prove that the equation Z^n =1 have exacly n roots given by Z


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 57902 by maxmathsup by imad last updated on 14/Apr/19

$p r o v e t h a t t h e e q u a t i o n Z^{n} = 1 h a v e e x a c l y n r o o t s g i v e n b y$ $Z_{k} = e^{i \frac{2 k π}{n}} k \in [[0, n - 1]]$
Commented by Abdo msup. last updated on 14/Apr/19

$l e t p u t Z = r e^{i θ} s o Z^{n} = 1 \Leftrightarrow r^{n} e^{i n θ} = e^{i 2 k π} (k \in Z) \Rightarrow$ $r^{n} = 1 a n d n θ = 2 k π \Rightarrow r = 1 a n d θ = \frac{2 k π}{n}$ $s o t h e r o o t s o f t h i s e q u a t i o n a r e Z_{k} = e^{\frac{i 2 k π}{n}}$ $k \in Z d u e t o Z_{- k} = {\bar{Z}}_{k} w e c a n t a k e k ⩾ 0$ $i f k ⩽ n - 1 t h e r e i s e x a c l y n r o o t s a n d k \in [[0, n - 1]]$ $i f k > n l e t d i v i d e k b y n \Rightarrow \exists (q, r) f r o m N /$ $k = q n + r a n d 0 ⩽ r ⩽ n - 1 \Rightarrow$ $Z_{k} = e^{i \frac{2 (q n + r) π}{n}} = e^{i 2 q π} e^{\frac{i 2 r π}{n}} = e^{\frac{i 2 r π}{n}} = Z_{r}$ $s o t h e r e i s e x a t l y n r o o t s Z_{k} g i v e n b y$ $Z_{k} = e^{\frac{i 2 k π}{n}} a n d k \in [[0, n - 1 ⌉] .$
	Answered by mr W last updated on 14/Apr/19

	$l e t Z = 1 (\cos θ + i \sin θ) = e^{θ i}$ $Z^{n} = \cos n θ + i \sin n θ = 1$ $\Rightarrow \cos n θ = 1, \sin n θ = 0$ $\Rightarrow n θ = 2 k π$ $\Rightarrow θ = \frac{2 k π}{n} w i t h 0 ⩽ \frac{k}{n} < 1 o r 0 ⩽ k ⩽ n - 1$ $\Rightarrow Z = e^{i \frac{2 k π}{n}} w i t h 0 ⩽ k ⩽ n - 1$
	Commented by mr W last updated on 14/Apr/19

	$0 ⩽ θ < 2 π$ $t h e r e a r e e x a c t l y n v a l u e s f o r θ i n$ $t h i s r a n g e, i . e . t h e r e a r e e x a c t l y n$ $r o o t s .$
	Commented by maxmathsup by imad last updated on 14/Apr/19

	$y e s s i r b u t w h a t t h e r e i s o n l y n r o o t s . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum