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

$$Findallthoserootsoftheequation$$$${\mathit{z}}^{12}-56{\mathit{z}}^6-512=0whoseimaginary$$$$partispositive.$$

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

$${\mathrm{z}}^6=\mathrm{y}$$$${\mathrm{y}}^2-56\mathrm{y}-512=0$$$$\mathrm{y}=\frac{56\pm 72}{2}=28\pm 36\Rightarrow \mathrm{y}=-8\vee \mathrm{y}=64$$$$\left(1\right):{\mathrm{z}}^6=-8=8(\cos \pi +\mathrm{i}.\sin \pi )$$$$\mathrm{z}=\rho (\cos \theta +\mathrm{i}.\sin \theta )\Rightarrow {\mathrm{z}}^6={\rho }^6(\cos \left(6\theta \right)+\mathrm{i}.\sin \left(6\theta \right))=8(\cos \pi +\mathrm{i}.\sin \pi )$$$$\Rightarrow \rho =\sqrt{2}$$$$\Rightarrow \cos \left(6\theta \right)=\cos \pi \therefore 6\theta =\pi +2\mathrm{k}\pi \Rightarrow {\theta }_{\mathrm{k}}=\frac{\pi +2\mathrm{k}\pi }{6},0⩽\mathrm{k}<6$$$$\Rightarrow \sin \left(6\theta \right)=\sin \pi \therefore {\theta }_{\mathrm{k}}=\frac{\pi +2\mathrm{k}\pi }{6},0⩽\mathrm{k}<6$$$$\Rightarrow {\mathrm{z}}_{\mathrm{k}}=\sqrt{2}(\cos {\theta }_{\mathrm{k}}+\mathrm{i}.\sin {\theta }_{\mathrm{k}})$$$${\mathrm{z}}_0=\frac{\sqrt{6}+\mathrm{i}\sqrt{2}}{2}\wedge {\mathrm{z}}_1=\mathrm{i}\sqrt{2}\wedge {\mathrm{z}}_2=\frac{-\sqrt{6}+\mathrm{i}\sqrt{2}}{2}$$$${\mathrm{z}}_3=\frac{-\sqrt{6}-\mathrm{i}\sqrt{2}}{2}\wedge {\mathrm{z}}_4=-\mathrm{i}\sqrt{2}\wedge {\mathrm{z}}_5=\frac{\sqrt{6}-\mathrm{i}\sqrt{2}}{2}$$$$\left(2\right):{\mathrm{z}}^6=64=64(\cos \left(2\pi \right)+\mathrm{i}.\sin \left(2\pi \right))$$$${\theta }_{\mathrm{k}}=\frac{2\mathrm{k}\pi }{6},0⩽\mathrm{k}<6$$$${\mathrm{z}}_{\mathrm{k}}=2(\cos {\theta }_{\mathrm{k}}+\mathrm{i}.\sin {\theta }_{\mathrm{k}})$$$${\mathrm{z}}_0=2\wedge {\mathrm{z}}_1=1+\mathrm{i}\sqrt{3}\wedge {\mathrm{z}}_2=-1+\mathrm{i}\sqrt{3}$$$${\mathrm{z}}_3=-2\wedge {\mathrm{z}}_4=-1-\mathrm{i}\sqrt{3}\wedge {\mathrm{z}}_5=1-\mathrm{i}\sqrt{3}$$$$\mathrm{so}\mathrm{the}\mathrm{roots}\mathrm{that}\mathrm{have}\mathrm{the}\mathrm{im}.\mathrm{positive}\mathrm{are}:$$$$\frac{\sqrt{6}+\mathrm{i}\sqrt{2}}{2},\mathrm{i}\sqrt{2},\frac{-6+\mathrm{i}\sqrt{2}}{2},1+\mathrm{i}\sqrt{3},-1+\mathrm{i}\sqrt{3}$$

Commented by ajfour last updated on 21/Aug/20

$$\frac{\sqrt{6}+\mathrm{i}\sqrt{2}}{2},\mathrm{i}\sqrt{2},\frac{-\sqrt{6}+\mathrm{i}\sqrt{2}}{2},1+\mathrm{i}\sqrt{3},-1+\mathrm{i}\sqrt{3}$$$$YesSir,thanksplentifully!$$

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