Year: 2017
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Complex numbers 2
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.1
Banger Comparisons: 2
The transformation $R$ in the complex plane is a rotation (anticlockwise) by an angle $\theta$ about the point represented by the complex number $a$.
The transformation $S$ in the complex plane is a rotation (anticlockwise) by an angle $\phi$ about the point represented by the complex number $b$.
\begin{questionparts}
\item
The point $P$ is represented by the complex number~$z$.
Show that the image of $P$ under $R$ is represented by
\[
\e^{{\mathrm i} \theta}z + a(1-\e^{{\rm i} \theta})\,.
\]
\item Show that
the transformation $SR$ (equivalent to $R$ followed by $S$)
is a rotation about the point represented by
$c$, where
\[
%\textstyle
c\,\sin
\tfrac12 (\theta+\phi)
= a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta
+ b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi
\,,
\]
provided $\theta+\phi \ne 2n\pi$ for any integer $n$.
What is the transformation $SR$ if $\theta +\phi = 2\pi$?
\item
Under what circumstances is $RS =SR$?
\end{questionparts}\begin{questionparts}
\item We can map $a \mapsto 0$, rotate the whole plane, then shift the plane back to $a$, so
$z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})$
\item $z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})$
\begin{align*}
e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\
\end{align*}
Therefore this is rotation by angle $\phi + \theta$ and about
\begin{align*}
\frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\
&= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\
&= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\
\end{align*}
as required.
If $\phi + \theta = 2\pi$, then $z \mapsto z + (b-a)(1 - e^{i\phi})$ which is a translation.
\item If $\phi + \theta \neq 2 \pi$ then $RS = ST$ if
\begin{align*}
&& a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta
+ b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi
+ a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\
&& a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta
+ b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\
&& a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\
\Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\
\Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\
\end{align*}
If $\phi + \theta \neq 2 \pi$ then $RS = ST$ if $b = a$ or $e^{i\phi} = e^{i \theta}$ ie rotation through the same angle.
\end{questionparts}
This was the least popular pure question being attempted by only just over a quarter of candidates, and was the least successful of all the questions scoring 5/20. Most candidates gave up after part (i), and some made much more of this first result, not being very succinct. Most could write down without difficulty, but then did not spot an easy way to move beyond this. The standard of algebra displayed was in general poor, in particular moving between complex and trigonometric forms.