Question

The amplitude of $\frac{1+i\sqrt{3}}{\sqrt{3}+i}$ is

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

Answer 1

The correct option is C

$\frac{\pi }{6}$

$\text{Let}z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\Rightarrow z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\times \frac{\sqrt{3}-i}{\sqrt{3}-i}\Rightarrow z=\frac{\sqrt{3}+2i-\sqrt{3}{i}^2}{3-i^2}\Rightarrow z=\frac{\sqrt{3}+\sqrt{3}+2i}{4}\Rightarrow z=\frac{2\sqrt{3}+2i}{4}\Rightarrow z=\frac{\sqrt{3}}{2}+\frac{1}{2}itan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=\frac{1}{\sqrt{3}}\Rightarrow \alpha =\frac{\pi }{6}$

Since, z lies in the first quadrant.

Therefore, arg (z) =$ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$