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:- $(a)\dfrac{\pi}{6}$

$(b)\dfrac{\pi}{3}$

$(c)\dfrac{\pi}{2}$

$(d)\dfrac{2\pi}{3}$

This question is based on COMPLEX NUMBERS..

Please post the solution in Detail...

#Algebra #ComplexNumbers

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

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

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

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

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

$\arg z = \boxed{\dfrac{\pi}{6}}$

Alan Enrique Ontiveros Salazar - 6 years, 5 months ago

Is the Answer Pi/6 ???

A Former Brilliant Member - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$