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x-iy-3-6-2-7-i-1-7-i-find-x-y-

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Question Number 129809 by mohammad17 last updated on 19/Jan/21
(x+iy)^3 =((−6+2(√7)i)/(1−(√7)i)) find x,y
(x+iy)3=−6+27i1−7ifindx,y
Answered by Olaf last updated on 20/Jan/21
(x+iy )^3 = ((−6+2(√7)i)/(1−(√7)i)) Let x+iy = ρe^(iθ) ∣((−6+2(√7)i)/(1−(√7)i))∣ = ((√(36+28))/( (√(1+7)))) = 2(√2) arg(((−6+2(√7)i)/(1−(√7)i))) = arg(−6+2(√7))−arg(1−(√7))+2kπ = arctan(−((√7)/3))−arctan(−(√7))+2kπ = arctan((√7))−arctan(((√7)/3))+2kπ = arctan((((√7)−((√7)/3))/(1+(√7)×((√7)/3))))+2kπ = arctan(((√7)/5))+2kπ ρ^3 e^(i3θ) = 2(√2)e^(iarctan(((√7)/5))+2kπ) ρ = (2)^(1/3) and θ = (1/3)arctan(((√7)/5))+(2/3)kπ x = ρcosθ and y = ρsinθ
(x+iy)3=−6+27i1−7iLetx+iy=ρeiθ∣−6+27i1−7i∣=36+281+7=22arg(−6+27i1−7i)=arg(−6+27)−arg(1−7)+2kπ=arctan(−73)−arctan(−7)+2kπ=arctan(7)−arctan(73)+2kπ=arctan(7−731+7×73)+2kπ=arctan(75)+2kπρ3ei3θ=22eiarctan(75)+2kπρ=23andθ=13arctan(75)+23kπx=ρcosθandy=ρsinθ

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