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Question Number 1439 by Rasheed Ahmad last updated on 04/Aug/15

A r e t h e r e t w o c o m p l e x n u m b e r s

w h i c h a r e c u b e o f o n e a n o t h e r ?

I f y e s w h a t a r e t h e y ?

Answered by 123456 last updated on 04/Aug/15

{\begin{cases} x = y^{3} \\ y = x^{3} \end{cases}

x = y^{3} = {(x^{3})}^{3} = x^{9}

x^{9} - x = 0

x (x^{8} - 1) = 0

x = 0 \lor x = \sqrt[8]{1}

S = {(x, y) \in C^{2}, k \in {0, 1, 2, 3, 4, 5, 6, 7}; (x, y) = (0, 0) \lor (x, y) = (e^{ι π k / 4}, e^{3 ı π k / 4})}

Commented by 123456 last updated on 04/Aug/15

(0, 0) \equiv (0, 0)

(1, 1) \equiv (1, 1)

(e^{ı π / 4}, e^{3 ı π / 4}) \equiv (\frac{\sqrt{2}}{2} + ı \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} + ı \frac{\sqrt{2}}{2})

(e^{ı π / 2}, e^{3 ı π / 2}) \equiv (ı, - ı)

(e^{3 ı π / 4}, e^{9 ı π / 4}) \equiv (- \frac{\sqrt{2}}{2} + ı \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} + ı \frac{\sqrt{2}}{2})

(e^{ı π}, e^{3 ı π}) \equiv (- 1, - 1)

(e^{5 ı π / 4}, e^{15 ı π / 4}) \equiv (- \frac{\sqrt{2}}{2} - ı \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} - ı \frac{\sqrt{2}}{2})

(e^{3 ı π / 2}, e^{9 ı π / 2}) \equiv (- ı, ı)

(e^{7 ı π / 4}, e^{21 ı π / 4}) \equiv (\frac{\sqrt{2}}{2} - ı \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} - ı \frac{\sqrt{2}}{2})

Commented by 123456 last updated on 15/Aug/15

n \in N^{*}

{\begin{cases} x^{n} = y^{3 n} \\ y^{3 n} = x^{9 n} \end{cases}

x^{n} = x^{9 n} \Leftrightarrow x^{n} (x^{8 n} - 1) = 0

{\begin{cases} x^{3 n} = y^{9 n} \\ y^{n} = x^{3 n} \end{cases}

y^{n} = y^{9 n} \Leftrightarrow y^{n} (y^{8 n} - 1) = 0

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