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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

1)findxy∈R2)findx,y∈Z(x+2yi)6=8i

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.

letz=x+2yiz6=8i=(2)6eπ2i⇒z=2e(π12+kπ3)i(k=0,1,2,...,5)k=0:z=2e(π12)i=3+12+3−12ix=3+12y=3−14k=1:z=2e(π12+π3)i=3−12+3+12ix=3−12y=3+14k=2:z=2e(π12+2π3)i=−1+ix=−1y=12k=3:z=2e(π12+3π3)i=−3−12+−3+12ix=−3+12y=−3−14k=4:z=2e(π12+4π3)i=−3+12+−3−12ix=−3−12y=−3+14k=5:z=2e(π12+5π3)i=1−ix=1y=−12forx,y∈Cx+2yi=zxandycannotbeuniquelydetermined.

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}

⇒(x−2yi)6=−8i⇒(x2+4y2)6=64⇒x2+4y2=2⇒x=2cos(θ)y=12sin(θ)θ∈[0,2π]⇒8(ei6θ)=8i⇒6θ=π2+2kπ⇒θ∈π12+kπ3,k∈{0,......5}x=2cos(θ),y=sin(θ)2,θ∈{π12+kπ3,0⩽k⩽5}

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