Edexcel F1 2018 June — Question 9 8 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeComplex conjugate properties
DifficultyModerate -0.3 This is a straightforward Further Maths question requiring algebraic manipulation of complex numbers (cross-multiplying and rearranging), then applying standard modulus and argument formulas to given values. While it's Further Maths content, the techniques are routine and the question is highly structured with the answer to part (a) provided for use in part (b), making it slightly easier than average overall.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)

9. Given that $$\frac { z - k \mathrm { i } } { z + 3 \mathrm { i } } = \mathrm { i } \text {, where } k \text { is a positive real constant }$$
  1. show that \(z = - \frac { ( k + 3 ) } { 2 } + \frac { ( k - 3 ) } { 2 } \mathrm { i }\)
  2. Using the printed answer in part (a),
    1. find an exact simplified value for the modulus of \(z\) when \(k = 4\)
    2. find the argument of \(z\) when \(k = 1\). Give your answer in radians to 3 decimal places, where \(- \pi < \arg z < \pi\)

9. Given that

$$\frac { z - k \mathrm { i } } { z + 3 \mathrm { i } } = \mathrm { i } \text {, where } k \text { is a positive real constant }$$
\begin{enumerate}[label=(\alph*)]
\item show that $z = - \frac { ( k + 3 ) } { 2 } + \frac { ( k - 3 ) } { 2 } \mathrm { i }$
\item Using the printed answer in part (a),
\begin{enumerate}[label=(\roman*)]
\item find an exact simplified value for the modulus of $z$ when $k = 4$
\item find the argument of $z$ when $k = 1$. Give your answer in radians to 3 decimal places, where $- \pi < \arg z < \pi$
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel F1 2018 Q9 [8]}}