Express the following complex in the form r(cos θ + i sin θ):
1 + i tan α
Solution
\[\text{Let } z = 1 + i\tan \alpha \]
\[ \because \tan \alpha\text { is periodic with period }π. \text { So, let us take } \]
\[\alpha \in [0,\frac{\pi}{2}) \cup ( \frac{\pi}{2}, \pi]\]
\[Case I: \]
\[\text { When } \alpha \in [0, \frac{\pi}{2})\]
\[z = 1 + i\tan \alpha \]
\[ \Rightarrow \left| z \right| = \sqrt{1 + \tan^2 \alpha}\]
\[ = \left| \sec \alpha \right| \left[ \because 0 < \alpha < \frac{\pi}{2} \right]\]
\[ = \sec \alpha\]
\[\text { Let } \beta \text { be an acute angle given by } \tan \beta = \left| \frac{Im (z)}{Re(z)} \right|\]
\[\tan \beta = \left| \tan \alpha \right|\]
\[ = \tan \alpha\]
\[ \Rightarrow \beta = \alpha \]
\[\text { As z lies in the first quadrant . Therefore}, \arg(z) = \beta = \alpha\]
\[\text { Thus, z in the polar form is given by } \]
\[z = \sec \alpha \left( \cos\alpha + i\sin \alpha \right)\]
\[\text{Case II }: \]
\[z = 1 + i \tan \alpha \]
\[ \Rightarrow \left| z \right| = \sqrt{1 + \tan^2 \alpha}\]
\[ = \left| \sec \alpha \right| \left[ \because \frac{\pi}{2} < \alpha < \pi \right]\]
\[ = - \sec \alpha\]
\[\text { Let } \beta \text { be an acute angle given by } \tan \beta = \left| \frac{Im (z)}{Re(z)} \right|\]
\[\tan \beta = \left| \tan \alpha \right|\]
\[ = - \tan \alpha\]
\[ \Rightarrow \tan \beta = \tan \left( \pi - \alpha \right)\]
\[ \Rightarrow \beta = \pi - \alpha\]
\[\text { As, z lies in the fourth quadrant } . \]
\[ \therefore \arg(z) = - \beta = \alpha - \pi\]
\[\text { Thus, z in the polar form is given by } \]
\[z = - \sec \alpha \left\{ \cos\left( \alpha - \pi \right) + i\sin \left( \alpha - \pi \right) \right\} \]
