# Express the Following Complex in the Form R(Cos θ + I Sin θ): 1 + I Tan α - Mathematics

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

#### Solution

$\text{Let } z = 1 + i\tan \alpha$

$\because \tan \alpha\text { is periodic with period }π. \text { So, let us take }$

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

$Case I:$

$\text { When } \alpha \in [0, \frac{\pi}{2})$

$z = 1 + i\tan \alpha$

$\Rightarrow \left| z \right| = \sqrt{1 + \tan^2 \alpha}$

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

$= \sec \alpha$

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

$\tan \beta = \left| \tan \alpha \right|$

$= \tan \alpha$

$\Rightarrow \beta = \alpha$

$\text { As z lies in the first quadrant . Therefore}, \arg(z) = \beta = \alpha$

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

$z = \sec \alpha \left( \cos\alpha + i\sin \alpha \right)$

$\text{Case II }:$

$z = 1 + i \tan \alpha$

$\Rightarrow \left| z \right| = \sqrt{1 + \tan^2 \alpha}$

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

$= - \sec \alpha$

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

$\tan \beta = \left| \tan \alpha \right|$

$= - \tan \alpha$

$\Rightarrow \tan \beta = \tan \left( \pi - \alpha \right)$

$\Rightarrow \beta = \pi - \alpha$

$\text { As, z lies in the fourth quadrant } .$

$\therefore \arg(z) = - \beta = \alpha - \pi$

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

$z = - \sec \alpha \left\{ \cos\left( \alpha - \pi \right) + i\sin \left( \alpha - \pi \right) \right\}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Exercise 13.4 | Q 3.1 | Page 57