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](https://www.tinkutara.com/question/Q187552.png)
$${z}^{\mathrm{9}} +{z}^{\mathrm{6}} +{z}^{\mathrm{3}} =−\mathrm{1}\:\Rightarrow{z}^{\mathrm{9}} +{z}^{\mathrm{6}} +{z}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{z}^{\mathrm{12}} −\mathrm{1}}{{z}^{\mathrm{3}} −\mathrm{1}}=\mathrm{0},\:{z}^{\mathrm{3}} \:\neq\mathrm{1}\Rightarrow{z}\neq\mathrm{1},\:{z}\neq−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$$\:\Rightarrow\:{z}^{\mathrm{12}} =\mathrm{1}={e}^{\mathrm{2}{k}\pi{i}} \\ $$$$\Rightarrow{z}={e}^{\frac{{k}}{\mathrm{6}}\pi{i}} ,\:{k}\in\left[\mathrm{0},\:\mathrm{11}\right] \\ $$$$\Rightarrow{z}_{\mathrm{0}} =\mathrm{1}\left(\mathrm{rejected}\right),\:{z}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i},\:{z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$${z}_{\mathrm{3}} ={i},\:{z}_{\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\left(\mathrm{rejected}\right),\:{z}_{\mathrm{5}} =−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${z}_{\mathrm{6}} =−\mathrm{1}\left(\mathrm{rejected}\:\mathrm{from}\:\mathrm{hypothesis}\right),\:{z}_{\mathrm{7}} =−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${z}_{\mathrm{8}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\left(\mathrm{rejected}\right),\:{z}_{\mathrm{9}} =−{i},\:{z}_{\mathrm{10}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$${z}_{\mathrm{11}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$
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}](https://www.tinkutara.com/question/Q187573.png)
$${z}^{\mathrm{9}} +{z}^{\mathrm{6}} +{z}^{\mathrm{3}} =−\mathrm{1}\Rightarrow{z}^{\mathrm{9}} +{z}^{\mathrm{6}} +{z}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{z}^{\mathrm{6}} \left({z}^{\mathrm{3}} +\mathrm{1}\right)+\left({z}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({z}^{\mathrm{3}} +\mathrm{1}\right)\left({z}^{\mathrm{6}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{z}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\:\vee\:{z}^{\mathrm{6}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{z}^{\mathrm{3}} =−\mathrm{1}={e}^{\left(\mathrm{2}{m}+\mathrm{1}\right)\pi{i}} \:\vee\:{z}^{\mathrm{6}} =−\mathrm{1}={e}^{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi{i}} \\ $$$$\Rightarrow{z}={e}^{\left(\frac{\mathrm{2}{m}+\mathrm{1}}{\mathrm{3}}\right)\pi{i}} ,\:{m}\in\left[\mathrm{0},\:\mathrm{2}\right]\:\vee\:{z}={e}^{\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{6}}\right)\pi{i}} ,\:{n}\in\left[\mathrm{0},\:\mathrm{5}\right] \\ $$$$\Rightarrow{z}=\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i};\:−\mathrm{1};\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right\} \\ $$$$\vee\:{z}=\left\{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i};\:{i};\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i};\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{i};\:−{i};\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{i}\right\} \\ $$