algebra precalculus - Show that the solutions of the equation $(1+x)^{2n}+(1-x)^{2n}=0$ are $x=pm i ~tan{frac{(2r-1)pi}{4n}}, ~~r=1,2,cdots n.$

Show that the solutions of the equation $(1+x)^{2n}+(1-x)^{2n}=0$ are $$x=\pm i~ \tan{\frac{(2r-1)\pi}{4n}}, ~~r=1,2,\cdots n.$$

Attempt

Clearly, $$\frac{1+x}{1-x}=(-1)^{1/2n}=(\cos{(2k+1)\pi}+i \sin{(2k+1)\pi})^{1/2n}=\cos{\frac{(2k+1)\pi}{2n}}+i \sin{\frac{(2k+1)\pi}{2n}}$$
(by de Moivre's formula)

By componendo dividendo and some simplification,

I am getting $$x=i\tan{\frac{2k+1}{4n}\pi}, ~~k=1,2,3,\cdots 2n$$

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 $\alpha$, $\tan\alpha=\tan(\alpha-\pi)$.

For $n\leq k<2n$:

$$\begin{align}i\tan\frac{2k+1}{4n}\pi &= i\tan\frac{2(k-2n)+1}{4n}{\pi} \\ &=-i\tan\frac{2(2n-k)-1}{4n}\pi \end{align}$$

So $k=n,n+1,\dots,2n-1$ corresponds to $r=2n-k$ and the minus sign.

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

$k=1,2,\dots,n-1$ corresponds to $r=k+1$ with the positive sign.