Question Number 211105 by peter frank last updated on 28/Aug/24
Answered by mm1342 last updated on 28/Aug/24
Commented by peter frank last updated on 28/Aug/24
Answered by Frix last updated on 28/Aug/24
![z^5 =e^(2πni) ; n∈Z z_n =e^(((2πn)/5)i) ⇔ Z_5 ={z_n =e^(((2πn)/5)i) ∣n∈Z} [Further we notice e^(((2πn)/5)i) =e^(((2(n+5k)π)/5)i) ; k∈Z] z_m z_n =e^(((2πm)/5)i) e^(((2πn)/5)i) =e^(((2π(m+n))/5)i) m, n ∈Z ⇒ z_m , z_n ∈Z_5 m+n∈Z ⇒ z_(m+n) ∈Z_5 z_1 =e^(((4π)/5)i) =z_2 ∈Z_5 [4=2n ⇔ n=2] z_2 =e^(((8π)/5)i) =z_4 ∈Z_5 [8=2n ⇔ n=4] z_1 z_2 =e^(((12π)/5)i) =z_6 ∈Z_5 [12=2n ⇔ n=6] [12=2(n+5k) ⇔ n=1∧k=1]](https://www.tinkutara.com/question/Q211112.png)
Commented by peter frank last updated on 28/Aug/24
