Friday, May 25, 2018

algebra precalculus - Show that the solutions of the equation (1+x)2n+(1x)2n=0 are x=pmi tanfrac(2r1)pi4n,  r=1,2,cdotsn.




Show that the solutions of the equation (1+x)2n+(1x)2n=0 are x=±i tan(2r1)π4n,  r=1,2,n.




Attempt




Clearly, 1+x1x=(1)1/2n=(cos(2k+1)π+isin(2k+1)π)1/2n=cos(2k+1)π2n+isin(2k+1)π2n


(by de Moivre's formula)



By componendo dividendo and some simplification,



I am getting x=itan2k+14nπ,  k=1,2,3,2n



How to get the desired result? I am getting 2n solutions and I understand that the solutions that I have obtained is equal to that of the desired. But how to to get the desired from the solution that I have obtained? Please provide me the mathematical steps.



Is there any other method of solving to get the answer directly.



Answer



For any α, tanα=tan(απ).



For nk<2n:



itan2k+14nπ=itan2(k2n)+14nπ=itan2(2nk)14nπ




So k=n,n+1,,2n1 corresponds to r=2nk and the minus sign.



k=2n corresponds to r=1 with the positive sign.



k=1,2,,n1 corresponds to r=k+1 with the positive sign.


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