Year: 2013
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Complex numbers 2
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show that $(z-\e^{i\theta})(z-\e^{-i\theta})=z^2 -2z\cos\theta +1\,$.
Write down the $(2n)$th roots of $-1$ in the form $\e^{i\theta}$, where
$-\pi <\theta \le \pi$, and deduce that
\[
z^{2n} +1
= \prod_{k=1}^n \left(z^2-2z \cos\left( \tfrac{(2k-1)\pi}{2n}\right) +1\right)
\,.
\]
Here, $n$ is a positive integer, and the $\prod$ notation denotes the product.
\begin{questionparts}
\item By substituting $z=i$ show that, when $n$ is even,
\[
\cos \left(\tfrac \pi {2n}\right)
\cos \left(\tfrac {3\pi} {2n}\right)
\cos \left(\tfrac {5\pi} {2n}\right)
\cdots
\cos \left(\tfrac{(2n-1) \pi} {2n}\right)
= {(-1\vphantom{\dot A})}^{\frac12 n} 2^{1-n}
\,.
\]
\item
Show that, when $n$ is odd,
\[
\cos^2 \left(\tfrac \pi {2n}\right)
\cos ^2 \left(\tfrac {3\pi} {2n}\right)
\cos ^2 \left(\tfrac {5\pi} {2n}\right)
\cdots
\cos ^2 \left(\tfrac{(n-2) \pi} {2n}\right)
= n 2^{1-n}
\,.
\]
You may use without proof
the fact that
$1+z^{2n}= (1+z^2)(1-z^2+z^4 - \cdots + z^{2n-2})\,$
when $n$ is odd.
\end{questionparts}\begin{align*}
&& (z-e^{i \theta})(z-e^{-i\theta}) &= z^2 - (e^{i\theta}+e^{-i\theta})z + 1 \\
&&&= z^2-2\cos \theta z + 1
\end{align*}
The $2n$th roots of $-1$ are $e^{\frac{i (2k+1)\pi}{2n}}, k \in \{-n, \cdots, n-1 \}$ or $e^{\frac{i k \pi}{2n}}, k \in \{-2n+1, -2n+3, \cdots, 2n-3, 2n-1 \}$
\begin{align*}
&& z^{2n}+1 &= (z-e^{-i(2n-1)/2n})\cdot (z-e^{-i(2n-3)/2n})\cdots (z-e^{i(2n-3)/2n})\cdot (z-e^{i(2n-1)/2n}) \\
&&&= \prod_{k=1}^n \left (z - e^{i \frac{2k-1}{2n}\pi} \right)\left (z - e^{-i \frac{2k-1}{2n}\pi} \right)\\
&&&= \prod_{k=1}^n \left (z^2 - 2z \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right)
\end{align*}
\begin{questionparts}
\item \begin{align*}
&& i^{2n} + 1 &= \prod_{k=1}^n \left (i^2 - 2i \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \\
\Rightarrow && (-1)^n + 1 &= (-1)^n2^ni^n\prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) \\
\Rightarrow && \prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) &= 2^{1-n}(-1)^{n/2} \tag{if $n\equiv 0\pmod{2}$}
\end{align*}
\item When $n$ is odd, we notice that two of the roots are $i$ and $-i$, if we exclude those, (ie by factoring out $z^2+1$, we see that
\begin{align*}
&& 1-z^2+z^4-\cdots + z^{2n-2} &= \prod_{k=1, 2k-1\neq n}^n \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \\
&&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=(n+1)/2}^{n} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\
&&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=1}^{(n-1)/2} \left (z^2+2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\
\Rightarrow && 1-i^2 + i^4 + \cdots + i^{2n-2} &= \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right) \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right)\\
\Rightarrow && n &= 2^{n-1} \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) \\
\Rightarrow && \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) &= n2^{1-n}
\end{align*}
\end{questionparts}
Just over two thirds of candidates attempted this with moderate success, approximately one third marks. Most succeeded with the opening result but even so, some lacked full explanation. Whilst most wrote down the correct form for the roots, few correctly expressed all the roots in the given range. Surprisingly, there was very limited understanding of the connection between the roots and the factors of 1 so the general result was not well answered. Conversely, part (i) was well-answered with the exception of those who did not deal with the powers of satisfactorily. Part (ii) was beyond most candidates mainly because they failed to cancel the factor 1. However, those that managed to deal with this aspect generally answered the whole question very well.