Year: 2016
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Complex numbers 2
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Let $\omega = \e^{2\pi {\rm i}/n}$, where $n$ is a positive integer.
Show that, for any complex number $z$,
\[
(z-1)(z-\omega) \cdots (z - \omega^{n-1}) = z^n -1\,.
\]
The points $X_0, X_1, \ldots\, X_{n-1}$ lie on a circle with centre $O$ and radius 1, and are the vertices of a regular polygon.
\begin{questionparts}
\item The point $P$ is equidistant from $X_0$ and $X_1$.
Show that, if $n$ is even,
\[
|PX_0| \times |PX_1 |\times \,\cdots\, \times |PX_{n-1}| = |OP|^n +1\,
,\]
where $|PX_ k|$ denotes the distance from $P$ to $X_k$.
Give the corresponding result when $n$ is odd. (There are two cases to consider.)
\item Show that
\[
|X_0 X_1|\times |X_0 X_2|\times \,\cdots\, \times |X_0 X_{n-1}| =n\,.
\]
\end{questionparts}Notice that $\omega^n = e^{2\pi i} = 1$, so $\omega$ is a root of $z^n - 1$, notice also that $(\omega^k)^n =1$ so therefore the $n$ roots are $1, \omega, \omega^2, \cdots, \omega^{n-1}$ and so $(z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)$. By considering the coefficient of $z^n$ we can see that $C = 1$.
\begin{questionparts}
\item $P$ lies on the perpendicular bisect of $1$ and $\omega$, so $p = re^{\pi i/n}$, where $r$ can be positive or negative, but $|r| = |OP|$.
\begin{align*}
&& |PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| &= |(p-1)(p-\omega) \cdots (p-\omega^{n-1})| \\
&&&= |p^n - 1| \\
&&&= |r^ne^{\pi i} - 1| \\
&&&= |-|OP|^n - 1| \tag{since $n$ even} \\
&&&= |OP|^n+1
\end{align*}
If $n$ is odd, depending on the sign of $r$ we get $|OP|^n+1$ or $||OP|^n-1|$.
\item $\,$ \begin{align*}
&& (z-\omega) \cdots(z-\omega^{n-1}) &= \frac{z^n-1}{z-1} \\
&&&= 1 + z +\cdots + z^{n-1} \\
&& |X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| &= |(1 - \omega)\cdots(1-\omega^{n-1})| \\
&&&= 1+1+1^2+\cdots + 1^{n-1} \\
&&&= n
\end{align*}
\end{questionparts}
This was the least popular question in the Pure section of the paper being attempted by less than 30% of the candidates, and whilst there were some very good solutions, the standard of attempts was generally not good, and in fact, only one of the Mechanics questions was less well-answered in the whole paper. In general, the stem was pretty well answered, but even here, having appreciated that each factor on the LHS was a factor of the RHS, frequently there was no consideration given to the, admittedly simple to obtain, scalar factor being one. In spite of the stem, most did not appreciate how to proceed with part (i), and so went little further.