Year: 2018
Paper: 3
Question Number: 6
Course: LFM Stats And Pure
Section: Complex Numbers (L8th)
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 The distinct points $A$, $Q$ and $C$ lie on a straight line in the Argand diagram, and represent the distinct complex numbers $a$, $q$ and $c$, respectively.
Show that $\dfrac {q-a}{c-a}$ is real and hence that $(c-a)(q^*-a^*) = (c^*-a^*)(q-a)\,$.
Given that $aa^* = cc^* = 1$, show further that
\[
q+ ac q^* = a+c
\,.
\]
\item
The distinct points $A$, $B$, $C$ and $D$ lie, in anticlockwise order, on the circle of unit radius with centre at the origin (so that, for example, $aa^* =1$). The lines $AC$ and $BD$ meet at $Q$.
Show that
\[
(ac-bd)q^* = (a+c)-(b+d)
\,,
\]
where $b$ and $d$ are complex numbers represented by the points $B$ and $D$ respectively, and show further that
\[
(ac-bd)
(q+q^*) =
(a-b)(1+cd) +(c-d)(1+ab)
\,.
\]
\item
The lines $AB$ and $CD$ meet at $P$, which represents the complex number $p$. Given that $p$ is real, show that $p(1+ab)=a+b\,$. Given further that $ac-bd \ne 0\,$, show that
\[
p(q+q^*) = 2
\,.
\]
\end{questionparts}\begin{questionparts}
\item $A$, $Q$, $C$ lie on a straight line if $q = \lambda a + (1-\lambda)c$ for some $\lambda \in \mathbb{R}$,
\begin{align*}
&& q &= \lambda a + (1-\lambda)c \\
\Leftrightarrow && q - a &= (\lambda - 1)a + (1-\lambda)c \\
\Leftrightarrow && q - a &= (\lambda - 1)(a-c) \\
\Leftrightarrow && \frac{q - a}{c-a} &= 1-\lambda \\
\end{align*}
therefore $\frac{q-a}{c-a} \in \mathbb{R}$
\begin{align*}
&& \frac{q-a}{c-a} & \in \mathbb{R} \\
\Leftrightarrow && \left (\frac{q-a}{c-a} \right)^* &= \frac{q-a}{c-a} \\
\Leftrightarrow && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\
\end{align*}
Given $aa^* = cc^* = 1$,
\begin{align*}
&& (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\
\Leftrightarrow && q^*(c-a) - \frac{c}{a}+1 &= q \frac{a-c}{ca} - \frac{a}{c}+1 \\
\Leftrightarrow && (c-a)\l q^* +\frac{q}{ca}\r &= \frac{c}{a} - \frac{a}{c} \\
&&&= \frac{c^2-a^2}{ac} \\
\Leftrightarrow && q^* +\frac{q}{ca} &= \frac{c+a}{ac} \\
\Leftrightarrow && q^*ac +q &= a+c
\end{align*}
\item Since $Q$ lies on $AC$ and $BD$ we must have
\begin{align*}
&&& \begin{cases}
q^*ac +q &= a+c \\
q^*bd +q &= b+d \\
\end{cases} \\
\Rightarrow && q^*(ac-bd) &= (a+c)-(b+d) \\
\Rightarrow && q(ac-bd) &= (b+d)ac-(a+c)bd \\
\Rightarrow && (q+q^*)(ac-bd) &= (a+c)(1-bd)+(b+d)(ac-1) \\
&&&=a-abd+c-bcd+abc-b+acd-d \\
&&&= a(1+cd)-b(1+cd)+c(1+ab)-d(1+ab) \\
&&&= (a-b)(1+cd)+(c-d)(1+ab)
\end{align*}
\item If $AB$ and $CD$ meet at $p$ we must have $p^*ab + p = a+b$, ie $p(1+ab) = a+b$ amd $p(1+cd) = c+d$, so
\begin{align*}
&& (q+q^*)(ac-bd) &= (a-b) \frac{c+d}{p} + (c-d) \frac{a+b}{p} \\
\Leftrightarrow && p(q+q^*)(ac-bd) &= (a-b)(c+d)+(c-d)(a+b) \\
&&&= ac+ad-bc-bd+ac+bc-ad-bd \\
&&&= 2(ac-bd) \\
\Leftrightarrow && p(q+q^*) &= 2
\end{align*}
\end{questionparts}
The least popular of the Pure questions being attempted by under 40% of the candidates, it was the second least successfully attempted question in the whole paper with a mean score of only just better than 5/20. Many alternative solutions were successfully offered for the very first result, but then it was not an uncommon pitfall that and were treated as scalar multiples of, and moreover, the * notation appeared to confuse some candidates. Those who moved onto part (ii), got started fairly well, but then found difficulties; those that used the fact that the points were on the unit circle, however, derived the final equation with ease, earning full marks at that stage. The few candidates who had the stamina to see the question through to part (iii) were generally very successful, though odd marks were lost when division by various factors was not justified as valid by their being non-zero.