Prove that the solutions to $(z+1)^7 = z^7$ have same real parts

Question

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.

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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.

Answer 1

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 !

Answer 2

$\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 $t$.

Answer 3

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 $(-1/2)$.