The monodromy representation of the projection map from a Fermat Curve

I have been trying to solve the following problem for quite some time now: Let X denote the Fermat curve of degree d in $\mathbb{P}^2$, defined by the homogenous polynomial $$x^d+y^d+z^d=0$$. Let $F:X \rightarrow \mathbb{P}^1$ be defined by $F([x:y:z]) = [x:y]$, show that F has d branch points, and find the d corresponding permutations.

This is what I have done so far, first, I've noted that for a given point in the image, the number of preimages to it are equal to the number of solutions in z to $z^d = -(x^d+y^d)$, and in general, this number is d, and we have ramification exactly when $x^d+y^d=0$. This gives that $x = \zeta y$, where $\zeta^d = -1$. So there are d ramification points, each of order d,so we know that the monodromy representation must consist of d cycles $\sigma_0,..,\sigma_d$ of length d in $S_d$, and further we must have that $\sigma_0\sigma_1\dotsm\sigma_d=(1)$.

Now, I thought of somehow getting a local coordinate and from this finding the permutations, but I can't seem to make it to work. Should I triangulate some surface (as was so cleverly done in my previous question here:Calculating Monodromy)?

Any help would be appreciated. Thank you.

Notice that for any $\xi$ s.t. $\xi^d=1$ the map $P_\xi:[x:y:z]\mapsto [\xi x:y:z]$ is a symmetry of your curve; the corresponding map $p_\xi:\mathbb{P}^1\to\mathbb{P}^1$ is $[x:y]\mapsto [\xi x:y]$ (so that $F\circ P_\xi=p_\xi\circ F$). As a result, all $\sigma_i$'s are equal to each other, hence the monodromy group is $C_d$.