Year: 2010
Paper: 3
Question Number: 3
Course: UFM Pure
Section: Complex numbers 2
About 80% of candidates attempted at least five questions, and well less than 20% made genuine attempts at more than six. Those attempting more than six questions fell into three camps which were those weak candidates who made very little progress on any question, those with four or five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th or even 8th questions as an "insurance policy" against a solution that seemed strong but wasn't, or else for entertainment!
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1545.2
Banger Comparisons: 3
For any given positive integer $n$, a number $a$ (which may be complex) is said to be a \textit{primitive $n$th root of unity} if $a^n=1$ and there is no integer $m$ such that $0 < m < n$ and $a^m = 1$.
Write down the two primitive 4th roots of unity.
Let ${\rm C}_n(x)$ be the polynomial such that the roots of the equation ${\rm C}_n(x)=0$ are the primitive $n$th roots of unity,
the coefficient of the highest power of $x$ is one and the equation has no repeated roots.
Show that ${\rm C}_4(x) = x^2+1\,$.
\begin{questionparts}
\item Find ${\rm C}_1(x)$, ${\rm C}_2(x)$, ${\rm C}_3(x)$, ${\rm C}_5(x)$ and ${\rm C}_6(x)$, giving your answers as
unfactorised polynomials.
\item Find the value of $n$ for which ${\rm C}_n(x) = x^4 + 1$.
\item Given that $p$ is prime, find an expression for ${\rm C}_p(x)$, giving your answer as an unfactorised polynomial.
\item Prove that there are no positive integers $q$, $r$ and $s$ such that ${\rm C}_q(x) \equiv {\rm C}_r(x) {\rm C}_s(x)\,$.
\end{questionparts}The primitive 4th roots of unity are $i$ and $-i$. (Since the other two roots of $x^4-1$ are also roots of $x^2-1$
${\rm C}_4(x) = (x-i)(x+i) = x^2+1$ as required.
\begin{questionparts}
\item $\,$ \begin{align*}
&& {\rm C}_1 (x) &= x-1 \\
&& {\rm C}_2 (x) &= x+1 \\
&& {\rm C}_3 (x) &= x^2+x+1 \\
&& {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\
&& {\rm C}_6 (x) &= x^2-x+1 \\
\end{align*}
\item Since $(x^4+1)(x^4-1) = x^8-1$ we must have $n \mid 8$. But $n \neq 1,2,4$ so $n = 8$.
\item ${\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1$
\item Suppose ${\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)$, then if $\omega$ is a primitive $q$th root of unity we must ${\rm C}_q(\omega) = 0$, but that means that one of ${\rm C}_r(\omega)$, ${\rm C}_s(\omega)$ is $0$. But that's only possible if $r$ or $s$ $=q$. If this were the case, then what would the other value be? There are no possible values, hence it's not possible.
\end{questionparts}
Just over half the candidates attempted this question with most scores being quarter, half or three quarters in equal shares. Most candidates understood the idea of the question, the definition of a primitive root, and many wrote the roots of unity in (modulus)-argument form or exponential form. Failure to present a logical argument in parts (ii) and (iv) was a common problem and C6(x) tripped up quite a few.