns must
not be in denominator and non-zero
terms on same side. = 0 seen or implied
by solution.
Rogue solutions; a2 = –81, b2 = – 4 .
Any rogue solutions must be discarded
before A1 awarded.
Both roots. Can use ± but not ±2 ± 9i
and not ± 2 – 9i. ±(2 – 9i), ±(–2 + 9i)
or ±2 m 9i are all acceptable. | Allow equating real and imaginary
considering (a+ib)(c+id)
(b4 – 77b2 – 324 = 0)
Factorised forms:
(b2 – 81)(b2 + 4)
DR requires evidence of solving
quadratic in a2 (or b2). Can be
implied by sight of all 4 solutions.
For follow through need to discard
rogue solutions
2 – 9i and – 2 + 9i without working
from quartic could score
B1M1M1B0A0A1
In this question you must show detailed reasoning.

Use an algebraic method to find the square roots of $-77 - 36\text{i}$. [6]

\hfill \mbox{\textit{OCR Further Pure Core AS 2020 Q1 [6]}}