Question

# Express the complex number $$\dfrac {(1 + \sqrt {3}i)^{2}}{\sqrt {3} - i}$$ in the form of $$a + ib$$. Hence, find the modulus and argument of the complex number

Solution

## $$z = \dfrac {(1 + \sqrt {3}i)^{2}}{(\sqrt {3} - i)}$$$$= \dfrac {1 + 3i^{2} + 2\sqrt {3} i}{(\sqrt {3} - i)}$$$$= \dfrac {1 - 3 + 2\sqrt {3} i}{\sqrt {3} - i}$$$$= \dfrac {-2 + 2\sqrt {3}i}{\sqrt {3} - i} \times \dfrac {\sqrt {3} + i}{\sqrt {3} + i}$$$$= \dfrac {-2\sqrt {3} + 6i - 2i + 2\sqrt {3} i^{2}}{3 - i^{2}}$$$$= \dfrac {-4\sqrt {3} + 4i}{3 + 1}$$$$= \dfrac {4(-\sqrt {3} + i)}{4}$$$$\therefore z = -\sqrt {3} + i$$Here, $$a = -\sqrt {3}, b = 1$$Modulus, $$|z| = \sqrt {a^{2} + b^{2}}$$$$= \sqrt {(-\sqrt {3})^{2} + (1)^{2}}$$$$= \sqrt {3 + 1} = \sqrt {4}$$$$= 2$$$$Argument = -\pi + \tan^{-1}\left |\dfrac {b}{a}\right |$$$$= -\pi + \tan^{-1} \left |\dfrac {1}{-\sqrt {3}}\right |$$$$= -\pi + \tan^{-1} \dfrac {1}{\sqrt {3}}$$$$= -\pi + \dfrac {\pi}{6} = \dfrac {-5\pi}{6}$$Mathematics

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