Find-all-those-roots-of-the-equation-z-12-56z-6-512-0-whose-imaginary-part-is-positive-

Question Number 109090 by ajfour last updated on 21/Aug/20

$${Find}\:{all}\:{those}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\:\boldsymbol{{z}}^{\mathrm{12}} −\mathrm{56}\boldsymbol{{z}}^{\mathrm{6}} −\mathrm{512}=\mathrm{0}\:\:{whose}\:{imaginary} \\ $$$${part}\:{is}\:{positive}. \\ $$

Answered by floor(10²Eta[1]) last updated on 21/Aug/20

z^6 =y y^2 −56y−512=0 y=((56±72)/2)=28±36⇒y=−8∨y=64 (1):z^6 =−8=8(cosπ+i.sinπ) z=ρ(cosθ+i.sinθ)⇒z^6 =ρ^6 (cos(6θ)+i.sin(6θ))=8(cosπ+i.sinπ) ⇒ρ=(√2) ⇒cos(6θ)=cosπ∴6θ=π+2kπ⇒θ_k =((π+2kπ)/6), 0≤k<6 ⇒sin(6θ)=sinπ∴θ_k =((π+2kπ)/6), 0≤k<6 ⇒z_k =(√2)(cosθ_k +i.sinθ_k ) z_0 =(((√6)+i(√2))/2)∧z_1 =i(√2)∧z_2 =((−(√6)+i(√2))/2) z_3 =((−(√6)−i(√2))/2)∧z_4 =−i(√2)∧z_5 =(((√6)−i(√2))/2) (2):z^6 =64=64(cos(2π)+i.sin(2π)) θ_k =((2kπ)/6), 0≤k<6 z_k =2(cosθ_k +i.sinθ_k ) z_0 =2∧z_1 =1+i(√3)∧z_2 =−1+i(√3) z_3 =−2∧z_4 =−1−i(√3)∧z_5 =1−i(√3) so the roots that have the im. positive are: (((√6)+i(√2))/2), i(√2), ((−6+i(√2))/2), 1+i(√3), −1+i(√3)

$$\mathrm{z}^{\mathrm{6}} =\mathrm{y} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{56y}−\mathrm{512}=\mathrm{0} \\ $$$$\mathrm{y}=\frac{\mathrm{56}\pm\mathrm{72}}{\mathrm{2}}=\mathrm{28}\pm\mathrm{36}\Rightarrow\mathrm{y}=−\mathrm{8}\vee\mathrm{y}=\mathrm{64} \\ $$$$\left(\mathrm{1}\right):\mathrm{z}^{\mathrm{6}} =−\mathrm{8}=\mathrm{8}\left(\mathrm{cos}\pi+\mathrm{i}.\mathrm{sin}\pi\right) \\ $$$$\mathrm{z}=\rho\left(\mathrm{cos}\theta+\mathrm{i}.\mathrm{sin}\theta\right)\Rightarrow\mathrm{z}^{\mathrm{6}} =\rho^{\mathrm{6}} \left(\mathrm{cos}\left(\mathrm{6}\theta\right)+\mathrm{i}.\mathrm{sin}\left(\mathrm{6}\theta\right)\right)=\mathrm{8}\left(\mathrm{cos}\pi+\mathrm{i}.\mathrm{sin}\pi\right) \\ $$$$\Rightarrow\rho=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\left(\mathrm{6}\theta\right)=\mathrm{cos}\pi\therefore\mathrm{6}\theta=\pi+\mathrm{2k}\pi\Rightarrow\theta_{\mathrm{k}} =\frac{\pi+\mathrm{2k}\pi}{\mathrm{6}},\:\mathrm{0}\leqslant\mathrm{k}<\mathrm{6} \\ $$$$\Rightarrow\mathrm{sin}\left(\mathrm{6}\theta\right)=\mathrm{sin}\pi\therefore\theta_{\mathrm{k}} =\frac{\pi+\mathrm{2k}\pi}{\mathrm{6}},\:\mathrm{0}\leqslant\mathrm{k}<\mathrm{6} \\ $$$$\Rightarrow\mathrm{z}_{\mathrm{k}} =\sqrt{\mathrm{2}}\left(\mathrm{cos}\theta_{\mathrm{k}} +\mathrm{i}.\mathrm{sin}\theta_{\mathrm{k}} \right) \\ $$$$\mathrm{z}_{\mathrm{0}} =\frac{\sqrt{\mathrm{6}}+\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge\mathrm{z}_{\mathrm{1}} =\mathrm{i}\sqrt{\mathrm{2}}\wedge\mathrm{z}_{\mathrm{2}} =\frac{−\sqrt{\mathrm{6}}+\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{z}_{\mathrm{3}} =\frac{−\sqrt{\mathrm{6}}−\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge\mathrm{z}_{\mathrm{4}} =−\mathrm{i}\sqrt{\mathrm{2}}\wedge\mathrm{z}_{\mathrm{5}} =\frac{\sqrt{\mathrm{6}}−\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right):\mathrm{z}^{\mathrm{6}} =\mathrm{64}=\mathrm{64}\left(\mathrm{cos}\left(\mathrm{2}\pi\right)+\mathrm{i}.\mathrm{sin}\left(\mathrm{2}\pi\right)\right) \\ $$$$\theta_{\mathrm{k}} =\frac{\mathrm{2k}\pi}{\mathrm{6}},\:\mathrm{0}\leqslant\mathrm{k}<\mathrm{6} \\ $$$$\mathrm{z}_{\mathrm{k}} =\mathrm{2}\left(\mathrm{cos}\theta_{\mathrm{k}} +\mathrm{i}.\mathrm{sin}\theta_{\mathrm{k}} \right) \\ $$$$\mathrm{z}_{\mathrm{0}} =\mathrm{2}\wedge\mathrm{z}_{\mathrm{1}} =\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\wedge\mathrm{z}_{\mathrm{2}} =−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}} \\ $$$$\mathrm{z}_{\mathrm{3}} =−\mathrm{2}\wedge\mathrm{z}_{\mathrm{4}} =−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\wedge\mathrm{z}_{\mathrm{5}} =\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{that}\:\mathrm{have}\:\mathrm{the}\:\mathrm{im}.\:\mathrm{positive}\:\mathrm{are}: \\ $$$$\frac{\sqrt{\mathrm{6}}+\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}},\:\mathrm{i}\sqrt{\mathrm{2}},\:\frac{−\mathrm{6}+\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}},\:\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}},\:−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}} \\ $$

Commented by ajfour last updated on 21/Aug/20

(((√6)+i(√2))/2), i(√2), ((−(√6)+i(√2))/2), 1+i(√3), −1+i(√3) Yes Sir, thanks plentifully!

$$\frac{\sqrt{\mathrm{6}}+\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}},\:\mathrm{i}\sqrt{\mathrm{2}},\:\frac{−\sqrt{\mathrm{6}}+\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}},\:\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}},\:−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}} \\ $$$${Yes}\:{Sir},\:\:{thanks}\:{plentifully}! \\ $$

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