prove-that-the-equation-Z-n-1-have-exacly-n-roots-given-by-Z-k-e-i-2kpi-n-k-0-n-1-

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 \dots