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# Roots of Unity

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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.

Sep 7, 2019