Show that the solutions of the equation (1+x)2n+(1−x)2n=0 are x=±i tan(2r−1)π4n, r=1,2,⋯n.
Attempt
Clearly, 1+x1−x=(−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(α−π).
For n≤k<2n:
itan2k+14nπ=itan2(k−2n)+14nπ=−itan2(2n−k)−14nπ
So k=n,n+1,…,2n−1 corresponds to r=2n−k and the minus sign.
k=2n corresponds to r=1 with the positive sign.
k=1,2,…,n−1 corresponds to r=k+1 with the positive sign.
No comments:
Post a Comment