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Question
Find the modulus and argument of the following complex number and hence express in the polar form:
\[\frac{- 16}{1 + i\sqrt{3}}\]
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Solution
\[ \frac{- 16}{1 + i\sqrt{3}}\]
\[\text { Rationalising the denominator }: \]
\[\frac{- 16}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}}\]
\[ \Rightarrow \frac{- 16 + 16\sqrt{3}i}{1 - 3 i^2} \]
\[ \Rightarrow \frac{- 16 + 16\sqrt{3}i}{4} \left( \because i^2 = - 1 \right)\]
\[ \Rightarrow - 4 + 4\sqrt{3}i\]
\[r = \left| z \right|\]
\[ = \sqrt{16 + 48}\]
\[ = 8\]
\[\text { Let } \tan \alpha = \left| \frac{Im(z)}{Re(z)} \right|\]
\[\text { Then }, \tan \alpha = \left| \frac{4\sqrt{3}}{- 4} \right|\]
\[ = \sqrt{3} \]
\[ \Rightarrow \alpha = \frac{\pi}{3}\]
\[\text { Since the point } \left( - 4, 4\sqrt{3} \right)\text { lies in the third quadrant, the argument is given by }\]
\[ \theta = \pi - \alpha\]
\[ = \pi - \frac{\pi}{3}\]
\[ = \frac{2\pi}{3}\]
\[\text { Polar form } = r\left( \cos \theta + i\sin \theta \right) \]
\[ = 8\left\{ cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right\}\]
