Question

The amplitude of $\frac{1+i\sqrt{3}}{\sqrt{3}+i}$ is
(a) $\frac{\mathrm{\pi }}{3}$
(b) $-\frac{\mathrm{\pi }}{3}$
(c) $\frac{\mathrm{\pi }}{6}$
(d) $-\frac{\mathrm{\pi }}{6}$

Answer 1

(c) $\frac{\pi }{6}$

$$\begin{gathered} \mathrm{Let}z=\frac{1+i\sqrt{3}}{\sqrt{3}+i} \\ \Rightarrow z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\times \frac{\sqrt{3}-i}{\sqrt{3}-i} \\ \Rightarrow z=\frac{\sqrt{3}+2i-\sqrt{3}{i}^2}{3-i^2} \\ \Rightarrow z=\frac{\sqrt{3}+\sqrt{3}+2i}{4} \\ \Rightarrow z=\frac{2\sqrt{3}+2i}{4} \\ \Rightarrow z=\frac{\sqrt{3}}{2}+\frac{1}{2}i \\ \tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ =\frac{1}{\sqrt{3}} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{6} \\ \mathrm{Since},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}. \\ \mathrm{Therefore},\mathrm{a}\mathrm{r}\mathrm{g}(z)={\tan }^{-1}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)=\frac{\mathrm{\pi }}{6} \end{gathered}$$