# The amplitude of

Question:

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

(a) $\frac{\pi}{3}$

(b) $-\frac{\pi}{3}$

(a) $\frac{\pi}{3}$

(b) $-\frac{\pi}{3}$

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

(d) $-\frac{\pi}{6}$

Solution:

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

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

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

$\Rightarrow z=\frac{\sqrt{3}+2 i-\sqrt{3} i^{2}}{3-i^{2}}$

$\Rightarrow z=\frac{\sqrt{3}+\sqrt{3}+2 i}{4}$

$\Rightarrow z=\frac{2 \sqrt{3}+2 i}{4}$

$\Rightarrow z=\frac{\sqrt{3}}{2}+\frac{1}{2} i$

$\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$=\frac{1}{\sqrt{3}}$

$\Rightarrow \alpha=\frac{\pi}{6}$

Since, $z$ lies in the first quadrant.

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