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Can someone help me in approaching and solving this math problem?

For any integer \(n\) the argument of \[z=\dfrac{(\sqrt{3}+i)^{4n+1}}{(1-i\sqrt{3})^{4n}}\] is:- \[(a)\dfrac{\pi}{6}\]




This question is based on COMPLEX NUMBERS..

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Note by Parag Zode
2 years, 5 months ago

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\(z=\dfrac{(2 cis (\frac{\pi}{6}))^{4n+1}}{(2 cis (-\frac{\pi}{3}))^{4n}}\)

\(z=\left(\dfrac{2 cis (\frac{\pi}{6})}{2 cis (-\frac{\pi}{3})}\right)^{4n} 2 cis (\frac{\pi}{6})\)

\(z=2(cis (\frac{\pi}{2}))^{4n} cis (\frac{\pi}{6})\)

\(z=2(i)^{4n} cis (\frac{\pi}{6})\)

\(z=2 cis (\frac{\pi}{6})\)

\(\arg z = \boxed{\dfrac{\pi}{6}}\) Alan Enrique Ontiveros Salazar · 2 years, 4 months ago

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Is the Answer Pi/6 ??? Abhineet Nayyar · 2 years, 5 months ago

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