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Question Number 187548 by Rupesh123 last updated on 18/Feb/23

Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

z^9 +z^6 +z^3 =−1 ⇒z^9 +z^6 +z^3 +1=0 ⇒((z^(12) −1)/(z^3 −1))=0, z^3 ≠1⇒z≠1, z≠−(1/2)±((√3)/2)i ⇒ z^(12) =1=e^(2kπi) ⇒z=e^((k/6)πi) , k∈[0, 11] ⇒z_0 =1(rejected), z_1 =((√3)/2)+(1/2)i, z_2 =(1/2)+((√3)/2)i z_3 =i, z_4 =−(1/2)+((√3)/2)i(rejected), z_5 =−((√3)/2)+(1/2)i z_6 =−1(rejected from hypothesis), z_7 =−((√3)/2)−(1/2)i z_8 =−(1/2)−((√3)/2)i(rejected), z_9 =−i, z_(10) =(1/2)−((√3)/2)i z_(11) =((√3)/2)−(1/2)i

z9+z6+z3=−1⇒z9+z6+z3+1=0⇒z12−1z3−1=0,z3≠1⇒z≠1,z≠−12±32i⇒z12=1=e2kπi⇒z=ek6πi,k∈[0,11]⇒z0=1(rejected),z1=32+12i,z2=12+32iz3=i,z4=−12+32i(rejected),z5=−32+12iz6=−1(rejectedfromhypothesis),z7=−32−12iz8=−12−32i(rejected),z9=−i,z10=12−32iz11=32−12i

Commented by Rupesh123 last updated on 18/Feb/23

Good job, bro!

Answered by Ar Brandon last updated on 18/Feb/23

z^9 +z^6 +z^3 =−1⇒z^9 +z^6 +z^3 +1=0 ⇒z^6 (z^3 +1)+(z^3 +1)=0 ⇒(z^3 +1)(z^6 +1)=0 ⇒z^3 +1=0 ∨ z^6 +1=0 ⇒z^3 =−1=e^((2m+1)πi) ∨ z^6 =−1=e^((2n+1)πi) ⇒z=e^((((2m+1)/3))πi) , m∈[0, 2] ∨ z=e^((((2n+1)/6))πi) , n∈[0, 5] ⇒z={(1/2)+((√3)/2)i; −1; (1/2)−((√3)/2)i} ∨ z={((√3)/2)+(1/2)i; i; −((√3)/2)+(1/2)i; −((√3)/2)−(1/2)i; −i; ((√3)/2)−(1/2)i}

z9+z6+z3=−1⇒z9+z6+z3+1=0⇒z6(z3+1)+(z3+1)=0⇒(z3+1)(z6+1)=0⇒z3+1=0∨z6+1=0⇒z3=−1=e(2m+1)πi∨z6=−1=e(2n+1)πi⇒z=e(2m+13)πi,m∈[0,2]∨z=e(2n+16)πi,n∈[0,5]⇒z={12+32i;−1;12−32i}∨z={32+12i;i;−32+12i;−32−12i;−i;32−12i}

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