Position and nature of singularities of an algebraic function (Ahlfors)

complex-analysis algebraic-geometry

Solution 1:

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.

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