Residues in singular points of complex function. I am asked to get the residues in the singular points of $f(z) = \frac{z^2 + 1}{z^2(z + 2)}$ . 
The problem is that I cant find what a singular point is for a complex function and how to get the residues in that point.
Any help will be much appreciated.
 A: A singular point is a (usually isolated) point where your function is not holomorphic. For your example, $z=0$ and $z=-2$ are singularities. More precisely, $z=0$ is a double pole and $z=-2$ is a simple pole. 
If you know nothing about how to compute residues, please read up on it in your textbook. For simple poles, there are a couple of fairly easy methods. For example, if $f = p/q$ and $q$ has a simple zero at $z=a$, then
$$
\operatorname{Res}\limits_{z=a} f(z) = \frac{p(a)}{q'(a)} = \lim_{z\to a} \frac{(z-a)p(z)}{q(z)}.
$$ 
In your example:
$$
\operatorname{Res}\limits_{z=-2} f(z) = \lim_{z\to -2} \frac{(z+2)(z^2+1)}{z^2(z+2)} = \frac54.
$$ 
For poles of higher order, things get a little more complicated. For a pole of order $k$, it turns out that
$$
\operatorname{Res}\limits_{z=a} f(z) = \frac{1}{(k-1)!} \lim_{z\to a} \frac{d^{k-1}}{dz^{k-1}} \big((z-a)^k f(z)\big).
$$
I'll leave it up to you to do the computation for the double ($k=2$) pole at $z=0$.
A: Singular points means what values of z make the denominator 0.  In other words, what satisfies $z^2(z-2)=0$?  (0 and 2).  Now, each of these points is a pole of a certain order.  Basically, $\lim_{z\to z_o} (z-z_o)^m f(z)$, where $z_o$ is your pole, is defined, and m is the order of the pole.  To find the residue of a pole of order 1, you can do $\lim_{z\to z_o} (z-z_o)f(z)$ or $\dfrac{F(z_o)}{G'(z_o)}$, where F is the portion of f(z) that's defined for the point $z_o$.  For a pole of order m, you need to do $\dfrac{F^{(m-1)}(z_o)}{(m-1)!}$.
