How to solve $(z-1)^n=(z+1)^n$

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I have problem with this equation because when I put $\left(\dfrac{z+1}{z-1}\right)^n=1$ I find solutions different than when I put $\left(\dfrac{z-1}{z+1}\right)^n=1$ So which one should I use ? For first one -icot(kπ/5) and icot(kπ/5) for the second one

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The approaches agree. Whether you argue$$\frac{z+1}{z-1}=e^{2\pi ik/n},\,k\in\Bbb Z\implies z=\frac{e^{2\pi ik/n}+1}{e^{2\pi ik/n}-1}=-i\cot\frac{\pi k}{n}$$or$$\frac{z-1}{z+1}=e^{2\pi il/n},\,l\in\Bbb Z\implies\frac{z+1}{z-1}=e^{2\pi ik/n},\,k=-l,$$from which we can continue with the above logic, the solution sets are the same.