Position and nature of singularities of an algebraic function (Ahlfors) I want to solve the following exercise, from Ahlfors' Complex Analysis text, page 306:

Determine the position and nature of the singularities of the algebraic function defined by $w^3-3wz+2z^3=0.$

Here is my solution attempt. I would appreciate your opinion (is it true?)
The critical points $\{c_k\}$ are the zeros of the leading coefficients as well as the zeros of the discriminant of the polynomial above, which I'll denote by $P(w,z)$ from now on. Since the leading coefficient $a_0(z)=1$ is non-vanishing, the critical points can only occur as zeros of the discriminant of $P(w,z)$ (which is the resultant of $P$ and $P_w)$. According to my calculations (i.e. obtaining the resultant by a sequence of polynomial divisions, starting with $P$ divided by $P_w$)  the discriminant is $3z^4-3z$
. It has zeros whenever $z_0=0$ or $z_k=\exp(2\pi ik/3)$ for $k \in \{1,2,3\}$. According to the analysis on page 304, all points $\{z_k \}_{k=0}^3$ are ordinary algebraic singularities.
We are left with examining the point $z=\infty$. Unfortunately, the book only finds a bound on the degree of the pole at infinity. Moreover, it is not even guarantee that it is a pole (apparently, it could also be an ordinary point). I have no idea how to determine the nature of the singular point at $\infty.$
To sum up, I have two questions:


*

*Is the part regarding the zeros of the discriminant true?

*How can I determine the nature at $z=\infty$?


Thanks.

Regarding 1, yes.
Regarding 2, we can show the following.

*

*$w$ cannot be bounded as $z$ tends to infinity, by dividing both sides of $w^3-4wz+3z^3=0$ by $z^3$.

*$\frac{w}{z}$ must be bounded as $z$ tends to infinity, by dividing both sides of $w^3-4wz+3z^3=0$ by $w^3$.

Therefore $z=\infty$ is an algebraic pole of order 1.
