Complex Numbers — MATHS STD 11 Science — Question

$\tan\alpha$ is periodic function with period $\pi$

So, let us take $\alpha$ lying in the interval $\Big[0,\frac{\pi}{2}\Big)\cup\Big(\frac{\pi}{2},\pi\Big].$

Case - 1: when $\alpha\in\Big[0,\frac{\pi}{2}\Big)$

$|\text{z}|=\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha$

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

$\tan\beta=|\tan\alpha|=\tan\alpha$

$\Rightarrow\beta=\alpha$

As z represented by a point in first quadrant.

$\therefore \ \text{arg(z)}=\beta=\alpha$

So polar form of z is $\sec\alpha\big(\cos\alpha+\text{i}\sin\alpha\big)$

Case - 2: when $\alpha\in\Big(\frac{\pi}{2},\pi\Big]$

$|\text{z}|=\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha$

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

$\tan\beta=|\tan\alpha|=-\tan\alpha=\tan(\pi-\alpha)$

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

As z represented by a point in fourth quadarnt.

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

So polar form of z is $-\sec\alpha\big(\cos(\alpha-\pi)+\text{i}\sin(\alpha-\pi)\big).$