Question

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

 tan α − i

Answer 1

\[ \text { Let }z = \tan \alpha - i \]

\[ \because \tan \alpha\text {  is periodic with period } \pi . \text { So, let us take } \]

\[\alpha \in [0, \frac{\pi}{2}) \cup ( \frac{\pi}{2}, \pi]\]

\[\text { Case I }: \]

\[z = \tan \alpha - i \]

\[ \Rightarrow \left| z \right| = \sqrt{\tan^2 + 1}\]

\[ = \left| \sec \alpha \right| \left[ \because 0 < \alpha < \frac{\pi}{2} \right]\]

\[ = \sec \alpha\]

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

\[\tan \beta = \frac{1}{\left| \tan \alpha \right|}\]

\[ = \left| \cot \alpha \right|\]

\[ = \cot \alpha\]

\[ = \tan \left( \frac{\pi}{2} - \alpha \right)\]

\[ \Rightarrow \beta = \frac{\pi}{2} - \alpha \]

\[\text { We can see that Re }(z) > 0 \text { and Im}(z) < 0 . \text { So, z lies in the fourth quadrant }. \]

\[ \therefore \arg(z) = - \beta = \alpha - \frac{\pi}{2}\]

\[\text { Thus, z in the polar form is given by }\]

\[z = \sec \alpha \left\{ \cos\left( \alpha - \frac{\pi}{2} \right) + i\sin \left( \alpha - \frac{\pi}{2} \right) \right\} \]

\[\text { Case II }: \]

\[z = \tan \alpha - i \]

\[ \Rightarrow \left| z \right| = \sqrt{\tan^2 + 1}\]

\[ = \left| \sec \alpha \right| \left[ \because \frac{\pi}{2} < \alpha < \pi \right]\]

\[ = - \sec \alpha\]

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

\[\tan \beta = \frac{1}{\left| \tan \alpha \right|}\]

\[ = \left| \cot \alpha \right|\]

\[ = - \cot \alpha\]

\[ = \tan \left( \alpha - \frac{\pi}{2} \right)\]

\[ \Rightarrow \beta = \alpha - \frac{\pi}{2}\]

\[\text{We can see that Re}(z) < 0 \text { and Im} (z) < 0 . So, z \text { lies in the third quadrant }. \]

\[ \therefore \arg(z) = \pi + \beta = \frac{\pi}{2} + \alpha\]

\[\text { Thus, z in the polar form is given by } \]

\[z = - \sec \alpha \left\{ \cos\left( \frac{\pi}{2} + \alpha \right) + i\sin \left( \frac{\pi}{2} + \alpha \right) \right\} \]