simplify w_n =(1+in)^n −(1−in)^n with n integr natural


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Question Number 66032 by mathmax by abdo last updated on 07/Aug/19

$s i m p l i f y w_{n} = {(1 + i n)}^{n} - {(1 - i n)}^{n} w i t h n i n t e g r n a t u r a l$
Commented by mr W last updated on 08/Aug/19

$= i 2 {(1 + n^{2})}^{\frac{n}{2}} \sin (n \tan^{- 1} n)$
Commented by mathmax by abdo last updated on 08/Aug/19
$we have 1+in =(√(1+n^2 ))e^(iarctann) ⇒(1+in)^n =(1+n^2 )^(n/2) e^(inarctan(n)) also 1−in =vonj(1+in) =(√(1+n^2 ))e^(−iarctan(n)) ⇒ (1−in)^n =(1+n^2 )^(n/2) e^(−inarctan)n)) ⇒ w_n =(1+n^2 )^(n/2) {2i sin(narctan(n)} =2i (1+n^2 )^(n/2) sin(narctan(n))$
$w e h a v e 1 + i n = \sqrt{1 + n^{2}} e^{i a r c t a n n} \Rightarrow {(1 + i n)}^{n} = {(1 + n^{2})}^{\frac{n}{2}} e^{i n a r c t a n (n)}$ $a l s o 1 - i n = v o n j (1 + i n) = \sqrt{1 + n^{2}} e^{- i a r c t a n (n)} \Rightarrow$ ${(1 - i n)}^{n} = {(1 + n^{2})}^{\frac{n}{2}} e^{- i n a r c t a n) n)} \Rightarrow$ $w_{n} = {(1 + n^{2})}^{\frac{n}{2}} {2 i s i n (n a r c t a n (n)} = 2 i {(1 + n^{2})}^{\frac{n}{2}} s i n (n a r c t a n (n))$
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