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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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