Year: 2013
Paper: 3
Question Number: 6
Course: LFM Stats And Pure
Section: Complex Numbers (L8th)
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let $z$ and $w$ be complex numbers. Use a diagram to show that
$\vert z-w \vert \le \vert z\vert + \vert w \vert\,.$
For any complex numbers $z$ and $w$, $E$ is defined by
\[
E = zw^* + z^*w +2 \vert zw \vert\,.
\]
\begin{questionparts}
\item
Show that
$\vert z-w\vert^2 = \left( \vert z \vert + \vert w\vert\right)^2 -E\,$,
and deduce that $E$ is real and non-negative.
\item Show that
$\vert 1-zw^*\vert^2 = \left ( 1 +\vert zw \vert \right)^2 -E\,$.
\end{questionparts}
Hence show that, if both $\vert z \vert >1$ and $\vert w \vert >1$, then
\[
\frac {\vert z-w\vert} {\vert 1-zw^*\vert }
\le \frac{\vert z \vert +\vert w\vert }{1+\vert z w \vert}\,.
\]
Does this inequality also hold if both $\vert z \vert <1$ and
$\vert w \vert <1$?\begin{questionparts}
\item $\,$ \begin{align*}
&& |z-w|^2 &= (z-w)(z^*-w^*) \\
&&&= zz^* - wz^*-zw^* + ww^* \\
&&&= |z|^2+|w|^2 - E + 2|zw| \\
&&&= (|z|+|w|)^2 - E \\
\Rightarrow && E &= (|z|+|w|)^2 - |z-w|^2 &\in \mathbb{R}
\end{align*}
and by the triangle inequality $|z|+|w| \geq |z-w|$, so $E \geq 0$
\item $\,$ \begin{align*}
&& |1-zw^*|^2 &= (1-zw^*)(1-z^*w) \\
&&&= 1 - zw^*-z^*w + |zw|^2 \\
&&&= 1 - E + 2|zw| + |zw|^2 \\
&&&= (1+|zw|)^2 - E
\end{align*}
\begin{align*}
&& \frac{|z-w|^2}{|1-zw^*|^2} &= \frac{(|z|+|w|)^2-E}{(1+|zw|)^2-E} \\
\Leftrightarrow && (1+|zw|^2)|z-w|^2 -E|z-w|^2 &= (|z|+|w|)^2|1-zw^*|^2-E|1-zw^*|^2\\
\Leftrightarrow && (1+|zw|^2)|z-w|^2-(|z|+|w|)^2|1-zw^*|^2 &= E(|z-w|^2-|1-zw^*|^2)\\
&&&= E(|z|^2-zw^*-z^*w+|w|^2-1+zw^*+z^*w-|z|^2|w|^2) \\
&&&= E(|z|^2+|w|^2-1-|z|^2|w|^2) \\
&&&= -E(1-|z|^2)(1-|w|^2) \\
&&&\leq 0 \\
\Leftrightarrow&& (1+|zw|^2)|z-w|^2& \leq (|z|+|w|)^2|1-zw^*|^2\\
\Leftrightarrow&& \frac{|z-w|^2}{|1-zw^*|^2} &\leq \frac{(|z|+|w|)^2}{(1+|zw|)^2}\\
\Leftrightarrow && \frac{|z-w|}{|1-zw^*|} &\leq \frac{(|z|+|w|)}{(1+|zw|)}\\
\end{align*}
Yes, this inequality holds if $|z|, |w|$ are the same side of $1$ and is reversed otherwise.
\end{questionparts}
About half attempted this with marginally more success than question 4. Many candidates tried to write expressions or similar and then tried to expand which involved a lot more work than dealing with conjugates directly. Some tried to use the cosine rule rather than the triangle inequality from the diagram. In general, the first result and parts (i) and (ii) were well done but only the strongest candidates did better than pick up the odd mark here and there in trying to obtain the inequality. A lot of mistakes were made mishandling inequalities, but even those who could do this correctly overlooked the necessity of substantiating that the square roots are positive and that the denominator is non-zero.