Pre-U Pre-U 9794/2 2018 June — Question 5 10 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2018
SessionJune
Marks10
TopicComplex Numbers Argand & Loci
TypePerpendicular bisector locus
DifficultyStandard +0.3 Part (i) involves routine complex number arithmetic (squaring, conjugate, solving simultaneous equations). Part (ii) requires recognizing that |z-u|=|z-v| represents the perpendicular bisector, sketching it, and finding an intersection—all standard techniques. This is slightly easier than average as it's methodical application of well-practiced procedures without requiring novel insight.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

5 The complex numbers \(u\) and \(v\) are given by \(u = 3 + 2 \mathrm { i }\) and \(v = 1 + 4 \mathrm { i }\).
  1. Given that \(a u ^ { 2 } + b v ^ { * } = 7 + 36 \mathrm { i }\) find the values of the real constants \(a\) and \(b\).
  2. Show the points representing \(u\) and \(v\) on an Argand diagram and hence sketch the locus given by \(| z - u | = | z - v |\). Find the point of intersection of this locus with the imaginary axis.

Question 5(i) and 5(ii)
(i)
- \(a(5+12\text{i}) + b(1-4\text{i}) = 7 + 36\text{i}\) B1 \(u^2 = 5+12\text{i}\) soi
- B1 \(v^* = 1-4\text{i}\) soi
- \(5a + b = 7\); \(12a - 4b = 36\) M1 Equate Re parts and Im parts. Must be using given equation
- \(12a - 4(7-5a) = 36\); \(32a - 28 = 36\); \(32a = 64\) M1 Dep on first M mark. Attempt to solve simultaneously, using valid method
- \(a = 2\), \(b = -3\) A1 Obtain \(a=2\) and \(b=-3\)
(ii)
- Plot \(u\) and \(v\) B1 Mark both points correctly on the Argand diagram. Some indication of scale needed
- Sketch perpendicular bisector B1 Attempt perpendicular bisector of their \(u\) and \(v\). Must be clearly intended as a perpendicular bisector e.g. drawn accurately or angle and mid-point clear on sketch
- Equation of perpendicular bisector is \(y = x+1\); using \(x=0\), gives \(y=1\) M1 Attempt to find point of intersection of perpendicular bisector with Im axis
- \(0 + \text{i}\) A1 Obtain \(0+\text{i}\). Allow just i, or coord equiv e.g. \((0,1)\). SR B1 for \(0+\text{i}\), or equiv, with no justification (includes scale drawing)
**Question 5(i) and 5(ii)**

**(i)**
- $a(5+12\text{i}) + b(1-4\text{i}) = 7 + 36\text{i}$ **B1** $u^2 = 5+12\text{i}$ soi
- **B1** $v^* = 1-4\text{i}$ soi
- $5a + b = 7$; $12a - 4b = 36$ **M1** Equate Re parts and Im parts. Must be using given equation
- $12a - 4(7-5a) = 36$; $32a - 28 = 36$; $32a = 64$ **M1** Dep on first M mark. Attempt to solve simultaneously, using valid method
- $a = 2$, $b = -3$ **A1** Obtain $a=2$ and $b=-3$

**(ii)**
- Plot $u$ and $v$ **B1** Mark both points correctly on the Argand diagram. Some indication of scale needed
- Sketch perpendicular bisector **B1** Attempt perpendicular bisector of their $u$ and $v$. Must be clearly intended as a perpendicular bisector e.g. drawn accurately or angle and mid-point clear on sketch
- Equation of perpendicular bisector is $y = x+1$; using $x=0$, gives $y=1$ **M1** Attempt to find point of intersection of perpendicular bisector with Im axis
- $0 + \text{i}$ **A1** Obtain $0+\text{i}$. Allow just i, or coord equiv e.g. $(0,1)$. **SR B1** for $0+\text{i}$, or equiv, with no justification (includes scale drawing)
5 The complex numbers $u$ and $v$ are given by $u = 3 + 2 \mathrm { i }$ and $v = 1 + 4 \mathrm { i }$.\\
(i) Given that $a u ^ { 2 } + b v ^ { * } = 7 + 36 \mathrm { i }$ find the values of the real constants $a$ and $b$.\\
(ii) Show the points representing $u$ and $v$ on an Argand diagram and hence sketch the locus given by $| z - u | = | z - v |$. Find the point of intersection of this locus with the imaginary axis.

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2018 Q5 [10]}}