Constructing Riemann surfaces

Riemann surfaces of algebraic functions are exposed in the textbook "Forster, Otto: Riemann surfaces. Chapter I, Sect. 8 Algebraic functions." (I refer to the German edition from 1977 and take over its notations.) Existence of these Riemann surfaces is proved by means of holomorphic coverings. After presenting the general theorem Forster explicitely calculates the example $\sqrt[2] {f(z)}$.

Your question asks for the Riemann surface $Y$ of $\sqrt[4] {f(z)}$ with $f(z) = z^4-1$, notably for its genus $g(Y)$.

Hence we work with compact Riemann surfaces. We start with the Riemann surface $X = \mathbb P^1$ and the polynomial

$$P(T) = T^4 - f(z) \in \mathscr M(X)[T]$$

over the field $\mathscr M(X)$ of meromorphic functions on projective space $\mathbb P^1.$ Note that the holomorphic function $f(z)$ is a polynomial of degree $4$. It can be considered a meromorphic function on $X$ with a pole of order $4$ at $\infty \in X$.

The polynomial $P(T) \in \mathscr M(X)[T]$ is irreducible, because its discriminant $\Delta \in \mathscr M(X)$ does not vanish identically. By the general theory (Theorem 8.9) a compact Riemann surface $Y$, a proper holomorphic covering

$$\pi:Y \longrightarrow X$$

of degree $4$ and a meromorphic function $F \in \mathscr M(Y)$ with $(\pi^*P)(F) = 0$ exist. The latter means

$$F^4(y) = f (p(y)) \ for \ all \ y \in Y.$$

It explains why the triple $(Y,\pi, F)$ is named the Riemann surface defined by $P(T)$ and why the algebraic function $F$ is named the $4$-th root of f. The triple $(Y, \pi, F)$ is uniquely determined up to canonical isomorphy.

We compute the genus $g(Y)$ by the Riemann-Hurwitz formula

$$g(Y) = (b/2) + n * (g(X) - 1) + 1$$

with $n = 4$, $g(X) = 0$ and $b$ the total branching order of $\pi$. The image of the branch divisor of $\pi$ is the set

$$A := \{1,-1, i, -i, \infty\} \subset X.$$

For each $a \in A, a \neq \infty$, on a small simply connected neighbourhood $U_a \subset X$ of $a$ we have $f(z) = (z - a) * g_a(z)$ with

$$g_a(z) = \prod_{x \in A - \{a \} } (z-x)$$

holomorphic without zeroes. Hence a holomorphic function $h_a$ in $U_a$ exists with $g_a(z) = h_a^4(z)$. And for $z \in U_a$ holds

$$f(z) = (z-a) * h_a^4(z).$$

With $U_a^* := U_a - \{a \}$ the restriction $\pi \ | \pi^{-1}(U_a^*) \longrightarrow U_a^*$ is an unbranched 4-fold covering and $\# \pi^{-1}(a) = 1$. The branching order of $\pi$ at $y \in \pi^{-1}(a)$ is $4 -1 = 3$.

On the other hand, on a small simply connected neighbourhood $U_{\infty} \subset X$ of $\infty \in X$ we have $f(z) = z^4*g_{\infty}(z)$ with $g_{\infty}(z)= 1 - (1/z^4)$ holomophic and without zeroes. Hence

$$f(z) = h_{\infty}^4(z)$$

with a meromorphic function $h_{\infty}$ on $U_{\infty}$. The restriction $\pi \ | \ \pi^{-1}(U_{\infty}) \longrightarrow U_{\infty}$ splits into $4$ disjoint coverings, each of degree 1. We obtain $\# \pi^{-1}(\infty) = 4$, and $\pi$ has branching order $0$ at every point $y\in \pi^{-1}(\infty)$.

Summing up we get $b = 4 * 3 = 12$ and $g(Y) = 6 -4 + 1 = 3$.