Question

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

• A. ${60}^o$
• B. ${120}^o$
• C. ${240}^o$
• D. ${300}^o$

Answer 1

The correct option is A ${60}^o$

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

By eativnalization

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

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

$=\frac{1+(-1)\left(3\right)-2i\sqrt{3}}{1+(-1)\left(3\right)}[\because i^2=-1]$

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

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

$\Rightarrow P(1,\sqrt{3})$

$\theta ={\tan }^{-1}\left|\frac{\sqrt{3}}{1}\right|={\tan }^{-1}\sqrt{3}={60}^o$

$\Rightarrow \theta ={60}^o$