Prove that the solutions to $(z+1)^7 = z^7$ have same real parts Prove that all the solutions to the equation $(z+1)^7=z^7$ have equal real parts, and find what this real part is equal to.

I found that $z=\frac{1}{e^{2\pi i/7}-1}, \frac{1}{e^{4\pi i/7}-1}, \frac{1}{e^{6\pi i/7}-1}, \frac{1}{e^{8\pi i/7}-1}, \frac{1}{e^{10\pi i/7}-1}, \frac{1}{e^{12\pi i/7}-1},$ but I'm not sure what to do next.
 A: Based on the proposed equation, one concludes that
\begin{align*}
(z + 1)^{7} = z^{7} & \Rightarrow |z + 1| = |z|\\\\
& \Rightarrow |z + 1|^{2} = |z|^{2}\\\\
& \Rightarrow z\overline{z} + z + \overline{z} + 1 = z\overline{z}\\\\
& \Rightarrow 2\operatorname{Re}(z) = -1\\\\
& \Rightarrow \operatorname{Re}(z) = -\frac{1}{2}
\end{align*}
and we are done.
Hopefully this helps !
A: Let $~\displaystyle w = \left(z + \frac{1}{2}\right).$
Then, you have that
$$\left(w + \frac{1}{2}\right)^7 = \left(w - \frac{1}{2}\right)^7.$$
This implies that
$$\left|w + \frac{1}{2}\right|^7 = \left|w - \frac{1}{2}\right|^7.$$
This implies that
$$\left|w + \frac{1}{2}\right| = \left|w - \frac{1}{2}\right|.$$
This implies that any satisfying value for $w$ must be equidistant between the two complex values $[(1/2) + i(0)]$ and $[-(1/2) + i(0)].$
This implies that any satisfying value for $w$ must be on the perpendicular bisector of the line segment running from $[(1/2) + i(0)]$ to $[-(1/2) + i(0)].$
This implies that any satisfying value for $w$ must have it's real portion equal to $0$.
Therefore, since $~\displaystyle w = \left(z + \frac{1}{2}\right),$ you must have that any (corresponding) satisfying value for $z$ must have its real portion equal to $(-1/2)$.
A: $\Re \frac 1 {e^{it}-1}= \Re \frac {e^{-it}-1} {|e^{-it}-1|^{2}}=\frac {\cos t -1} {2-2\cos t}=-\frac 1  2 $ for any real number $t$.
