Question

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

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

Answer 1

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

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

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

\[ \Rightarrow \frac{1 + 3i + 2i + 6 i^2}{1 - 9 i^2} \]

\[ \Rightarrow \frac{- 5 + 5i}{10} \left( \because i^2 = - 1 \right)\]

\[ \Rightarrow \frac{- 1}{2} + \frac{i}{2}\]

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

\[ = \sqrt{\frac{1}{4} + \frac{1}{4}}\]

\[ = \frac{1}{\sqrt{2}}\]

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

\[\text {Then }, \tan \alpha = \left| \frac{\frac{1}{2}}{\frac{- 1}{2}} \right|\]

\[ = 1 \]

\[ \Rightarrow \alpha = \frac{\pi}{4}\]

\[\text { Since point } \left( \frac{- 1}{2}, \frac{1}{2} \right) \text { lies in the second quadrant, the argument is given by }\]

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

\[ = \pi - \frac{\pi}{4}\]

\[ = \frac{3\pi}{4}\]

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

\[ = \frac{1}{\sqrt{2}}\left( cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right)\]