complex zeros of difference of $n$th powers of affine polynomials

Question

I am trying to find the general complex solutions to the equation $$(az-b)^n=(cz-d)^n,$$

assuming $a,b,c,d\in\mathbb{C}\setminus\lbrace 0\rbrace$ and all are distinct, which is naturally satisfied by $$z_0=\dfrac{b-d}{a-c}$$ with (I think) multiplicity $1$ but as this is equivalent to the zeros of a complex polynomial of degree $n$, there are $n-1$ solutions missing, by the fundamental theorem of algebra. I tried to adjoin a root of unity $\mathrm{e}^{2\pi i n^{-1}}$ to $z_0$ to obtain further solutions, but that failed. How do I find the remaining solutions as a function of $a,b,c,d$?

Answer 1

Thanks to the comment, I can give the solution here. Any $z_j$ is given by

$$z_j=\frac{d-b e^{\frac{2\pi i j}{n}}}{c-a e^{\frac{2\pi i j}{n}}}$$

for $j\in\lbrace 0,...,n-1\rbrace$ and I thank you for the cake.