| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Challenging +1.2 This is a multi-part complex numbers question requiring squaring complex numbers, using modulus-argument form, and working with arguments in geometric progressions. Part (i) involves algebraic manipulation with surds (5 marks), part (ii)(a) uses a given tan result to find an argument (2 marks), and part (ii)(b) requires understanding that arguments multiply under powers and solving a linear congruence modulo 2π (3 marks). While it requires multiple techniques and careful work with exact forms, the individual steps are standard Further Maths procedures without requiring novel insight—moderately above average difficulty. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02i Quadratic equations: with complex roots4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((a + ib)^2 = 2 + 2i \Leftrightarrow a^2 - b^2 = 2\) and \(ab = 1\) | M1 | Squaring & equating Re/Im parts |
| \(a^2 - \frac{1}{a^2} - 2 = 0 \Rightarrow a^4 - 2a^2 - 1 = 0 \Rightarrow (a^2-1)^2 = 2\) (or by the quadratic formula) | M1 | |
| Subst'. for \(b\) (say) and solving a quadratic in \(a^2\) | A1 | |
| \(a = \sqrt{2+1}\) (AG) MUST note that \(a^2 > 0\) to explain choice of +ve sq.rt. | M1, A1 | |
| Similarly, \(b = \sqrt{\sqrt{2}-1}\) from \(b^4 + 2b^2 - 1 = 0\) or \(b = \frac{1}{a}\) | M1, A1 | |
| [5] | ||
| (ii)(a) \(z_2 = -\sqrt{2+1} + i\sqrt{\sqrt{2}-1}\) | M1 | |
| \(\arg(z_2) = \pi - \tan^{-1}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\) | M1 | Attempt incl'. rationalising denom'. |
| \(= \pi - \tan^{-1}(\sqrt{2}-1) = \pi - \frac{1}{8}\pi = \frac{7}{8}\pi\) | A1 | |
| [2] | ||
| (ii)(b) \(\arg(z_2^n) = \frac{7}{8}n\pi\) | B1 | |
| \(= (2k + \frac{1}{4})\pi\) | M1 | |
| \(n = \frac{16k+2}{7}\), giving least \(n = 14\) | A1 | |
| [Condone lack of convincing explanation that this IS the least such \(n\).] | [3] |
**(i)** $(a + ib)^2 = 2 + 2i \Leftrightarrow a^2 - b^2 = 2$ and $ab = 1$ | M1 | Squaring & equating Re/Im parts
$a^2 - \frac{1}{a^2} - 2 = 0 \Rightarrow a^4 - 2a^2 - 1 = 0 \Rightarrow (a^2-1)^2 = 2$ (or by the quadratic formula) | M1 |
Subst'. for $b$ (say) and solving a quadratic in $a^2$ | A1 |
$a = \sqrt{2+1}$ **(AG) MUST note that $a^2 > 0$ to explain choice of +ve sq.rt.** | M1, A1 |
Similarly, $b = \sqrt{\sqrt{2}-1}$ from $b^4 + 2b^2 - 1 = 0$ or $b = \frac{1}{a}$ | M1, A1 |
| [5] |
**(ii)(a)** $z_2 = -\sqrt{2+1} + i\sqrt{\sqrt{2}-1}$ | M1 |
$\arg(z_2) = \pi - \tan^{-1}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}}$ | M1 | Attempt incl'. rationalising denom'.
$= \pi - \tan^{-1}(\sqrt{2}-1) = \pi - \frac{1}{8}\pi = \frac{7}{8}\pi$ | A1 |
| [2] |
**(ii)(b)** $\arg(z_2^n) = \frac{7}{8}n\pi$ | B1 |
$= (2k + \frac{1}{4})\pi$ | M1 |
$n = \frac{16k+2}{7}$, giving least $n = 14$ | A1 |
[Condone lack of convincing explanation that this IS the least such $n$.] | [3] |
---The complex number $z_1$ is such that $z_1 = a + ib$, where $a$ and $b$ are positive real numbers.
\begin{enumerate}[label=(\roman*)]
\item Given that $z_1^2 = 2 + 2i$, show that $a = \sqrt{\sqrt{2} + 1}$ and find the exact value of $b$ in a similar form. [5]
\end{enumerate}
The complex number $z_2$ is such that $z_2 = -a + ib$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item \begin{enumerate}[label=(\alph*)]
\item Determine $\arg z_2$ as a rational multiple of $\pi$.
[You may use the result $\tan(\frac{1}{8}\pi) = \sqrt{2} - 1$.] [2]
\item The point $P_n$ in an Argand diagram represents the complex number $z_2^n$, for positive integers $n$. Find the least value of $n$ for which $P_n$ lies on the half-line with equation
$$\arg(z) = \frac{1}{4}\pi.$$ [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q12 [10]}}