Question 9 - A-Level Maths

Edexcel FP1 2014 January — Question 9 8 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	January
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex Numbers Arithmetic
Type	Linear equations in z and z*
Difficulty	Standard +0.3 This is a straightforward Further Maths FP1 question requiring substitution of z = x + iy and z* = x - iy, then equating real and imaginary parts to solve simultaneous equations. While it's a Further Maths topic, the technique is mechanical and requires no insight beyond standard manipulation, making it slightly easier than average overall.
Spec	4.02c Complex notation: z, z*, Re(z), Im(z), \|z\|, arg(z)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide

9. Given that $z = x + \mathrm { i } y$, where $x \in \mathbb { R } , y \in \mathbb { R }$, find the value of $x$ and the value of $y$ such that $$( 3 - i ) z ^ { * } + 2 i z = 9 - i$$ where $z ^ { * }$ is the complex conjugate of $z$.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
$(3-i)z^* + 2iz = 9 - i$
$(3-i)(x-iy) + 2i(x+iy) = 9 - i$	M1	Substituting $z = x + iy$ and $z^* = x - iy$ into $(3-i)z^* + 2iz = 9 - i$
$3x - 3iy - ix - y + 2ix - 2y = 9 - i$	A1	Multiplies out $(3-i)(x-iy)$ correctly. This mark can be implied by correct later working. Equating either real or imaginary parts.
Re part: $3x - y - 2y = 9$	M1, A1	One set of correct equations
Im part: $-3y - x + 2x = -1$	A1
$3x - 3y = 9$
$x - 3y = -1$
$2x = 10 \Rightarrow x = 5$	ddM1	Attempt to solve simultaneous equations to find one of $x$ or $y$
$x - 3y = -1 \Rightarrow 5 - 3y = -1 \Rightarrow y = 2$	A1	Either $x = 5$ or $y = 2$
$\{z = 5 + 2i\}$	A1	Both $x = 5$ and $y = 2$

| $(3-i)z^* + 2iz = 9 - i$ | | |
| $(3-i)(x-iy) + 2i(x+iy) = 9 - i$ | M1 | Substituting $z = x + iy$ and $z^* = x - iy$ into $(3-i)z^* + 2iz = 9 - i$ |
| $3x - 3iy - ix - y + 2ix - 2y = 9 - i$ | A1 | Multiplies out $(3-i)(x-iy)$ correctly. This mark can be implied by correct later working. Equating either real or imaginary parts. |
| **Re part:** $3x - y - 2y = 9$ | M1, A1 | One set of correct equations |
| **Im part:** $-3y - x + 2x = -1$ | A1 | |
| $3x - 3y = 9$ | | |
| $x - 3y = -1$ | | |
| $2x = 10 \Rightarrow x = 5$ | ddM1 | Attempt to solve simultaneous equations to find one of $x$ or $y$ |
| $x - 3y = -1 \Rightarrow 5 - 3y = -1 \Rightarrow y = 2$ | A1 | Either $x = 5$ or $y = 2$ |
| $\{z = 5 + 2i\}$ | A1 | Both $x = 5$ and $y = 2$ |

---

Show LaTeX source

9. Given that $z = x + \mathrm { i } y$, where $x \in \mathbb { R } , y \in \mathbb { R }$, find the value of $x$ and the value of $y$ such that

$$( 3 - i ) z ^ { * } + 2 i z = 9 - i$$

where $z ^ { * }$ is the complex conjugate of $z$.\\

\hfill \mbox{\textit{Edexcel FP1 2014 Q9 [8]}}

This paper (10 questions)

View full paper

Q1 5 Q2 7 Q3 4 Q4 5 Q5 8 Q6 9 Q7 6 Q8 12 Q9 8 Q10 11

Answer	Marks	Guidance
\((3-i)z^* + 2iz = 9 - i\)
\((3-i)(x-iy) + 2i(x+iy) = 9 - i\)	M1	Substituting \(z = x + iy\) and \(z^* = x - iy\) into \((3-i)z^* + 2iz = 9 - i\)
\(3x - 3iy - ix - y + 2ix - 2y = 9 - i\)	A1	Multiplies out \((3-i)(x-iy)\) correctly. This mark can be implied by correct later working. Equating either real or imaginary parts.
Re part: \(3x - y - 2y = 9\)	M1, A1	One set of correct equations
Im part: \(-3y - x + 2x = -1\)	A1
\(3x - 3y = 9\)
\(x - 3y = -1\)
\(2x = 10 \Rightarrow x = 5\)	ddM1	Attempt to solve simultaneous equations to find one of \(x\) or \(y\)
\(x - 3y = -1 \Rightarrow 5 - 3y = -1 \Rightarrow y = 2\)	A1	Either \(x = 5\) or \(y = 2\)
\(\{z = 5 + 2i\}\)	A1	Both \(x = 5\) and \(y = 2\)