Year: 2019
Paper: 3
Question Number: 6
Course: LFM Stats And Pure
Section: Complex Numbers (L8th)
There was a significant rise in the total entry this year with an increase of nearly 8.5% on 2018. One question was attempted by over 90%, two others were very popular, and three further questions were attempted by 60% or more. No question was generally avoided and even the least popular attracted more than 10% of the candidates. 88% restricted themselves to attempting no more than 7 questions, and only a handful, but not the very best, scored strongly attempting more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The point $P$ in the Argand diagram is represented by the the complex number $z$, which satisfies
$$zz^* - az^* - a^*z + aa^* - r^2 = 0.$$
Here, $r$ is a positive real number and $r^2 \neq a^*a$. By writing $|z - a|^2$ as $(z - a)(z - a)^*$, show that the locus of $P$ is a circle, $C$, the radius and the centre of which you should give.
\begin{questionparts}
\item The point $Q$ is represented by $\omega$, and is related to $P$ by $\omega = \frac{1}{z}$. Let $C'$ be the locus of $Q$. Show that $C'$ is also a circle, and give its radius and centre.
If $C$ and $C'$ are the same circle, show that
$$(|a|^2 - r^2)^2 = 1$$
and that either $a$ is real or $a$ is imaginary. Give sketches to indicate the position of $C$ in these two cases.
\item Suppose instead that the point $Q$ is represented by $\omega$, where $\omega = \frac{1}{z^*}$. If the locus of $Q$ is $C$, is it the case that either $a$ is real or $a$ is imaginary?
\end{questionparts}\begin{align*}
&& |z-a|^2 &= (z-a)(z-a)^* \\
&&&= (z-a)(z^*-a^*) \\
&&&= zz^*-az^*-a^*z+aa^* \\
&&&= r^2
\end{align*}
Therefore the locus of $P$ is a circle centre $a$ radius $r$.
\begin{questionparts}
\item \begin{align*}
&& 0 &= zz^* - az^* - a^*z + aa^* - r^2 \\
&&&= \frac{1}{\omega \omega^{*}} - \frac{a}{\omega^*} - \frac{a^*}{\omega} + aa^*-r^2 \\
\Rightarrow && 0 &= 1-a\omega-a^*\omega^*+(|a|^2-r^2)\omega\omega^* \\
\Rightarrow && 0 &= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)-\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)+ \frac{1}{|a|^2-r^2} \\
&&&= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}-\frac{|a|^2}{(|a|^2-r^2)^2}+ \frac{1}{|a|^2-r^2} \\
&&&=\omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}- \frac{r^2}{(|a|^2-r^2)^2}
\end{align*}
Therefore $\displaystyle \left|\omega-\left ( \frac{a^*}{|a|^2-r^2}\right)\right|^2 = \frac{r^2}{(|a|^2-r^2)^2}$ ie $\omega$ lies on a circle centre $\frac{a^*}{|a|^2-r^2}$, radius $\frac{r}{||a|^2-r^2|}$.
If these are the same circle then $r = \frac{r}{||a|^2-r^2|} \Rightarrow (|a|^2-r^2)^2 = 1$ and $a = \frac{a^*}{|a|^2-r^2} \Rightarrow a = \pm a^*$, ie $a$ is purely real or imaginary.
\item This is the same story, except we end up with centre $\frac{a}{|a|^2-r^2}$, so we do not end up with the same conditions
\end{questionparts}
Half the candidates tried this question, but it was one of the four least successfully attempted. The majority successfully demonstrated that the locus of P was a circle with the correct centre and radius, but few made further significant progress. They usually substituted for z in terms of w but then failed to rearrange into the necessary form. If they did achieve this, then they were able to score most of the marks up until the very last part which required careful justification to earn full marks.