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Question Number 82877 by M±th+et£s last updated on 25/Feb/20

1)find xy∈R 2)find x,y∈Z (x+2yi)^6 =8i

$$\left.\mathrm{1}\right){find}\:{xy}\in{R} \\ $$$$\left.\mathrm{2}\right){find}\:{x},{y}\in{Z} \\ $$$$\left({x}+\mathrm{2}{yi}\right)^{\mathrm{6}} =\mathrm{8}{i} \\ $$

Commented by mr W last updated on 25/Feb/20

let z=x+2yi z^6 =8i=((√2))^6 e^((π/2)i) ⇒z=(√2)e^(((π/(12))+((kπ)/3))i) (k=0,1,2,...,5) k=0: z=(√2)e^(((π/(12)))i) =(((√3)+1)/2)+(((√3)−1)/2)i x=(((√3)+1)/2) y=(((√3)−1)/4) k=1: z=(√2)e^(((π/(12))+(π/3))i) =(((√3)−1)/2)+(((√3)+1)/2)i x=(((√3)−1)/2) y=(((√3)+1)/4) k=2: z=(√2)e^(((π/(12))+((2π)/3))i) =−1+i x=−1 y=(1/2) k=3: z=(√2)e^(((π/(12))+((3π)/3))i) =((−(√3)−1)/2)+((−(√3)+1)/2)i x=−(((√3)+1)/2) y=−(((√3)−1)/4) k=4: z=(√2)e^(((π/(12))+((4π)/3))i) =((−(√3)+1)/2)+((−(√3)−1)/2)i x=−(((√3)−1)/2) y=−(((√3)+1)/4) k=5: z=(√2)e^(((π/(12))+((5π)/3))i) =1−i x=1 y=−(1/2) for x,y ∈C x+2yi=z x and y can not be uniquely determined.

$${let}\:{z}={x}+\mathrm{2}{yi} \\ $$$${z}^{\mathrm{6}} =\mathrm{8}{i}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}} {e}^{\frac{\pi}{\mathrm{2}}{i}} \\ $$$$\Rightarrow{z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}+\frac{{k}\pi}{\mathrm{3}}\right){i}} \:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{5}\right) \\ $$$${k}=\mathrm{0}: \\ $$$${z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}\right){i}} =\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}{i} \\ $$$${x}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${k}=\mathrm{1}: \\ $$$${z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}+\frac{\pi}{\mathrm{3}}\right){i}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}}{i} \\ $$$${x}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${k}=\mathrm{2}: \\ $$$${z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right){i}} =−\mathrm{1}+{i} \\ $$$${x}=−\mathrm{1} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${k}=\mathrm{3}: \\ $$$${z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}+\frac{\mathrm{3}\pi}{\mathrm{3}}\right){i}} =\frac{−\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}+\frac{−\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}}{i} \\ $$$${x}=−\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}} \\ $$$${y}=−\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${k}=\mathrm{4}: \\ $$$${z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}+\frac{\mathrm{4}\pi}{\mathrm{3}}\right){i}} =\frac{−\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}}+\frac{−\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}{i} \\ $$$${x}=−\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${y}=−\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${k}=\mathrm{5}: \\ $$$${z}=\sqrt{\mathrm{2}}{e}^{\left(\frac{\pi}{\mathrm{12}}+\frac{\mathrm{5}\pi}{\mathrm{3}}\right){i}} =\mathrm{1}−{i} \\ $$$${x}=\mathrm{1} \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${for}\:{x},{y}\:\in\mathbb{C} \\ $$$${x}+\mathrm{2}{yi}={z} \\ $$$${x}\:{and}\:{y}\:{can}\:{not}\:{be}\:{uniquely}\:{determined}. \\ $$

Answered by mind is power last updated on 25/Feb/20

⇒(x−2yi)^6 =−8i ⇒(x^2 +4y^2 )^6 =64 ⇒x^2 +4y^2 =2 ⇒x=(√2)cos(θ) y=(1/(√2))sin(θ) θ∈[0,2π] ⇒8(e^(i6θ) )=8i ⇒6θ=(π/2)+2kπ⇒θ∈(π/(12))+((kπ)/3),k∈{0,......5} x=(√2)cos(θ),y=((sin(θ))/(√2)),θ∈{(π/(12))+((kπ)/3),0≤k≤5}

$$\Rightarrow\left({x}−\mathrm{2}{yi}\right)^{\mathrm{6}} =−\mathrm{8}{i} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)^{\mathrm{6}} =\mathrm{64} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}}{cos}\left(\theta\right) \\ $$$${y}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{sin}\left(\theta\right) \\ $$$$\theta\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\Rightarrow\mathrm{8}\left({e}^{{i}\mathrm{6}\theta} \right)=\mathrm{8}{i} \\ $$$$\Rightarrow\mathrm{6}\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\Rightarrow\theta\in\frac{\pi}{\mathrm{12}}+\frac{{k}\pi}{\mathrm{3}},{k}\in\left\{\mathrm{0},......\mathrm{5}\right\} \\ $$$${x}=\sqrt{\mathrm{2}}{cos}\left(\theta\right),{y}=\frac{{sin}\left(\theta\right)}{\sqrt{\mathrm{2}}},\theta\in\left\{\frac{\pi}{\mathrm{12}}+\frac{{k}\pi}{\mathrm{3}},\mathrm{0}\leqslant{k}\leqslant\mathrm{5}\right\} \\ $$$$ \\ $$

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