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Nov 2018 p32 q9
1958
(a) (i) Without using a calculator, express the complex number \(\frac{2 + 6i}{1 - 2i}\) in the form \(x + iy\), where \(x\) and \(y\) are real.
(ii) Hence, without using a calculator, express \(\frac{2 + 6i}{1 - 2i}\) in the form \(r(\cos \theta + i \sin \theta)\), where \(r > 0\) and \(-\pi < \theta \leq \pi\), giving the exact values of \(r\) and \(\theta\).
(b) On a sketch of an Argand diagram, shade the region whose points represent complex numbers \(z\) satisfying both the inequalities \(|z - 3i| \leq 1\) and \(\text{Re } z \leq 0\), where \(\text{Re } z\) denotes the real part of \(z\). Find the greatest value of \(\arg z\) for points in this region, giving your answer in radians correct to 2 decimal places.
Solution
(a)(i) Multiply the numerator and denominator by the conjugate of the denominator: \((1 - 2i)(1 + 2i) = 1 + 4 = 5\). The expression becomes \(\frac{(2 + 6i)(1 + 2i)}{5} = \frac{2 + 4i + 6i + 12i^2}{5} = \frac{2 + 10i - 12}{5} = \frac{-10 + 10i}{5} = -2 + 2i\).
(a)(ii) The modulus \(r\) is \(\sqrt{(-2)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}\). The argument \(\theta\) is \(\arctan\left(\frac{2}{-2}\right) = \arctan(-1) = \frac{3\pi}{4}\).
(b) The region is a circle centered at \(3i\) with radius 1. The greatest value of \(\arg z\) occurs at the intersection of the circle and the imaginary axis, which is at \(z = i\). The argument is \(\arg z = \frac{\pi}{2} + \sin^{-1}\left(\frac{1}{3}\right) \approx 1.91\).