The argument of $\frac{1+i \sqrt{3}}{\sqrt{3}+i}, i=\sqrt{-1}$ is

The argument of $\frac{1+i \sqrt{3}}{\sqrt{3}+i}, i=\sqrt{-1}$ is
  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{6}$
  4. $\frac{\pi}{2}$

Solution

$\begin{aligned} & \text { Let } z=\frac{1+i \sqrt{3}}{\sqrt{3}+i} \\ = & \frac{(1+i \sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} \\ \therefore \quad z & =\frac{\sqrt{3}}{2}+\frac{1}{2} i\end{aligned}$ Argument of $\begin{aligned} z & =\tan ^{-1}\left(\frac{b}{a}\right) \\ & =\tan ^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \\ & =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ & =\frac{\pi}{6} \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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