AQA Further Paper 2 2021 June — Question 8 6 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeRegion shading with multiple inequalities
DifficultyChallenging +1.8 This is a Further Maths loci problem requiring students to find the geometric interpretation of two conditions (perpendicular bisector and circle), determine their intersection constraints, and express the result in a specific form with prime numbers. It demands visualization, algebraic manipulation of modulus equations, and optimization thinking, but follows a relatively standard Further Maths approach to loci intersection problems.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

8 The complex number \(z\) satisfies the equations $$\left| z ^ { * } - 1 - 2 i \right| = | z - 3 |$$ and $$| z - a | = 3$$ where \(a\) is real.
Show that \(a\) must lie in the interval \([ 1 - s \sqrt { t } , 1 + s \sqrt { t } ]\), where \(s\) and \(t\) are prime numbers.
[0pt] [6 marks]

Question 8:
AnswerMarks Guidance
Let \(z = x + iy\); \(x - iy - 1 - 2i =
\((x-1)^2 + (y+2)^2 = (x-3)^2 + y^2\)M1 Expresses \(z\) and \(z^*\) in terms of \(x\) and \(y\) and substitutes
\(-2x + 1 + 4y + 4 = -6x + 9 \Rightarrow y = 1 - x\)A1 Deduces correct linear Cartesian equation
\((x-a)^2 + y^2 = 9\)B1 Correct Cartesian equation for second equation; circle centre \((a,0)\) radius \(3\)
Solving simultaneously: \((x-a)^2 + (1-x)^2 = 9\)
AnswerMarks Guidance
\(2x^2 + (-2a-2)x + (a^2 - 8) = 0\)M1 Solves Cartesian equations simultaneously
\(\Delta \geq 0\): \((2a+2)^2 - 4(2)(a^2-8) \geq 0\)
\(a^2 - 2a - 17 \leq 0\)
AnswerMarks Guidance
\((a-1)^2 \leq 18\)E1 States discriminant is non-negative or creates inequality
\(1 - \sqrt{18} \leq a \leq 1 + \sqrt{18}\)
AnswerMarks Guidance
\(a \in [1-3\sqrt{2},\, 1+3\sqrt{2}]\)R1 Completes rigorous argument for correct range
## Question 8:

Let $z = x + iy$; $|x - iy - 1 - 2i| = |x + iy - 3|$

$(x-1)^2 + (y+2)^2 = (x-3)^2 + y^2$ | M1 | Expresses $z$ and $z^*$ in terms of $x$ and $y$ and substitutes

$-2x + 1 + 4y + 4 = -6x + 9 \Rightarrow y = 1 - x$ | A1 | Deduces correct linear Cartesian equation

$(x-a)^2 + y^2 = 9$ | B1 | Correct Cartesian equation for second equation; circle centre $(a,0)$ radius $3$

Solving simultaneously: $(x-a)^2 + (1-x)^2 = 9$

$2x^2 + (-2a-2)x + (a^2 - 8) = 0$ | M1 | Solves Cartesian equations simultaneously

$\Delta \geq 0$: $(2a+2)^2 - 4(2)(a^2-8) \geq 0$

$a^2 - 2a - 17 \leq 0$

$(a-1)^2 \leq 18$ | E1 | States discriminant is non-negative or creates inequality

$1 - \sqrt{18} \leq a \leq 1 + \sqrt{18}$

$a \in [1-3\sqrt{2},\, 1+3\sqrt{2}]$ | R1 | Completes rigorous argument for correct range
8 The complex number $z$ satisfies the equations

$$\left| z ^ { * } - 1 - 2 i \right| = | z - 3 |$$

and

$$| z - a | = 3$$

where $a$ is real.\\
Show that $a$ must lie in the interval $[ 1 - s \sqrt { t } , 1 + s \sqrt { t } ]$, where $s$ and $t$ are prime numbers.\\[0pt]
[6 marks]\\

\hfill \mbox{\textit{AQA Further Paper 2 2021 Q8 [6]}}