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 $-3/2$.

$\left(\dfrac{z+3}{z}\right)=(1)^n= e^{\dfrac{2\pi i k}{n}}$ where k=0,1,2.....(n-1)

$\left(1+\dfrac{3}{z}\right)=(1)^n= \alpha^k$ where $\alpha=e^{\dfrac{2\pi i}{n}}$

$\implies z=\dfrac{3}{\alpha^k-1} ;k\neq0$

roots of your equation will be $z=\dfrac{3}{\alpha-1},\dfrac{3}{\alpha^2-1},.......\dfrac{3}{\alpha^{n-1}-1}$

but we know, for $n^{th}$ roots unity$\implies$ $\alpha^{n-1}=\dfrac{\alpha^n}{\alpha}=\dfrac{1}{\alpha};\alpha^{n-2}=\dfrac{1}{\alpha^2}$and so on

thus roots of your equation will be $z=\dfrac{3}{\alpha-1},\dfrac{3}{\alpha^2-1},.......\dfrac{3\alpha^2}{1-\alpha^2},\dfrac{3\alpha}{1-\alpha}$