Question

Argument of complex number $\frac{1+\sqrt{3}i}{\sqrt{3}+i}$ is

• A. $\frac{\pi }{3}$
• B. $\frac{5\pi }{6}$
• C. $-\frac{5\pi }{6}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is D $\frac{\pi }{6}$

Let, $z=\frac{1+\sqrt{3}i}{\sqrt{3}+i}$
$\Rightarrow z=\frac{1+\sqrt{3}i}{\sqrt{3}+i}\times \frac{\sqrt{3}-i}{\sqrt{3}-i}\Rightarrow z=\frac{\sqrt{3}-i+3i+\sqrt{3}}{3+1}\Rightarrow z=\frac{\sqrt{3}+i}{2}$
Now, $\tan \alpha =\left|\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|=\frac{\pi }{6}$
Clearly, $z$ lies in first quadrant.
$\therefore \arg \left(z\right)=\alpha =\frac{\pi }{6}$