Year: 2011
Paper: 3
Question Number: 8
Course: LFM Stats And Pure
Section: Complex Numbers (L8th)
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The complex numbers $z$ and $w$ are related by
\[
w= \frac{1+\mathrm{i}z}{\mathrm{i}+z}\,.
\]
Let $z=x+\mathrm{i}y$ and $w=u+\mathrm{i}v$,
where $x$, $y$, $u$ and $v$ are real.
Express $u$ and $v$ in terms of $x$ and $y$.
\begin{questionparts}
\item
By setting $x=\tan(\theta/2)$, or otherwise,
show that if the locus of $z$ is the real axis $y=0$, $-\infty < x < \infty$,
then the locus of $w$ is the circle $u^2+v^2=1$ with one point
omitted.
\item Find the locus of $w$ when the locus of $z$ is
the line segment $y=0$, $-1 < x < 1\,$.
\item Find the locus of $w$ when the locus of $z$ is
the line segment $x=0$, $-1 < y < 1\,$.
\item Find the locus of $w$ when the locus of $z$ is
the line $y=1$, $-\infty < x < \infty\,$.
\end{questionparts}\begin{align*}
w &= \frac{1+iz}{i+z} \\
&= \frac{1-y+ix}{x+i(1+y)} \\
&= \frac{((1-y)+ix)(x-i(1+y))}{x^2+(1+y)^2} \\
&= \frac{x(1-y)+x(1+y)}{x^2+(1+y)^2}+\frac{x^2+y^2-1}{x^2+(1+y)^2}i \\
&= \frac{2x}{x^2+(y+1)^2}+\frac{x^2+y^2-1}{x^2+(1+y)^2}i
\end{align*}
Therefore $u = \frac{2x}{x^2+(y+1)^2}, v = \frac{x^2+y^2-1}{x^2+(1+y)^2}$
\begin{questionparts}
\item Suppose $z = \tan(\theta/2) = t$ then $u = \frac{2t}{t^2+1} = \sin \theta, v = \frac{t^2-1}{t^2+1} = \cos \theta$, ie $u+iv$ is the unit circle, where $-\frac{\pi}{2} < \theta/2 < \frac{\pi}{2}$ or $-\pi < \theta < \pi$, ie excluding the point $(\sin \pi, \cos \pi) = (0,1)$.
\item When $-1 < x < 1$ we have $-\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}$ ie $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, ie the lower half of the unit circle.
\item When $x = 0, -1 < y < 1$ we have $u = 0, v = \frac{y^2-1}{(1+y)^2}$ which is the negative imaginary axis.
\item We have $u = \frac{2t}{t^2+4}, v = \frac{t^2}{t^2+4}$, ie $u^2 + v^2 = v$, ie $u^2+(v-\frac12)^2 = \frac12^2$, so a circle centre $\frac12i$ radius $\frac12$, missing out $(0,1)$
\end{questionparts}
The response rate of this was similar to question 4, but with success rate similar to question 2. Most students did reasonably well getting half to three quarters of the marks by finding u and v and doing part (i), and then getting hold of (ii) and (iii) or not. Part (iv) rightly discriminated the strong candidates from the generality. A few alternative methods were tried but mostly they had their limitations. Details like omitted points from loci and the negative sign that arises when using the cosine double angle formula frequently lost marks.