# 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...

Note by Parag Zode
3 years, 5 months ago

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

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

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

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

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

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

- 3 years, 5 months ago