simplify-A-n-x-1-ix-n-1-ix-n-x-from-C-

Question Number 142115 by mathmax by abdo last updated on 26/May/21

simplify A_{n} (x) = {(1 + ix)}^{n} + {(1 - ix)}^{n} x from C

Answered by mathmax by abdo last updated on 27/May/21

let x = {re}^{i θ} \Rightarrow 1 + ix = 1 + {ire}^{i θ} = 1 + ir (\cos θ + isin θ)

= 1 - rsin θ + ircos θ and

1 - ix = 1 - {ire}^{i θ} = 1 - ir (\cos θ + isin θ)

= 1 + rsin θ - ir \cos θ

∣ 1 + ix ∣= \sqrt{{(1 - rsin θ)}^{2} + r^{2} \cos^{2} θ} = \sqrt{1 - 2 rsin θ + r^{2}} \Rightarrow

1 + ix = \sqrt{1 - 2 rsin θ + r^{2}} e^{iarctan (\frac{rcos θ}{1 - rsin θ})} \Rightarrow

{(1 + ix)}^{n} = {(1 - 2 rsin θ + r^{2})}^{\frac{n}{2}} e^{inarctan (\frac{rcos θ}{1 - rsin θ})}

∣ 1 - ix ∣= \sqrt{{(1 + rsin θ)}^{2} + r^{2} \cos^{2} θ} = \sqrt{1 + 2 rsin θ + r^{2}}

\Rightarrow 1 - ix = \sqrt{1 + 2 rsin θ + r^{2}} e^{- iarctan (\frac{rcos θ}{1 + rsin θ})} \Rightarrow

{(1 - ix)}^{n} = {(1 + 2 rsin θ + r^{2})}^{\frac{n}{2}} e^{- in \arctan (\frac{rcos θ}{1 + rsin θ})}

\Rightarrow A_{n} = {(1 - 2 rsin θ + r^{2})}^{\frac{n}{2}} e^{inarctan (\frac{rcos θ}{1 - rsin θ})}

+ {(1 + 2 rsin θ + r^{2})}^{\frac{n}{2}} e^{- inarctan (\frac{rcos θ}{1 + rsin θ})}

Answered by mathmax by abdo last updated on 27/May/21

another way

A_{n} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(ix)}^{k} + \sum_{k = 0}^{n} C_{n}^{k} {(- ix)}^{k}

= \sum_{k = 0}^{n} C_{n}^{k} (i^{k} + {(- i)}^{k}) x^{k}

= \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} 2 {(- 1)}^{p} x^{2 p} = 2 \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} {(- 1)}^{p} x^{2 p}