Confusion about Poles A very basic question as usual, i was doing a question, say find the isolated singularities and classify them. $$\dfrac{z^{3}+1}{z^{2}(z-1)}$$
There are two singularities at $z=0$ and $z = 1$, so i compute the one at $z=0$ first. By partial fractions, $$f(z) = \dfrac{-1}{z^{2}}+\dfrac{2}{z-1}$$
So the laurent series about $z=0$ should be $$-\dfrac{1}{z^2}-2-2z-2z^2-...$$ for $0<|z|<1$. And hence it should be a pole of order 2 at $z=0$. But due to my weak concept, i cannot understand one thing, i believe we can also expand the above function in the region $1<|z| < \infty$ and then it will become something different (essential singularity?)? Which part of the concept am i missing here? thanks 
 A: A rational function has no essential singularities.
Partial fractions give
$$
\frac{z^3 + 1}{z^2(z - 1)} = -\frac{1}{z^2} - \frac{1}{z} + \frac{2}{z - 1} + 1
$$
Therefore, $z=0$ is a double pole and $z=1$ is a simple pole.
But that can also be gleaned directly from the denominator $z^2(z - 1)$.
In the partial fraction expression, note that near $z=0$ the function $\dfrac{2}{z - 1} + 1$ is holomorphic and so its Laurent series does not include terms of negative degree. Similarly for $-\dfrac{1}{z^2} - \dfrac{1}{z}$ near $z=1$.
A: First of all, your partial fractions do not look correct. Moreover, expanding in powers of $z$ will give you a Laurent's series of the form $\sum_{k=-\infty}^{\infty}a_kz^k$ with $a_{-2}\ne 0$ and $a_{-k}=0$ for all $k>2$, thereby suggesting a pole of order $2$ at $z=0$. 
But expanding $f(z)$ in powers of $(z-1)$ will give you a series of the form $\sum_{k=-\infty}^{\infty}a_k(z-1)^k$ with $a_{-1}\ne 0$ and $a_{-k}=0$ for all $k>1$, there suggesting a simple pole at $z=1$.
The first series gives the behaviour of $f(z)$ at $z=0$ while the second one gives the behaviour of it at $z=1$.
