2002 Paper 3 Q8

Year: 2002
Paper: 3
Question Number: 8

Course: LFM Stats And Pure
Section: Complex Numbers (L8th)

Difficulty: 1700.0 Banger: 1469.7

complex numbers unit modulus argument Ptolemy's theorem unit circle geometric proof modulus triangle inequality

Problem

Four complex numbers \(u_1\), \(u_2\), \(u_3\) and \(u_4\) have unit modulus, and arguments \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\), respectively, with \(-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi\). Show that \[ \arg \l u_1 - u_2 \r = \tfrac{1}{2} \l \theta_1 + \theta_2 -\pi \r + 2n\pi \] where \(n = 0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,\). Deduce that \[ \arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r = \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi \] for some integer \(n\). Prove that \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]

No solution available for this problem.

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Difficulty Rating: 1700.0

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Banger Rating: 1469.7

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Problem source

Four complex numbers $u_1$, $u_2$, $u_3$ and $u_4$ 
have unit modulus, and arguments $\theta_1$, 
$\theta_2$, $\theta_3$ and $\theta_4$, 
respectively, with $-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi$.
Show that
\[
\arg \l u_1 - u_2 \r  = \tfrac{1}{2} \l  \theta_1 + \theta_2 -\pi \r + 2n\pi
\]
where $n =  0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,$.
Deduce that
\[
\arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r 
= \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi 
\]
for some integer $n$.
Prove that
\[
| \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r |
= | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;.
\]

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