The roots of the equation $z^n=(z+3)^n$ 
Prove that the roots of the equation $z^n=(z+3)^n$ are collinear

Taking modulus on both sides, $$\vert z-0\vert =\vert z-3\vert$$
This represents the perpendicular bisector of line joining $x=0$ and $x=3$
That was easy. But I tried to solve it using algebra:
$$\frac{z+3}{z}=1^{\frac{1}{n}}=\cos{\frac{2k\pi}{n}}+i\sin{\frac{2k\pi}{n}}$$
After simplifying, I got $$z=\frac{3i ( \cos{\frac{k\pi}{n} }+i\sin{ \frac{k\pi}{n}})}{2\sin{ \frac{k\pi}{n}}}$$
What should I do next?

Hint:the roots have the same real part which is $-3/2$.
