I know that if $P \in \mathbb{C}[z]$ is a monic polynomial with distinct roots,
$$P(z)=(z-a_1) \ldots (z-a_k),$$
then the set
$$S= \left\{ (z,w)\in \mathbb{C}^2 : w^2=P(z) \right\}$$
is a connected Riemann surface. Now I want to consider the homogenization of the polynomial $w^2-P(z)$ and consider, for $k \geq 2$,
$$\bar{S}=\left\{[z,w,t]\in \mathbb{C}P^2 : w^2t^{k-2}=(z-a_1t)\ldots (z-a_kt)\right\}.$$
My goal is to show that $\bar{S}$ is a Riemann surface iff $k \leq 3$.
For $k=1$, we get
$$\bar{S}=\left\{[z,w,t]\in \mathbb{C}P^2 : w^2=(z-a_1t)t\right\},$$
which is a non-degenerate quadric, and then a Riemann surface.
Similarly, for $k=2$ we get
$$\bar{S}=\left\{[z,w,t]\in \mathbb{C}P^2 : w^2=(z-a_1t)(z-a_2t)\right\},$$
another non-degenerate quadric.
For $k=3$, we get the condition
$$w^2t=(z-a_1t)(z-a_2t)(z-a_3t),$$
and I guess the idea here would be to show that there’s no solution to $f=\partial f/\partial z=\partial f/\partial t=0$ where $f(z,w,t)=w^2t-(z-a_1t)(z-a_2t)(z-a_3t)$, which I haven’t tried yet. Is there a simpler way?
For $k>3$, I’m out of ideas. Any help would be appreciated.