Year: 2018
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Complex numbers 2
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Use De Moivre's theorem to show that, if $\sin\theta\ne0$\,, then
\[
\frac{
\left(\cot \theta + \rm{i}\right)^{2n+1}
-\left(\cot \theta - \rm{i}\right)^{2n+1}}{2\rm{i}}
=
\frac{\sin \left(2n+1\right)\theta}
{\sin^{2n+1}\theta}
\,,
\]
for any positive integer $n$.
Deduce that the solutions of the equation
\[
\binom{2n+1}{1}x^{n}-\binom{2n+1}{3}x^{n-1}
+\cdots +
\left(-1\right)^{n}=0
\]
are
\[x=\cot^{2}\left(\frac{m\pi}{2n+1}\right)
\]
where $ m=1$, $2$, $\ldots$ , $n\,$.
\item
Hence show that
\[
\sum_{m=1}^n
\cot^{2}\left(\frac{m\pi}{2n+1}\right)
=\frac{n\left(2n-1\right)}{3}.
\]
\item
Given that $0<\sin \theta <\theta <\tan \theta$ for $0 < \theta < \frac{1}{2}\pi$, show that
\[
\cot^{2}\theta<\frac{1}{\theta^{2}}<1+\cot^{2}\theta.
\]
Hence show that
\[
\sum^\infty_{m=1}
\frac{1}{m^2}=
\frac{\pi^2}{6}\,.\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
\frac{\left(\cot \theta + i\right)^{2n+1}
-\left(\cot \theta - i\right)^{2n+1}}{2i} &= \frac{1}{\sin^{2n+1} \theta}\frac{\left(\cos \theta + i \sin \theta \right)^{2n+1}
-\left(\cos\theta - i\sin \theta\right)^{2n+1}}{2i} \\
&= \frac{1}{\sin^{2n+1} \theta} \frac{e^{i(2n+1) \theta} - e^{-i(2n+1) \theta} }{2i} \\
&=\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta}
\end{align*}
Notice that:
\begin{align*}
(\cot \theta + i)^{2n+1} - (\cot \theta - i)^{2n+1} &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}(i)^k \cdot \cot^{2n+1-k} \theta - \sum_{k=0}^{2n+1} \binom{2n+1}{k}(-i)^k \cdot \cot^{2n+1-k} \theta \\
&= \sum_{k=0}^{2n+1} \binom{2n+1}{k} \l i^k - (-i)^k \r \cdot \cot^{2n+1-k} \theta \\
&= \sum_{l=0}^{n} \binom{2n+1}{2l+1} \l i^{2l+1} - (-i)^{2l+1} \r \cdot \cot^{2n+1-(2l+1)} \theta \\
&= \sum_{l=0}^{n} \binom{2n+1}{2l+1} 2i \cdot \cot^{2(n-l)} \theta \\
&= 2i\sum_{l=0}^{n} \binom{2n+1}{2l+1} \cot^{2(n-l)} \theta \\
\end{align*}
Therefore if $\theta$ satisfies $\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} = 0$ then $z = \cot^2 \theta$ satisfies the equation. But $\theta = \frac{m \pi}{2n+1}, m = 1, 2, \ldots, n$ are $n$ distinct all the roots must be $\cot^2 \frac{m \pi}{2n+1}$.
\item Notice that the sum of the roots will be $\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{(2n+1)\cdot 2n \cdot (2n-1)}{3! \cdot (2n+1)} = \frac{n \cdot (2n-1)}{3}$ and so \[
\sum_{m=1}^n
\cot^{2}\left(\frac{m\pi}{2n+1}\right)
=\frac{n\left(2n-1\right)}{3}.
\]
\item For $0 < \theta < \frac{1}{2}\pi$, \begin{align*}
&& 0 < \sin \theta < \theta < \tan \theta \\
\Leftrightarrow && 0 < \cot \theta < \frac{1}{\theta} < \frac{1}{\sin \theta} \\
\Leftrightarrow && 0 < \cot^2 \theta < \frac{1}{\theta^2} < \cosec^2 \theta = 1 + \cot^2 \theta\\
\end{align*}
Therefore
\begin{align*}
&& \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} < \sum_{n=1}^N \frac{(2N+1)^2}{n^2 \pi^2} < N + \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} \\
\Rightarrow && \frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \sum_{n=1}^N \frac{1}{n^2 \pi^2} < \frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\
\Rightarrow && \lim_{N \to \infty}\frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} < \lim_{N \to \infty}\frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\
\Rightarrow && \frac{1}{6} \leq \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} \leq \frac16 \\
\Rightarrow && \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6}
\end{align*}
\end{questionparts}
Marginally more popular than question 6, the mean score was 8/20. The first part of (i) created no problems, but far fewer got further, although those that expanded cot almost always managed to succeed, even if they did not always quite deal with the sin 0 condition. Most that attempted part (ii) scored all the marks, though some forgot to divide 2 1 2 1 by 3 1. Almost all that tried part (iii) proved the first result but applying the results to obtain the conclusion proved harder. A common mistake was to sum over rather than and there were quite often mistakes in the algebra for the last part.