Year: 2009
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Complex numbers 2
No solution available for this problem.
The vast majority of candidates (in excess of 95%) attempted at least five questions, and nearly a quarter attempted more than six questions, though very few doing so achieved high scores (about 2%). Most attempting more than six questions were submitting fragmentary answers, which, as the rubric informed candidates, earned little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1473.1
Banger Comparisons: 2
Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert
= 2\sin\frac12 (\beta-\alpha)\,$ for $0<\alpha<\beta<2\pi\,$.
Hence show that
\[
\big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert
\;
\big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert
+
\big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert
\;
\big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert
=
\big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert
\;
\big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert
\,,
\]
where $0<\alpha<\beta<\gamma<\delta<2\pi$.
Interpret this result as a theorem about cyclic quadrilaterals.
About a third of the candidates attempted this, though with less success than any of its predecessors. Attempts were mostly "all or nothing". Some candidates thought that the cyclic quadrilateral property had to be that opposite angles are supplementary, as the only property that they knew.