Year: 2021
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Complex numbers 2
The total entry was a marginal increase from that of 2019, that of 2020 having been artificially reduced. Comfortably more than 90% attempted one of the questions, four others were very popular, and a sixth was attempted by 70%. Every question was attempted by at least 10% of the candidature. 85% of candidates attempted no more than 7 questions, though very nearly all the candidates made genuine attempts on at most six questions (the extra attempts being at times no more than labelling a page or writing only the first line or two). Generally, candidates should be aware that when asked to "Show that" they must provide enough working to fully substantiate their working, and that they should follow the instructions in a question, so if it says "Hence", they should be using the previous work in the question in order to complete the next part. Likewise, candidates should be careful when dividing or multiplying, that things are positive, or at other times non-zero.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let
\[
z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}},
\]
where $\theta$ and $\phi$ are real, and $\theta - \phi \neq 2n\pi$ for any integer $n$. Show that
\[
z = i\cot\!\bigl(\tfrac{1}{2}(\phi - \theta)\bigr)
\]
and give expressions for the modulus and argument of $z$.
\item The distinct points $A$ and $B$ lie on a circle with radius $1$ and centre $O$. In the complex plane, $A$ and $B$ are represented by the complex numbers $a$ and $b$, and $O$ is at the origin. The point $X$ is represented by the complex number $x$, where $x = a + b$ and $a + b \neq 0$. Show that $OX$ is perpendicular to $AB$.
If the distinct points $A$, $B$ and $C$ in the complex plane, which are represented by the complex numbers $a$, $b$ and $c$, lie on a circle with radius $1$ and centre $O$, and $h = a + b + c$ represents the point $H$, then $H$ is said to be the \textit{orthocentre} of the triangle $ABC$.
\item The distinct points $A$, $B$ and $C$ lie on a circle with radius $1$ and centre $O$. In the complex plane, $A$, $B$ and $C$ are represented by the complex numbers $a$, $b$ and $c$, and $O$ is at the origin.
Show that, if the point $H$, represented by the complex number $h$, is the orthocentre of the triangle $ABC$, then either $h = a$ or $AH$ is perpendicular to $BC$.
\item The distinct points $A$, $B$, $C$ and $D$ (in that order, anticlockwise) all lie on a circle with radius $1$ and centre $O$. The points $P$, $Q$, $R$ and $S$ are the orthocentres of the triangles $ABC$, $BCD$, $CDA$ and $DAB$, respectively. By considering the midpoint of $AQ$, show that there is a single transformation which maps the quadrilateral $ABCD$ onto the quadrilateral $QRSP$ and describe this transformation fully.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& z &= \frac{e^{i \theta} + e^{i \phi}}{e^{i \theta} - e^{i \phi}} \\
&&&= \frac{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})}{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})} \\
&&&= \frac{(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})/2}{i(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})/2i} \\
&&&= \frac{\cos \frac12(\theta-\phi)}{i \sin \frac12(\theta-\phi)} \\
&&&= -i \cot \tfrac12(\theta-\phi)
\end{align*}
Therefore $|z| = \cot \tfrac12(\theta-\phi)$ and $\arg z = \frac{\pi}{2}$
\item Since $a,b$ lie on the unit circle, wlog $a = e^{i \theta}, b = e^{i\phi}$. Not that the line $OX$ has vector $a+b$ and $AB$ has vector $b-a$ and not their ratio has argument $\frac{\pi}{2}$ and hence they are perpendicular.
\item $AH$ has vector $h - a = (a+b+c) - a = b+c$ which we've already established is perpendicular to $c-b$ which is the vector for $BC$ (unless $b+c = 0$ in which case $h = a$).
\item $p = a +b+c, q = b+c+d$ etc. The midpoint of $AQ = \frac12(a+b+c+d)$ which is the same as the midpoint of $BR$, $CS$ and $DP$. Therefore we could say the transformation is reflection in the point $\frac12(a+b+c+d)$
\end{questionparts}
This was the least successfully attempted Pure question with a mean score under 6/20. It was less than 4% more popular than question 4. The first part of this question was generally well attempted, with a significant number of candidates being able to correctly verify the algebraic identity utilising a number of different approaches. There were some very neat solutions, but candidates who multiplied throughout by the complex conjugate and managed to keep track of the ensuing algebra were also often successful. Candidates must make sure that when they are trying to show a given result that they fully justify their solution – in this case some candidates missed out several steps of working and so did not gain full credit. Many candidates recognised that the form of z meant that the number was purely imaginary, but only a few candidates gained full credit for this part of the question with many omitting the modulus signs on the cot term for the modulus, or omitting the second possible angle. Some candidates were confused by the angles present in the given form of z and gave the argument as . In part (ii), the approach using the result from part (i) often did not score full marks due to the fact candidates would divide by quantities without explaining why they were non-zero. Some attempted this question with vector methods without clearly setting up that they were treating a,b as vectors rather than complex numbers. They were often unclear as to whether they were actually considering vectors, or considering complex numbers, which was particularly apparent in attempts to take the dot product of vectors without including the "dot" symbol. A number of candidates attempted to work out the gradients of the two line segments and show they multiplied to give -1: unfortunately, none recognised that a number of special cases were not taken care of with this method (cases where the lines were horizontal and vertical) and so did not score highly. Some candidates took a geometrical approach which needed to be fully explained to be convincing. For part (iii), many were more successful than for (ii): they recognised that part (ii) could be applied to give the result, and those who did generally gained full, or nearly full, credit. Vector approaches and considering the gradients of the line segments were used again in this part, with some candidates repeating the work they had done in the previous part, with the same pitfalls. Many omitted the case "if b+c=0 then h=a". Part (iv) was not attempted by a significant proportion. Of those who did attempt it, a significant number gained full credit. The most common mistake for this part of the question was candidates giving the transformation as "reflection through a point", which did not gain them credit as this is not considered to be a "Single transformation" as requested (each point is reflected through a different line). Another common mistake was the miscalculation of the midpoint of AQ as (b+c+d-a)/2 or as (a+b+c+d)/4.