Question

The argument of \( \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \) is:

• A. $\dfrac{\pi}{3}$
• B. $\dfrac{2\pi}{3}$
• C. $\dfrac{4\pi}{3}$
• D. $-\dfrac{2\pi}{3}$

Hint:

Always simplify complex expressions by multiplying with conjugate and find quadrant for correct angle.

Answer 1

The Correct Option is D

Let $z = \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$
Multiply numerator and denominator by the conjugate of denominator:
\[ z = \dfrac{(1 - i\sqrt{3})(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})} = \dfrac{(1 - i\sqrt{3})^2}{1 + 3} = \dfrac{1 - 2i\sqrt{3} - 3}{4} = \dfrac{-2 - 2i\sqrt{3}}{4} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i \]
Now find the argument: $\arg(z) = \tan^{-1}\left(\dfrac{-\sqrt{3}/2}{-1/2}\right) = \tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$
But the complex number lies in the third quadrant $\Rightarrow$ Argument = $-\dfrac{2\pi}{3}$