find-x-y-in-R-x-yi-3-1-15-i-5-3-i-

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

\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

t h a n k y o u s i r \dots

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