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Question Number 203919 by Samuel12 last updated on 02/Feb/24


	Answered by Mathspace last updated on 02/Feb/24

	$1 = e^{2 i k π} l e t z = {r e}^{i θ}$ $z^{5} = 1 \Leftrightarrow r^{5} e^{i 5 θ} = e^{i 2 k π} \Leftrightarrow$ $r = 1 a n d θ = \frac{2 k π}{5}$ $s o t h e r o o t s a r e z_{k} = e^{i \frac{2 k π}{5}}$ $a n d k \in [[0, 4]]$ $z_{0} = 1, z_{1} = e^{\frac{i 2 π}{5}}$ $z_{2} = e^{i \frac{4 π}{5}} = - e^{- \frac{i π}{5}} z_{3} = e^{i \frac{6 π}{5}} = - e^{\frac{i π}{5}}$ $z_{4} = e^{i \frac{8 π}{5}} = e^{i \frac{10 π - 2 π}{5}} = e^{- i \frac{2 π}{5}}$ $e^{\frac{i π}{5}} = c o s (\frac{π}{5}) + i s i n (\frac{π}{5})$ $a n d c o s (\frac{π}{5}) = \frac{1 + \sqrt{5}}{4} . . . . .$
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