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

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

$${simplify}\:\:{w}_{{n}} =\left(\mathrm{1}+{in}\right)^{{n}} −\left(\mathrm{1}−{in}\right)^{{n}} \:{with}\:{n}\:{integr}\:{natural} \\ $$

Commented by mr W last updated on 08/Aug/19

=i 2(1+n^2 )^(n/2) sin (n tan^(−1) n)

$$={i}\:\mathrm{2}\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \mathrm{sin}\:\left({n}\:\mathrm{tan}^{−\mathrm{1}} {n}\right) \\ $$

Commented by mathmax by abdo last updated on 08/Aug/19

we have 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))

$${we}\:{have}\:\mathrm{1}+{in}\:=\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }{e}^{{iarctann}} \:\Rightarrow\left(\mathrm{1}+{in}\right)^{{n}} =\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \:{e}^{{inarctan}\left({n}\right)} \\ $$$${also}\:\mathrm{1}−{in}\:={vonj}\left(\mathrm{1}+{in}\right)\:=\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }{e}^{−{iarctan}\left({n}\right)} \:\:\Rightarrow \\ $$$$\left(\mathrm{1}−{in}\right)^{{n}} \:=\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} {e}^{\left.−\left.{inarctan}\right){n}\right)} \:\Rightarrow \\ $$$${w}_{{n}} =\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \left\{\mathrm{2}{i}\:{sin}\left({narctan}\left({n}\right)\right\}\:=\mathrm{2}{i}\:\left(\mathrm{1}+{n}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \:{sin}\left({narctan}\left({n}\right)\right)\right. \\ $$

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