Ars Mathematica: Solución al problema 88

1 + \sqrt{3} i = 2_{60}; 1 - \sqrt{3} i = 2_{- 60}

(1 + \sqrt{3} i)^{n} - (1 - \sqrt{3} i)^{n} = {(2_{60})}^{n} - {(2_{- 60})}^{n} =

= 2^{n} (c o s (60 n) + i s e n (60 n)) - 2^{n} (c o s (- 60 n) + i s e n (- 60 n)) =

= 2^{n} (c o s (60 n) + i s e n (60 n) - c o s (- 60 n) - i s e n (- 60 n)) =

= 2^{n} (2 i s e n (60 n)

= 2^{n + 1} i s e n (60 n) (*)

n = 6 k

(*) = 2^{n + 1} i (s e n (60 \cdot 6 k)) = 2^{n + 1} i s e n (360 k) = 0

n = 6 k + 1

\begin{matrix} (*) & = & 2^{n + 1} i (s e n (60 \cdot (6 k + 1))) = 2^{n + 1} i s e n (360 k + 60) = \end{matrix}

= 2^{n + 1} i s e n 60 = 2^{n + 1} i \frac{\sqrt{3}}{2} = 2^{n} \sqrt{3} i

n = 6 k + 2

(*) = 2^{n + 1} i (s e n (60 \cdot (6 k + 2))) = 2^{n + 1} i s e n (360 k + 120) =

{= 2}^{n + 1} i s e n 120 = 2^{n + 1} i \frac{\sqrt{3}}{2} = 2^{n} \sqrt{3} i

n = 6 k + 3

(*) = 2^{n + 1} i (s e n (60 \cdot (6 k + 3))) = 2^{n + 1} i s e n (360 k + 180) = 0

n = 6 k + 4

(*) = 2^{n + 1} i (s e n (60 \cdot (6 k + 4))) = 2^{n + 1} i s e n (360 k + 240) =

{= 2}^{n + 1} i s e n 240 = - 2^{n + 1} i \frac{\sqrt{3}}{2} = - 2^{n} \sqrt{3} i

n = 6 k + 5

(*) = 2^{n + 1} i (s e n (60 \cdot (6 k + 5))) = 2^{n + 1} i s e n (360 k + 300) =

{= 2}^{n + 1} i s e n 300 = - 2^{n + 1} i \frac{\sqrt{3}}{2} = - 2^{n} \sqrt{3} i

Ars Mathematica