Confusion with the definition of a pole The question given to me is to determine the order of poles for $$f(z) = \frac{3z - 1}{z^2 + 2z + 5}$$
I know the singularities are $(-1 + 2i)$ and $(-1 -2i)$ by factorizing the denominator. Since I wanted to understand what a pole really was, To expand f(z) at $z_0$ = $(-1 + 2i)$ using Laurent series, I rearranged the expression. I got $$\frac{3z - 1}{(z - (-1 - 2i))}\left(1 - \frac{-1 + 2i}{z}\right)^{-1}\cdot\frac{1}{z}$$ OR $$\frac{3z - 1}{z(z - (-1 - 2i))}\left(1 - \frac{z}{-1 + 2i}\right)^{-1}\cdot\frac{-1}{-1 + 2i}$$
depending on which one converges.
If the first one converges, then doesn't that mean $(-1 + 2i)$ isn't a pole because the principal part contains infinite terms?
EDIT:
I have been told that my rearrangement is incorrect, which is sensible to me. Here's the corrected rearrangement:
Take w = z - (-1 + 2i). The question now becomes $$\frac{3(w + (-1 + 2i)) - 1}{(w+4i)w}$$. This can be rearranged as $$\frac{3(w + (-1 + 2i)) - 1}{w}\cdot\left(1+\frac{w}{4i}\right)^{-1}\cdot\frac{1}{4i}$$ if |w| < |4i| or $$\frac{3(w + (-1 + 2i)) - 1}{w}\cdot\left(1+\frac{4i}{w}\right)^{-1}\cdot\frac{1}{w}$$ if $|4i| < |w|$. But in this case, the expansion would only consist of an infinite principal part, meaning that $w = 0$ i.e. $z = -1 + 2i$ is not a pole. So, I would conclude that the answer to whether a complex function has a pole might depend on the annulus considered but that still doesn't make much sense to me.
 A: To decide if a singularity is removable, a pole or essential, we must expand the function in a Laurent series that converges in a deleted neighborhood of $z_0$. That is, the Laurent series must converge for all $z$ in the disk $0 < |z-z_0| < R$ for some $R$.
The answer to your question is, therefore, no. To make the "other series" to converge, we must choose an annulus of convergence which is "far away" from our point of interest, $-1+2i$.
A: I like to think of the "order" of the poles as simply how many terms exist in the Laurent series with the said singularity. For example, consider the Laurent series
\begin{equation*}
f(z) = \sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n} + \sum_{n=0}^{\infty} a_n(z-z_0)^n.
\end{equation*}
If $b_n=0$ for $n=\{k,...,n\}$ then $f$ has a pole at the singularity $z_0$, of the order $k$. 
Without answering the question directly and allowing you to toy with it some more, I hope this helps your general understanding of poles.
In this case, the factorization of the denominator should give the answer fairly quickly.
A: Suppose we wanted to construct the Laurent series centered at $z_0 = -1 + 2i$ This can get very tricky but its a good exercise. I'll just get you started:
\begin{align}
\frac{3z-1}{z^2 - 2z +5} &= \left(\frac{1}{z-(-1+2i)}\right)\frac{3z-1}{(4i + (z-(-1+2i))}\\
&= \left(\frac{1}{z-(-1+2i)}\right)\frac{3(z-(-1+2i)) - 1 + 3(-1+2i)}{4i\left(1 - \frac{z-(-1+2i)}{-4i}\right)}\\
&=\left(\frac{1}{z-(-1+2i)}\right)\left[-3\sum_{n=0}^\infty \left(\frac{z-(-1+2i)}{-4i}\right)^{n+1} + \frac{-4+2i}{4i}\sum_{n=0}^\infty \left(\frac{z-(-1+2i)}{-4i}\right)^{n}\right]
\end{align}
Provided $|z-(-1+2i)|<|4i| = 4$. (Look over this carefully as it may not be exactly right, however, I do commend it to your consideration.) The the main goal is to get everything in term of $z - z_0$. I leave it to you to collect terms so that it is in the form @MathDoer2320 presents. Then ask what what is the largest $k$ for which $b_k$ is nonzero and that will give you the order of the pole. For more fun: what does the Laurent series look like when $|z-(-1+2i)| > 4$? Does this suggest a different order for the pole?
A: Right, consider this characterization of a pole: for a function $f$, analytic on $0<|z-z_0|<r$ for some $r<0$, $z_0$ is a pole of order $n$ if $$\lim_{z\rightarrow z_0} |z-z_0|^n|f(z)| \qquad \text{exists (i.e. $\ne \infty$) and is nonzero}.$$ 
A pole is a local property and is determined in a neighborhood of the pole. In general, I think, the Laurent series only defined in a region bounded away from the point in question can't tell us much about that point. The Luarent series is only helpful in a neighborhood of the point. A very good discussion on poles (and zeros) is found in Ahlfors, Complex Analysis, in the third edition its in ch. 4 sec. 3.2. 
A: I consulted with my professor and I can answer my own question now. I also extend thanks to @JMcB.
A function can only be expressed as a Laurent series when it is analytic on the annulus $r_1 < |z - z_0| < r_2$. If the annulus contains a singularity, a function cannot be analytic on that annulus.
A function f being analytic in a punctured disk $0 < |z - z_0| < r$ is a necessary condition for existence of pole. Let's consider three puncture disks centered at the singularity $z_0 = (-1 + 2i)$:


*

*$0 < |z - (-1 + 2i)| < 3.9$ -- This disk does not contain the points
$(-1 + 2i)$ and $(-1 - 2i)$. 

*$0 < |z - (-1 + 2i)| < 4.0$ -- This
disk does not contain the points $(-1 + 2i)$ and $(-1 - 2i)$. 

*$0 < |z - (-1 + 2i)| < 4.1$ -- This disk does not contain the points $(-1
+ 2i)$, but contains the point $(-1 - 2i)$. You cannot expand the function in Laurent series in this annulus.


The circle $|z - (-1 + 2i)| = |4i| = 4$ contains the point $(-1 - 2i)$. For $r > 4$, $|z - (-1 + 2i)| > |4i|$. Any punctured disk with $r >= 4$ will contain the singularity $(-1 -2i)$, so the function will not be analytic there.
We can say that $(-1 + 2i)$ is a pole of order $1$ because the function is analytic in the punctured disk $0 < |z - (-1 + 2i)| < r$ for some small $r$ ($r < 4$) and it can be expanded in Laurent series. The principal part of this Laurent series will only contain the term with power of $-1$, so it has the order $1$.
Similarly, $(-1 - 2i)$ is also a pole of order $1$ because the function is analytic in the punctured disk $0 < |z - (-1 - 2i)| < r$ for some small $r$ ($r < 4$ again).
