Laurent series at $z = 0$. I have the following two functions: $f(z) = z^{\frac{2}{3}}$ and $f(z) = (z+1)^{-\frac{4}{3}}$.
For the first one I used Taylor's theorem and there is no Laurent series for the first one. It's simply $z^{\frac{2}{3}}$. Can I just say since the negative part of the Laurent series is $0$ it's removable singularity. For the second one: $$ f(z) = 1 - \frac{4}{3}z + \frac{14}{9}z^2 - \cdots$$ It's just essential singularity .
If the Laurent series is a Taylor series, you have a removable singularity (ie you can define the function f at z_0 to be analytic).
If the Laurent series has finitely many negative powers of (z-z_0), then it is a pole (at z_0). This is because we can write a function f as f(z) = g(z)/(z-z_0)^m, where g is analytic at z_0, g(z_0) is not 0, and m is the order of the pole.
Otherwise, it is an essential singularity.