Question

If $Z=\frac{1-\sqrt{3}i}{1+\sqrt{3}i}$ then find $arg\left(z\right).$

• A. $-\frac{2\pi }{3}$
• B. $\frac{2\pi }{5}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

Answer: (D)

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

Consider the following question.

$Z=\frac{1-\sqrt{3}i}{1+\sqrt{3i}}$

$=\frac{(1-\sqrt{3}i)}{(1+\sqrt{3i})}\times \frac{(1-\sqrt{3}i)}{(1-\sqrt{3}i)}$

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

$=\frac{1+3i^2-2\sqrt{3}}{4}$

$=\frac{1-3-2\sqrt{3}}{4}$

$=\frac{-1-i\sqrt{3}}{2}$

$=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

$\arg (-\frac{1}{2}-i\frac{\sqrt{3}}{2})=-\pi +\frac{\pi }{3}$

$=\frac{2\pi }{3}$

Hence, this is the required answer.