find x,y in R (x+yi)^3 =((1+(√(15)) i)/((√5) − (√3) i))


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Question Number 59588 by aliesam last updated on 12/May/19

$f i n d x, y i n R$ ${(x + y i)}^{3} = \frac{1 + \sqrt{15} i}{\sqrt{5} - \sqrt{3} i}$
Commented by maxmathsup by imad last updated on 12/May/19
$⇒x+iy =^3 (√((1+(√(15))i)/((√5)−(√3)i))) let solve Z^3 =((1+(√(15))i)/((√5)−(√3)i)) we have ∣1+(√(15))i∣ =(√(1+15))=4 and ∣(√5)−(√3)i∣ =(√(5 +3))=2(√(2 ))⇒∣((1+(√(15))i)/((√5)−(√3)i))∣=(4/(2(√2))) =(√(2 )) 1+(√(15))i =4{(1/4)+((√(15))/4)i} =r e^(iθ) ⇒r=4 and cosθ=(1/4) and sinθ =((√(15))/4) ⇒ tanθ =(√(15)) ⇒θ =arctan((√(15))) also (√5)−(√3)i =2(√2){((√5)/(2(√2))) −((√3)/(2(√2)))i}=r^′ e^(iθ′) ⇒ r^′ =2(√2) and θ^′ =arctan(((−(√3))/(√5))) ⇒arg(((1+(√(15))i)/((√5)−(√3)i)))≡ arctan((√(15))) +arctan(((√3)/(√5)))[2π] ⇒ ⇒((1+(√(15))i)/((√5)−(√3)i)) =(√2) e^(i(arctan((√(15)))+arctan(((√3)/(√5))))) =Z^3 ⇒ Z =^6 (√2)e^((1/3)(arctan((√(15)))+arctan(((√3)/(√5))))i) =^6 (√2){cos((1/3)(arctan((√(15)))+arctan(((√3)/(√5)))))+i sin((1/3)(arctan((√(15)))+arctan(((√3)/(√5)))) ⇒ x =^6 (√2)cos((1/3)(arctan((√(15)))+arctan(((√3)/(√5))))) and y =^6 (√2)sin((1/3)(arctan((√(15))) +arctan(((√3)/(√5))))) .$
$\Rightarrow x + i y =^{3} \sqrt{\frac{1 + \sqrt{15} i}{\sqrt{5} - \sqrt{3} i}} l e t s o l v e Z^{3} = \frac{1 + \sqrt{15} i}{\sqrt{5} - \sqrt{3} i} w e h a v e$ $∣ 1 + \sqrt{15} i ∣ = \sqrt{1 + 15} = 4 a n d ∣ \sqrt{5} - \sqrt{3} i ∣ = \sqrt{5 + 3} = 2 \sqrt{2} \Rightarrow∣ \frac{1 + \sqrt{15} i}{\sqrt{5} - \sqrt{3} i} ∣= \frac{4}{2 \sqrt{2}} = \sqrt{2}$ $1 + \sqrt{15} i = 4 {\frac{1}{4} + \frac{\sqrt{15}}{4} i} = r e^{i θ} \Rightarrow r = 4 a n d c o s θ = \frac{1}{4} a n d s i n θ = \frac{\sqrt{15}}{4} \Rightarrow$ $t a n θ = \sqrt{15} \Rightarrow θ = a r c t a n (\sqrt{15}) a l s o \sqrt{5} - \sqrt{3} i = 2 \sqrt{2} {\frac{\sqrt{5}}{2 \sqrt{2}} - \frac{\sqrt{3}}{2 \sqrt{2}} i} = r^{^{'}} e^{i θ^{'}} \Rightarrow$ $r^{^{'}} = 2 \sqrt{2} a n d θ^{^{'}} = a r c t a n (\frac{- \sqrt{3}}{\sqrt{5}}) \Rightarrow a r g (\frac{1 + \sqrt{15} i}{\sqrt{5} - \sqrt{3} i}) \equiv a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}}) [2 π] \Rightarrow$ $\Rightarrow \frac{1 + \sqrt{15} i}{\sqrt{5} - \sqrt{3} i} = \sqrt{2} e^{i (a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}}))} = Z^{3} \Rightarrow$ $Z =^{6} \sqrt{2} e^{\frac{1}{3} (a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}})) i}$ $=^{6} \sqrt{2} {c o s (\frac{1}{3} (a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}}))) + i s i n (\frac{1}{3} (a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}})) \Rightarrow$ $x =^{6} \sqrt{2} c o s (\frac{1}{3} (a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}}))) a n d$ $y =^{6} \sqrt{2} s i n (\frac{1}{3} (a r c t a n (\sqrt{15}) + a r c t a n (\frac{\sqrt{3}}{\sqrt{5}}))) .$
	Answered by tanmay last updated on 12/May/19

	$\frac{(1 + \sqrt{15} i) (\sqrt{5} + \sqrt{3} i)}{8}$ $\frac{(2 + 2 \sqrt{15} i) (\sqrt{5} + \sqrt{3} i)}{16}$ $\frac{(5 - 3 + 2 \sqrt{15} i) (\sqrt{5} + \sqrt{3} i)}{16}$ $\frac{{(\sqrt{5} + \sqrt{3} i)}^{2} (\sqrt{5} + \sqrt{3} i)}{16}$ $\frac{{(\sqrt{5} + \sqrt{3} i)}^{3}}{{(2^{\frac{4}{3}})}^{3}}$ $s o x = \frac{\sqrt{5}}{2^{\frac{4}{3}}} a n d y = \frac{\sqrt{3}}{2^{\frac{4}{3}}}$
	Commented by MJS last updated on 12/May/19

	$your solution is z_{1} = \frac{\sqrt{5} \sqrt[3]{4}}{4} + \frac{\sqrt{3} \sqrt[3]{4}}{4} i but there are$ $2 more : z_{2} = z_{1} (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) = \frac{\sqrt[3]{4}}{8} (3 - \sqrt{5}) - \frac{\sqrt{3} \sqrt[3]{4}}{8} (1 + \sqrt{5}) i$ $and z_{3} = z_{1} (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) = - \frac{\sqrt[3]{4}}{8} (3 + \sqrt{5}) - \frac{\sqrt{3} \sqrt[3]{4}}{8} (1 - \sqrt{5}) i$
	Commented by tanmay last updated on 12/May/19

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