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Your question is not very clear. However, I guess you are asking for the branch points of the cover of $\mathbb{C}$ defined by $f(z, \alpha)=0$.

In this case, let us assume for the sake of symplicity that $f(z, \alpha)$ is monic in $z$; then

$f(z, \alpha)=z^n + f_{n-1}(\alpha)z^{n-1}+ \cdots + f_0(\alpha)$.

The Riemann surface $X \subset \mathbb{C}^2$ defined by $f(z, \alpha)=0$ is a cover $X \to \mathbb{C}$ of degree $n$, defined by $(z, \alpha) \to \alpha$.

The branch points of this cover are precisely the points $\bar{\alpha}$ such that $f(z, \bar{\alpha})$ has a multiple root. This is equivalent to require that $f(z, \bar{\alpha})$ and $f_z(z, \bar{\alpha})$ have a common root.

In other words, the branch points are the roots of the equation

$\textrm{Disc}_z(f)=0,$

where $\textrm{Disc}_z(f)$ denotes is the discriminant of $f$ with respect to $z$, i.e. the resultant of $f$ and $\partial f / \partial z$ computed with respect to $z$.

For instance, the branch points of the Riemann surface defined by

$z^3+f_1(\alpha)z+f_0(\alpha)=0$

are the roots of

$4 f_1(\alpha)^3+27 f_0(\alpha)^2=0$.

1

Your question is not very clear. However, I guess you are asking for the branch points of the cover of $\mathbb{C}$ defined by $f(z, \alpha)=0$.

In this case, let us assume for the sake of symplicity that $f(z, \alpha)$ is monic in $z$; then

$f(z, \alpha)=z^n + f_{n-1}(\alpha)z^{n-1}+ \cdots + f_0(\alpha)$.

The Riemann surface $X \subset \mathbb{C}^2$ defined by $f(z, \alpha)=0$ is a cover $X \to \mathbb{C}$ of degree $n$, defined by $(z, \alpha) \to \alpha$.

The branch points of this cover are precisely the points $\bar{\alpha}$ such that $f(z, \bar{\alpha})$ has a multiple root. This is equivalent to require that $f(z, \bar{\alpha})$ and $f_z(z, \bar{\alpha})$ have a common root.

In other words, the branch points are the roots of the equation

$\textrm{Disc}_z(f)=0,$

where $\textrm{Disc}_z(f)$ denotes the resultant of $f$ and $\partial f / \partial z$ computed with respect to $z$.

For instance, the branch points of the Riemann surface defined by

$z^3+f_1(\alpha)z+f_0(\alpha)=0$

are the roots of

$4 f_1(\alpha)^3+27 f_0(\alpha)^2=0$.