simplify A_n =(2+i(√5))^n +(2−i(√5))^n and B_n =(2+i(√5))^n −(2−i(√5))^n


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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
$we have ∣2+i(√5)∣ =(√(2^2 +((√5))^2 ))=(√(4+5))=3 ⇒ 2+i(√5)=3((2/3) +i((√5)/3)) =3 e^(iarctan(((√5)/2))) ⇒(2+i(√5))^n =3^n e^(inarctan(((√5)/2))) and (2−i(√5))^n =3^n e^(−inarctan(((√5)/2))) ⇒ A_n =3^n { e^(inarctan(((√5)/2))) +e^(−inarctan(((√5)/2))) } =3^n (2cos(n arctan(((√5)/2)))) =2.3^n cos(n arctan(((√5)/2))) also B_n =3^n { e^(inarctan(((√5)/2))) −e^(−inarctan(((√5)/2))) } =2i 3^n sin(narctan(((√5)/2)))$
$w e h a v e ∣ 2 + i \sqrt{5} ∣ = \sqrt{2^{2} + {(\sqrt{5})}^{2}} = \sqrt{4 + 5} = 3 \Rightarrow$ $2 + i \sqrt{5} = 3 (\frac{2}{3} + i \frac{\sqrt{5}}{3}) = 3 e^{i a r c t a n (\frac{\sqrt{5}}{2})} \Rightarrow {(2 + i \sqrt{5})}^{n} = 3^{n} e^{i n a r c t a n (\frac{\sqrt{5}}{2})}$ $a n d {(2 - i \sqrt{5})}^{n} = 3^{n} e^{- i n a r c t a n (\frac{\sqrt{5}}{2})} \Rightarrow$ $A_{n} = 3^{n} {e^{i n a r c t a n (\frac{\sqrt{5}}{2})} + e^{- i n a r c t a n (\frac{\sqrt{5}}{2})}} = 3^{n} (2 c o s (n a r c t a n (\frac{\sqrt{5}}{2})))$ $= 2 . 3^{n} c o s (n a r c t a n (\frac{\sqrt{5}}{2})) a l s o$ $B_{n} = 3^{n} {e^{i n a r c t a n (\frac{\sqrt{5}}{2})} - e^{- i n a r c t a n (\frac{\sqrt{5}}{2})}}$ $= 2 i 3^{n} s i n (n a r c t a n (\frac{\sqrt{5}}{2}))$
	Answered by mr W last updated on 06/Aug/19

	$2 + i \sqrt{5} = 3 (\frac{2}{3} + i \frac{\sqrt{5}}{3}) = 3 (\cos α + i \sin α)$ $w i t h α = \tan^{- 1} \frac{\sqrt{5}}{2}$ $2 - i \sqrt{5} = 3 (\frac{2}{3} - i \frac{\sqrt{5}}{3}) = 3 (\cos α - i \sin α)$ $A_{n} = 3^{n} (\cos n α + i \sin n α) + 3^{n} (\cos n α - i \sin n α)$ $\Rightarrow A_{n} = 2 \times 3^{n} \cos n α$ $B_{n} = 3^{n} (\cos n α + i \sin n α) - 3^{n} (\cos n α - i \sin n α)$ $\Rightarrow B_{n} = 2 \times 3^{n} \sin n α i$
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