The roots of the equation $z^n=(1+z)^n$... The complex roots of the equation 

$$z^{n}=(1+z)^{n}$$

$A.$ are vertices of a regular polygon
$B.$ lie on a circle
$C.$ are collinear
$D.$ none of these
Don't know where to start.. Please help.!
NOTE:  $z$ is a complex number.
 A: An easier method, stolen from comment to an answer from a similar thread.
If $z$ satisfies $z^n=(1+z)^n$, then $|z|^n=|1+z|^n$ and: $$|z|=|z+1|$$
The geometric interpretation is that the distance (in the complex plane) from $0$ to $z$ equals the distance from $-1$ to $z$. The locus of points with the same distance from $0$ and $-1$ is the vertical line: $$\mathrm{Re}\ z = -\frac12$$ so every solution to the original equation lies on this line.
A: Rewrite as $\left(\frac z{1+z}\right)^n = 1$.  Now from this you can work out the value of $\frac z{1+z}$, and note in particular that they lie on the unit circle.  And $\frac z{1+z}$ is a Möbius transform, so you should be able to figure out how its inverse maps the unit circle, remembering that Möbius transforms map circles and lines to circles and lines.
A: Clearly, $z\ne0,$
So dividing either sides by $z^n,$ we have $$\left(1+\frac1z\right)^n=1$$
Using  de Moivre's formula & Euler's formula, $$1+\frac1z=e^{\frac{2m\pi i}n}=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n$$ where $m$ is any integer
$$\implies \frac1z=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n-1=i2\sin\frac{m\pi}n\cos\frac{m\pi}n-2\sin^2\frac{m\pi}n$$
$$=2i\sin \frac{m\pi}n\left(\cos\frac{m\pi}n+i\sin \frac{m\pi}n\right)$$ (using $\cos2A=1-2\sin^2A$)
$$z=\frac1{2i\sin \frac{m\pi}n\left(\cos\frac{m\pi}n+i\sin \frac{m\pi}n\right)}=\frac{{\cos\frac{m\pi}n-i\sin \frac{m\pi}n}}{2i\sin \frac{m\pi}n}=-\frac12-i\frac{\cot\frac{m\pi}n}2$$
