One of the 15^{text{th}} roots of -1 is | vedclass

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(C) The $n^{	ext{th}}$ roots of a complex number $z = r(\cos 	heta + i \sin 	heta)$ are given by $z_k = r^{1/n} \operatorname{cis} \left( \frac{	h

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$\operatorname{cis} \frac{13 \pi}{15}$

Vedclass · Answer

(C) The $n^{	ext{th}}$ roots of a complex number $z = r(\cos 	heta + i \sin 	heta)$ are given by $z_k = r^{1/n} \operatorname{cis} \left( \frac{	heta + 2k\pi}{n} ight)$ for $k = 0, 1, \dots, n-1$. For $z = -1$,we have $r = 1$ and $	heta = \pi$. Thus,the $15^{	ext{th}}$ roots are $\operatorname{cis} \left( \frac{\pi + 2k\pi}{15} ight)$ for $k = 0, 1, \dots, 14$. For $k = 6$,the root is $\operatorname{cis} \left( \frac{\pi + 12\pi}{15} ight) = \operatorname{cis} \left( \frac{13\pi}{15} ight)$.

One of the $15^{\text{th}}$ roots of $-1$ is

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