Question

Find the modulus and argument of the following complex number and hence express in the polar form:

 \[\frac{- 16}{1 + i\sqrt{3}}\]

Answer 1

\[ \frac{- 16}{1 + i\sqrt{3}}\]

\[\text { Rationalising the denominator }: \]

\[\frac{- 16}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}}\]

\[ \Rightarrow \frac{- 16 + 16\sqrt{3}i}{1 - 3 i^2} \]

\[ \Rightarrow \frac{- 16 + 16\sqrt{3}i}{4} \left( \because i^2 = - 1 \right)\]

\[ \Rightarrow - 4 + 4\sqrt{3}i\]

\[r = \left| z \right|\]

\[ = \sqrt{16 + 48}\]

\[ = 8\]

\[\text { Let } \tan \alpha = \left| \frac{Im(z)}{Re(z)} \right|\]

\[\text { Then }, \tan \alpha = \left| \frac{4\sqrt{3}}{- 4} \right|\]

\[ = \sqrt{3} \]

\[ \Rightarrow \alpha = \frac{\pi}{3}\]

\[\text { Since the point } \left( - 4, 4\sqrt{3} \right)\text {  lies in the third quadrant, the argument is given by }\]

\[ \theta = \pi - \alpha\]

\[ = \pi - \frac{\pi}{3}\]

\[ = \frac{2\pi}{3}\]

\[\text { Polar form } = r\left( \cos \theta + i\sin \theta \right) \]

                            \[ = 8\left\{ cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right\}\]