Find the principal argument of the complex number z = 1 + √(3)i 1 - √(3)i — JEE Mathematics

Simplify $z$ by multiplying the numerator and denominator by the conjugate of the denominator:

$$z = \frac{(1 + \sqrt{3}i)(1 + \sqrt{3}i)}{(1 - \sqrt{3}i)(1 + \sqrt{3}i)} = \frac{1 + 2\sqrt{3}i - 3}{1 + 3} = \frac{-2 + 2\sqrt{3}i}{4} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$$

The complex number lies in the second quadrant ($Re(z) < 0, Im(z) > 0$).

$$\theta = \pi - \tan^{-1}\left| \frac{Im(z)}{Re(z)} \right| = \pi - \tan^{-1}\left| \frac{\sqrt{3}/2}{-1/2} \right| = \pi - \tan^{-1}(\sqrt{3}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

Answer: $\frac{2\pi}{3}$