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Question Number 211105 by peter frank last updated on 28/Aug/24

Answered by mm1342 last updated on 28/Aug/24

z=z_1 z_2 =cos((12π)/5)+isin((12π)/5)  =cos((2π)/5)+isin((2π)/5)=e^(((2π)/5)i ) ⇒z^5 =e^(2πi) =1  ⇒z^5 −1=0  ✓

z=z1z2=cos12π5+isin12π5=cos2π5+isin2π5=e2π5iz5=e2πi=1z51=0

Commented by peter frank last updated on 28/Aug/24

thank you

thankyou

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]

z5=e2πni;nZzn=e2πn5iZ5={zn=e2πn5inZ}[Furtherwenoticee2πn5i=e2(n+5k)π5i;kZ]zmzn=e2πm5ie2πn5i=e2π(m+n)5im,nZzm,znZ5m+nZzm+nZ5z1=e4π5i=z2Z5[4=2nn=2]z2=e8π5i=z4Z5[8=2nn=4]z1z2=e12π5i=z6Z5[12=2nn=6][12=2(n+5k)n=1k=1]

Commented by peter frank last updated on 28/Aug/24

thank you

thankyou

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