1-find-xy-R-2-find-x-y-Z-x-2yi-6-8i-

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

1) f i n d x y \in R

2) f i n d x, y \in Z

{(x + 2 y i)}^{6} = 8 i

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

l e t z = x + 2 y i

z^{6} = 8 i = {(\sqrt{2})}^{6} e^{\frac{π}{2} i}

\Rightarrow z = \sqrt{2} e^{(\frac{π}{12} + \frac{k π}{3}) i} (k = 0, 1, 2, \dots, 5)

k = 0 :

z = \sqrt{2} e^{(\frac{π}{12}) i} = \frac{\sqrt{3} + 1}{2} + \frac{\sqrt{3} - 1}{2} i

x = \frac{\sqrt{3} + 1}{2}

y = \frac{\sqrt{3} - 1}{4}

k = 1 :

z = \sqrt{2} e^{(\frac{π}{12} + \frac{π}{3}) i} = \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} + 1}{2} i

x = \frac{\sqrt{3} - 1}{2}

y = \frac{\sqrt{3} + 1}{4}

k = 2 :

z = \sqrt{2} e^{(\frac{π}{12} + \frac{2 π}{3}) i} = - 1 + i

x = - 1

y = \frac{1}{2}

k = 3 :

z = \sqrt{2} e^{(\frac{π}{12} + \frac{3 π}{3}) i} = \frac{- \sqrt{3} - 1}{2} + \frac{- \sqrt{3} + 1}{2} i

x = - \frac{\sqrt{3} + 1}{2}

y = - \frac{\sqrt{3} - 1}{4}

k = 4 :

z = \sqrt{2} e^{(\frac{π}{12} + \frac{4 π}{3}) i} = \frac{- \sqrt{3} + 1}{2} + \frac{- \sqrt{3} - 1}{2} i

x = - \frac{\sqrt{3} - 1}{2}

y = - \frac{\sqrt{3} + 1}{4}

k = 5 :

z = \sqrt{2} e^{(\frac{π}{12} + \frac{5 π}{3}) i} = 1 - i

x = 1

y = - \frac{1}{2}

f o r x, y \in C

x + 2 y i = z

x a n d y c a n n o t b e u n i q u e l y d e t e r m i n e d .

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

\Rightarrow {(x - 2 y i)}^{6} = - 8 i

\Rightarrow {(x^{2} + 4 y^{2})}^{6} = 64

\Rightarrow x^{2} + 4 y^{2} = 2

\Rightarrow x = \sqrt{2} c o s (θ)

y = \frac{1}{\sqrt{2}} s i n (θ)

θ \in [0, 2 π]

\Rightarrow 8 (e^{i 6 θ}) = 8 i

\Rightarrow 6 θ = \frac{π}{2} + 2 k π \Rightarrow θ \in \frac{π}{12} + \frac{k π}{3}, k \in {0, \dots \dots 5}

x = \sqrt{2} c o s (θ), y = \frac{s i n (θ)}{\sqrt{2}}, θ \in {\frac{π}{12} + \frac{k π}{3}, 0 ⩽ k ⩽ 5}