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Nov 2015 p31 q9
1979
The complex number 3 - i is denoted by u. Its complex conjugate is denoted by u*.
On an Argand diagram with origin O, show the points A, B and C representing the complex numbers u, u* and u* - u respectively. What type of quadrilateral is OABC?
Showing your working and without using a calculator, express \(\frac{u^*}{u}\) in the form x + iy, where x and y are real.
By considering the argument of \(\frac{u^*}{u}\), prove that \(\arctan\left(\frac{3}{4}\right) = 2 \arctan\left(\frac{1}{3}\right)\).
Solution
(i) Plot the points on the Argand diagram: A at (3, -1), B at (3, 1), and C at (0, 2). The quadrilateral OABC is a parallelogram because opposite sides are parallel and equal in length.
(ii) Substitute \(u = 3 - i\) and \(u^* = 3 + i\). Then \(\frac{u^*}{u} = \frac{3+i}{3-i}\). Multiply numerator and denominator by the conjugate of the denominator: \(\frac{(3+i)(3+i)}{(3-i)(3+i)} = \frac{9 + 6i + i^2}{9 + 1} = \frac{8 + 6i}{10} = \frac{4}{5} + \frac{3}{5}i\).
(iii) The argument of \(\frac{u^*}{u}\) is \(\arg(u^*) - \arg(u) = \arctan\left(\frac{3}{4}\right)\). Using the identity \(\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)\) when \(ab < 1\), set \(a = \frac{1}{3}\) and \(b = \frac{1}{3}\). Then \(2 \arctan\left(\frac{1}{3}\right) = \arctan\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) = \arctan\left(\frac{3}{4}\right)\).