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For any integer $ n$, the argument of $z=\large\frac{(\sqrt 3+i)^{4n+1}}{(1-\sqrt 3 i)^{4n}}$ is ?

$\begin{array}{1 1} (A) \large\frac{\pi}{6}\\ (B) \large\frac{\pi}{3}\\ (C) \large\frac{\pi}{2}\\ (D) \large\frac{2\pi}{3}\end{array} $

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Toolbox:
  • $r(\cos\theta+i \sin\theta)^n=r(\cos n \theta+i \sin n\theta)$
  • $arg\big(\large\frac{z_1}{z_2}\big)=$$arg(z_1)-arg(z_2)$
  • $r(\cos(2n\pi+\theta)+i \sin(2n\pi+\theta))=r(\cos\theta+i \sin\theta)$
$\sqrt 3+i=2(\cos\large\frac{\pi}{6}$$+i \sin\large\frac{\pi}{6})$
$1-\sqrt3 i=2(\cos(-\large\frac{\pi}{3}$$)+i \sin(-\large\frac{\pi}{3}$$))$
$(\sqrt 3+i)^{4n+1}=2^{4n+1}[\cos((4n+1)\large\frac{\pi}{6}$)$+i \sin((4n+1)\large\frac{\pi}{6}$$)]$
$arg((\sqrt 3+i)^{4n+1})=\large\frac{(4n+1)\pi}{6}$
$(1-\sqrt 3i)^{4n}=2^{4n}[\cos(-4n\large\frac{\pi}{3})$$+isin(-4n\large\frac{\pi}{3})]$
$arg((1-\sqrt3 i)^{4n})=\large-\frac{4n\pi}{3}$
$arg\bigg(\large\frac{(\sqrt 3+i)^{4n+1}}{(1-\sqrt 3i)^{4n}}\bigg)=\frac{(4n+1)\pi}{6}+\frac{4n\pi}{3}$
$=\large\frac{(12n+1)\pi}{6}=$$(2n\pi+\large\frac{\pi}{6})$
$=\large\frac{\pi}{6}$
answered Jul 28, 2013 by rvidyagovindarajan_1
edited May 26, 2014 by rohanmaheshwari0831_1
 

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