show that the root of (x+1)^7 = (x-1)^7 are given by -+cotrπ/7, r=1,2,3

Question

show that the root of $(x+1)^7 = (x-1)^7$ are given by $\pm cot{\dfrac{rπ}{7}}, r=1,2,3$ ? how is the solution of the problem ? why the value of $r$ is $1,2,3$ and why it is not equal to $1,2,3,4,5,6$ ?

Answer 1

Since $x=1$ isn't a root of this equation, it's equivalent to $z^7=1$ with $z:=\frac{x+1}{x-1}\implies x=\frac{z+1}{z-1}$. Writing $z=\exp\frac{2r\pi i}{7}$ with an integer $r$ satisfying $-3\le r\le 3$,$$x=\frac{\exp\frac{2r\pi i}{7}+1}{\exp\frac{2r\pi i}{7}-1}=\frac{\exp\frac{r\pi i}{7}+\exp\frac{-r\pi i}{7}}{\exp\frac{r\pi i}{7}-\exp\frac{-r\pi i}{7}}=\frac{2\cos\frac{r\pi}{7}}{2i\sin\frac{r\pi}{7}}=-i\cot\frac{r\pi}{7}.$$The case $r=0$ doesn't work because the cotangent wouldn't be finite. There's one error in your statement of the result: the roots are $\pm\color{limegreen}{i}\cot\frac{r\pi}{7}$ with $1\le r\le 3$.