Question Number 66032 by mathmax by abdo last updated on 07/Aug/19
$${simplify}\:\:{w}_{{n}} =\left(\mathrm{1}+{in}\right)^{{n}} −\left(\mathrm{1}−{in}\right)^{{n}} \:{with}\:{n}\:{integr}\:{natural} \\ $$
Commented by mr W last updated on 08/Aug/19
$$={i}\:\mathrm{2}\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \mathrm{sin}\:\left({n}\:\mathrm{tan}^{−\mathrm{1}} {n}\right) \\ $$
Commented by mathmax by abdo last updated on 08/Aug/19
$${we}\:{have}\:\mathrm{1}+{in}\:=\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }{e}^{{iarctann}} \:\Rightarrow\left(\mathrm{1}+{in}\right)^{{n}} =\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \:{e}^{{inarctan}\left({n}\right)} \\ $$$${also}\:\mathrm{1}−{in}\:={vonj}\left(\mathrm{1}+{in}\right)\:=\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }{e}^{−{iarctan}\left({n}\right)} \:\:\Rightarrow \\ $$$$\left(\mathrm{1}−{in}\right)^{{n}} \:=\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} {e}^{\left.−\left.{inarctan}\right){n}\right)} \:\Rightarrow \\ $$$${w}_{{n}} =\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \left\{\mathrm{2}{i}\:{sin}\left({narctan}\left({n}\right)\right\}\:=\mathrm{2}{i}\:\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \:{sin}\left({narctan}\left({n}\right)\right)\right. \\ $$