Proving that all solutions of $(1+z)^n=z^n$ have $\mathrm{Re}(z)=-\frac{1}{2}$ 
Question:
Prove that all solutions of $(1+z)^n=z^n$ have $\mathrm{Re}(z)=-\frac{1}{2}$

I've rearranged the problem to
$$\left(\frac{1+z}{z}\right)^n=1$$
i.e.
$$\left(1+\frac{1}{z}\right)^n=1$$
so I know that in each solution, the value $1+\frac{1}{z}$ takes on an $n$th root of unity ($\omega^0,\omega^1,\dots,\omega^{n-1}$), and hence each solution has the form
$$z=\frac{1}{\omega^i-1},\quad \text{where } i\in\{0,\dots,n-1\}$$
But how can I use this here? I was thinking of using either the fact that $\sum_i\omega^i=0$ or that $z^n+\frac{1}{z^n}=2\cos (n\arg z)$, but I'm not sure how to involve them.
 A: As a one-way implication with $n \not = 0$: 
$$(1+z)^n=z^n \implies \left|(1+z)^n\right|=\left|z^n\right|\implies \left|1+z\right|^n=\left|z\right|^n \implies \left|1+z\right|=\left|z\right| $$ and this is only true if $\Re(z)=-\frac12$, since you can rewrite it as $\left|z-(-1)\right|=\left|z-0\right|$ and take the perpendicular bisector on the Argand diagram 
A: Hint
$$\left(\frac{1+z}{z}\right)^n=1\iff \frac{1+z}{z}=e^{\frac{2ik\pi}{n}},\quad k=0,...,n.$$
Isolate $z$ and compute $$\Re(z):=\frac{z+\bar z}{2}.$$
A: Since $|z+1|,\ |z| \geq 0$, we have:
$$\displaystyle{(1+z)^n=z^n \Rightarrow |(1+z)^n|=|z^n| \Rightarrow {|1+z|}^n=|z|^n \Rightarrow |z+1|=|z| \Rightarrow {|1+z|}^2=|z|^2 \Rightarrow (1+z)(1+\bar z)=z\bar z \Rightarrow z+\bar z=-1 \Rightarrow \Re z =-\frac{1}{2}.}$$
A: From $(1+z)^n=z^n$ we derive $|1+z|=|z|$. If $z=x+iy$ (with $x,y \in \mathbb R$), then we get
$x^2+y^2=(x+1)^2+y^2$. This gives $2x+1=0$.
A: Put $\;w:=1+\frac1z\;$ , so we have to solve
$$w_k^n=1=e^{2k\pi i}\implies w_k=e^{2k\pi i/n}\implies$$
$$1+\frac1{z_k}=e^{2k\pi i/n}\implies z_k=\frac1{e^{2k\pi i/n}-1}=\frac{e^{-2k\pi i/n}-1}{2\left(1-\cos\frac{2k\pi}n\right)}$$
and the real part is
$$\frac{\cos\frac{2k\pi}n-1}{2\left(1-\cos\frac{2k\pi}n\right)}\;\;\ldots$$
