Is my solution correct?

Question: Find all \(z\), such that \((z+1)^7=z^7\).

My answer: Since \(z\ne 0\), we can divide by \( z^7\) on both sides to get \(\left(\frac{z+1}{z}\right)^7=1\). If we let \(w=\frac{z+1}{z}\), then \(w\) is equal to the \(7^{\text{th}}\) roots of unity. We can manipulate \(w=\frac{z+1}{z}\) by multiplying both sides by \(z\), then subtracting \(z\) from both sides to get \(1=zw-z\). We can then divide both sides by \(w-1\) to get \(z=\frac{1}{w-1}\).

The seven roots of unity are \(1\), \(e^{\frac{2i\pi}{7}}\), \(e^{\frac{4i\pi}{7}}\), \(e^{\frac{6i\pi}{7}}\), \(e^{\frac{-6i\pi}{7}}\), \(e^{\frac{-4i\pi}{7}}\), \(e^{\frac{-2i\pi}{7}}\). \(z=\frac{1}{1-1}\) if \(w=1\) is not defined, and we can see that it doesn't work for \((z+1)^7=z^7\), either, so the solutions for \(z\) are \(\frac{1}{e^{\frac{2i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{4i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{6i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{-6i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{-4i\pi}{7}}-1}\), and \(\frac{1}{e^{\frac{-2i\pi}{7}}-1}\), respectively.

Davis Sep 7, 2019