simplify-A-n-2-i-5-n-2-i-5-n-and-B-n-2-i-5-n-2-i-5-n-

Question Number 65918 by mathmax by abdo last updated on 05/Aug/19

s i m p l i f y A_{n} = {(2 + i \sqrt{5})}^{n} + {(2 - i \sqrt{5})}^{n}

a n d B_{n} = {(2 + i \sqrt{5})}^{n} - {(2 - i \sqrt{5})}^{n}

Commented by mathmax by abdo last updated on 06/Aug/19

w e h a v e ∣ 2 + i \sqrt{5} ∣ = \sqrt{2^{2} + {(\sqrt{5})}^{2}} = \sqrt{4 + 5} = 3 \Rightarrow

2 + i \sqrt{5} = 3 (\frac{2}{3} + i \frac{\sqrt{5}}{3}) = 3 e^{i a r c t a n (\frac{\sqrt{5}}{2})} \Rightarrow {(2 + i \sqrt{5})}^{n} = 3^{n} e^{i n a r c t a n (\frac{\sqrt{5}}{2})}

a n d {(2 - i \sqrt{5})}^{n} = 3^{n} e^{- i n a r c t a n (\frac{\sqrt{5}}{2})} \Rightarrow

A_{n} = 3^{n} {e^{i n a r c t a n (\frac{\sqrt{5}}{2})} + e^{- i n a r c t a n (\frac{\sqrt{5}}{2})}} = 3^{n} (2 c o s (n a r c t a n (\frac{\sqrt{5}}{2})))

= 2 . 3^{n} c o s (n a r c t a n (\frac{\sqrt{5}}{2})) a l s o

B_{n} = 3^{n} {e^{i n a r c t a n (\frac{\sqrt{5}}{2})} - e^{- i n a r c t a n (\frac{\sqrt{5}}{2})}}

= 2 i 3^{n} s i n (n a r c t a n (\frac{\sqrt{5}}{2}))

Answered by mr W last updated on 06/Aug/19

2 + i \sqrt{5} = 3 (\frac{2}{3} + i \frac{\sqrt{5}}{3}) = 3 (\cos α + i \sin α)

w i t h α = \tan^{- 1} \frac{\sqrt{5}}{2}

2 - i \sqrt{5} = 3 (\frac{2}{3} - i \frac{\sqrt{5}}{3}) = 3 (\cos α - i \sin α)

A_{n} = 3^{n} (\cos n α + i \sin n α) + 3^{n} (\cos n α - i \sin n α)

\Rightarrow A_{n} = 2 \times 3^{n} \cos n α

B_{n} = 3^{n} (\cos n α + i \sin n α) - 3^{n} (\cos n α - i \sin n α)

\Rightarrow B_{n} = 2 \times 3^{n} \sin n α i

Facebook Tweet Pin