Question

Express the following complex in the form r(cos θ + i sin θ):

\[\frac{1 - i}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}}\]

Answer 1

\[\text { Let z } = \frac{1 - i}{cos\frac{\pi}{3} + i sin\frac{\pi}{3}}\]

\[ = \frac{1 - i}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}\]

\[ = \frac{2 - 2i}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}}\]

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

\[ = \frac{2 - 2\sqrt{3} - 2i(1 + \sqrt{3})}{4}\]

\[ = \frac{\left( 1 - \sqrt{3} \right) + i( - 1 - \sqrt{3})}{2}\]

\[ = \frac{\left( 1 - \sqrt{3} \right)}{2} + i\frac{( - 1 - \sqrt{3})}{2}\]

\[\text { Now,} z = \frac{\left( 1 - \sqrt{3} \right)}{2} + i\frac{( - 1 - \sqrt{3})}{2}\]

\[ \Rightarrow \left| z \right| = \sqrt{\left( \frac{1 - \sqrt{3}}{2} \right)^2 + \left( \frac{- 1 - \sqrt{3}}{2} \right)^2}\]

\[ = \sqrt{\left( \frac{1 + 3 - 2\sqrt{3}}{4} \right) + \left( \frac{1 + 3 + 2\sqrt{3}}{4} \right)}\]

\[ = \sqrt{\frac{8}{4}}\]

\[ = \sqrt{2}\]

\[\text { Let } \beta \text { be an acute angle given by } \tan\beta = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|} .\text {  Then }, \]

\[\tan\beta = \frac{\left| \frac{1 + \sqrt{3}}{2} \right|}{\left| \frac{1 - \sqrt{3}}{2} \right|} = \left| \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \right| = \left| \frac{\tan\frac{\pi}{4} + \tan\frac{\pi}{3}}{1 - \tan\frac{\pi}{4}\tan\frac{\pi}{3}} \right|\]

\[ \Rightarrow \tan\beta = \left| \tan\left( \frac{\pi}{4} + \frac{\pi}{3} \right) \right| = \left| \tan\frac{7\pi}{12} \right|\]

\[ \Rightarrow \beta = \frac{7\pi}{12}\]

\[\text { Clearly, z lies in the fourth quadrant . Therefore}  , \arg\left( z \right) = - \frac{7\pi}{12}\]

\[\text { Hence, the polar form of z is } \]

\[\sqrt{2}\left( \cos\frac{7\pi}{12} - \sin\frac{7\pi}{12} \right)\]