Solve $(x+1)^n=(x-1)^n$, assuming $x$ is a complex number and $n>0$.

Hint: $x = 1$ is not a solution, so we might assume $x \neq 1$ and re-write this as:

$$\left(\frac{x + 1}{x - 1} \right)^n = 1$$

It boils down to just listing the $n$th roots of $1$ and computing $x$.

Your equation is equivalent to $$ \left(\frac{x+1}{x-1}\right)^n = 1 \tag{1}$$ and since the inverse function of $f(x)=\frac{x+1}{x-1}$ is given by $\frac{y-1}{y+1}$, we have: $$ x = \frac{\exp\frac{2\pi k i}{n}+1}{\exp\frac{2\pi k i}{n}-1}=-i\cot\frac{\pi k}{n},\qquad k\in\left[1,\ldots,n-1\right].\tag{2}$$

For $z^n=w$, one can write $w=we^{i2 \pi \ell}$ for integer $\ell$. Then, solving for $z$ reveals that

$$z=w^{1/n}e^{i2 \pi \ell/n}$$

for $\ell =0, 1, 2,\cdots n-1.$

Now for $z=x+1$ and $w=(x-1)^n$ we have that

$$x+1=(x-1)e^{i2 \pi \ell/n} \implies x = \frac{e^{i2 \pi \ell/n}+1}{e^{i2 \pi \ell/n}-1}=-i\cot{\pi\ell/n}\,\,\text{for}\,\,\ell =1,2,\cdots n-1$$