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

$$\mathrm{simplify}{\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right)={(1+\mathrm{ix})}^{\mathrm{n}}+{(1-\mathrm{ix})}^{\mathrm{n}}\mathrm{x}\mathrm{from}\mathrm{C}$$

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

$$\mathrm{let}\mathrm{x}={\mathrm{re}}^{\mathrm{i}\theta }\Rightarrow 1+\mathrm{ix}=1+{\mathrm{ire}}^{\mathrm{i}\theta }=1+\mathrm{ir}(\cos \theta +\mathrm{isin}\theta )$$$$=1-\mathrm{rsin}\theta +\mathrm{ircos}\theta \mathrm{and}$$$$1-\mathrm{ix}=1-{\mathrm{ire}}^{\mathrm{i}\theta }=1-\mathrm{ir}(\cos \theta +\mathrm{isin}\theta )$$$$=1+\mathrm{rsin}\theta -\mathrm{ir}\cos \theta$$$$\mid 1+\mathrm{ix}\mid =\sqrt{{(1-\mathrm{rsin}\theta )}^2+{\mathrm{r}}^2{\cos }^2\theta }=\sqrt{1-2\mathrm{rsin}\theta +{\mathrm{r}}^2}\Rightarrow$$$$1+\mathrm{ix}=\sqrt{1-2\mathrm{rsin}\theta +{\mathrm{r}}^2}{\mathrm{e}}^{\mathrm{iarctan}\left(\frac{\mathrm{rcos}\theta }{1-\mathrm{rsin}\theta }\right)}\Rightarrow$$$${(1+\mathrm{ix})}^{\mathrm{n}}={(1-2\mathrm{rsin}\theta +{\mathrm{r}}^2)}^{\frac{\mathrm{n}}{2}}{\mathrm{e}}^{\mathrm{inarctan}\left(\frac{\mathrm{rcos}\theta }{1-\mathrm{rsin}\theta }\right)}$$$$\mid 1-\mathrm{ix}\mid =\sqrt{{(1+\mathrm{rsin}\theta )}^2+{\mathrm{r}}^2{\cos }^2\theta }=\sqrt{1+2\mathrm{rsin}\theta +{\mathrm{r}}^2}$$$$\Rightarrow 1-\mathrm{ix}=\sqrt{1+2\mathrm{rsin}\theta +{\mathrm{r}}^2}{\mathrm{e}}^{-\mathrm{iarctan}\left(\frac{\mathrm{rcos}\theta }{1+\mathrm{rsin}\theta }\right)}\Rightarrow$$$${(1-\mathrm{ix})}^{\mathrm{n}}={(1+2\mathrm{rsin}\theta +{\mathrm{r}}^2)}^{\frac{\mathrm{n}}{2}}{\mathrm{e}}^{-\mathrm{in}\arctan \left(\frac{\mathrm{rcos}\theta }{1+\mathrm{rsin}\theta }\right)}$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}={(1-2\mathrm{rsin}\theta +{\mathrm{r}}^2)}^{\frac{\mathrm{n}}{2}}{\mathrm{e}}^{\mathrm{inarctan}\left(\frac{\mathrm{rcos}\theta }{1-\mathrm{rsin}\theta }\right)}$$$$+{(1+2\mathrm{rsin}\theta +{\mathrm{r}}^2)}^{\frac{\mathrm{n}}{2}}{\mathrm{e}}^{-\mathrm{inarctan}\left(\frac{\mathrm{rcos}\theta }{1+\mathrm{rsin}\theta }\right)}$$

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

$$\mathrm{another}\mathrm{way}$$$${\mathrm{A}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\mathrm{ix}\right)}^{\mathrm{k}}+\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-\mathrm{ix})}^{\mathrm{k}}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}({\mathrm{i}}^{\mathrm{k}}+{(-\mathrm{i})}^{\mathrm{k}}){\mathrm{x}}^{\mathrm{k}}$$$$=\sum _{\mathrm{p}=0}^{\left[\frac{\mathrm{n}}{2}\right]}{\mathrm{C}}_{\mathrm{n}}^{2\mathrm{p}}2{(-1)}^{\mathrm{p}}{\mathrm{x}}^{2\mathrm{p}}=2\sum _{\mathrm{p}=0}^{\left[\frac{\mathrm{n}}{2}\right]}{\mathrm{C}}_{\mathrm{n}}^{2\mathrm{p}}{(-1)}^{\mathrm{p}}{\mathrm{x}}^{2\mathrm{p}}$$

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