Question

For any integer $n$, $arg\left[\frac{{(\sqrt{3}+i)}^{4n+1}}{{(1-i\sqrt{3})}^{4n}}\right]$ equals

• A. $\frac{\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{6}$
• C. $\frac{2\mathrm{\pi }}{3}$
• D. $\frac{5\mathrm{\pi }}{6}$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{6}$

Explanation for the correct option:

Find the value of given expression:

Given that, $z=\frac{{\left(\sqrt{3}+i\right)}^{4n+1}}{{\left(1-i\sqrt{3}\right)}^{4n}}$

Multiply and divide numerator and denominator by $2$

$$\begin{gathered} =\frac{2{\left({\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{i}{2}}\right)}^{4n+1}}{2{\left({\displaystyle \frac{1}{2}}-i{\displaystyle \frac{\sqrt{3}}{2}}\right)}^{4n}} \\ =\frac{2^{4n+1}\left(\cos {\displaystyle \frac{\mathrm{\pi }}{6}}+i\sin {\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{2^{4n}\left(\cos {\displaystyle \frac{\mathrm{\pi }}{3}}-i\sin {\displaystyle \frac{\mathrm{\pi }}{3}}\right)} \\ =\frac{2{\left(e^{{\displaystyle \raisebox{1ex}{$i\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\right)}^{4n+1}}{{\left(e^{{\displaystyle \raisebox{1ex}{$-i\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^{4n}} \\ =\frac{2e^{{\displaystyle \raisebox{1ex}{$i(4n+1)\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}}{e^{{\displaystyle \raisebox{1ex}{$-i4n\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ =2e^{\raisebox{1ex}{$i(12n+1)\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.} \\ =2e^{2ni\mathrm{\pi }}{e}^{\raisebox{1ex}{$i\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.} \\ =2e^{\raisebox{1ex}{$i\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.} \end{gathered}$$

$\mathbf{\therefore }\mathit{a}\mathit{r}\mathit{g}\mathbf{}\mathit{z}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{6}}$

Hence, Option ‘B’ is Correct.