Recall that isolated singularities fall into three cases. These are points in the complex plane where a function runs into trouble, but is fine in a punctured disk around that point. The three cases are removable, poles, and essential singularities.
These can be defined by, in a deleted ball B_r(z_0) \setminus\{z_0\}, \begin{aligned} &\text{(Case I)} &&\sum_{n \ge 0}a_n(z-z_0)^n \\ &\text{(Case II)} &&\sum_{n \ge -N}a_n(z-z_0)^n, \quad N>1, a_N \ne 0 \\ &\text{(Case III)} && \sum_{n=-\infty}^\infty a_n (z-z_0)^n \end{aligned} Example: f(z) = \sin (z) /z is a function f:\mathbb C^* \to \mathbb C with a singularity at z=0. We can expand \sin into its Taylor series to show that \sin z = \sum_{n=0}\frac{(-1)^nz^{2n+1}}{(2n+1)!} \implies f(z) = \frac 1 z\left(z - \frac {z^3}{3!}+ \cdots\right). The powers of z are all non-negative which means the function is analytic on the deleted ball, so the singularity is removable. If we wanted to define an entire function, we could say \hat f = f on \mathbb C^* and \hat f(0) = a_0=1. Also, the residue is 0.
Example: f(z) = \frac{z^2-1}{z^2-1}\ne 1 which is f : \mathbb C \setminus \{\pm 1\} \to \mathbb C. There are (clearly) two removable singularities at \pm 1. We can’t equate this function to 1, but we could define \hat f : \mathbb C \to \mathbb C with \hat f(z) = 1. This ‘fixes’ the singularity in such a way that you could never tell it was there.
Example: f(z) = 1/z^4 is again defined on \mathbb C^*. There is an isolated singularity at z=0 and this is a pole of order 4.
Example: f(z) = 1/z^4 + 1/z^2 is the same story. The pole is still of order 4; we don’t care about the higher order terms.
Example: f(z) = \sinh (z^2)/z^7, with f : \mathbb C^* \to \mathbb C. Remember that \sinh z = \sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!}. Then, f(z) = \frac 1 {z^7}\sum_{n \ge 0} \frac{z^{4n+2}}{(2n+1)!} = \sum_{n \ge 0}\frac{z^{4n-5}}{(2n+1)!} = \frac 1 {z^5} + \frac 1 z \frac 1 {3!}+\cdots We have a pole of order 5 and the residue is 1/3! in this case.
Example: f(z) = e^{1/z}. In this case, f(z) = \sum_{n=0}^\infty \frac{(1/z)^n}{n!} = 1 + \frac 1 z + \frac 1 {z^2}\frac 1 {2!} + \cdots so we have an essential singularity at z=0 and the residue is 1.
(This is the big version.)
Theorem. Suppose f has an essential singularity at z_0. Let R > 0 be arbitrarily small. Then, on B_R(z_0) \setminus\{z_0\}, the function f attains every value in \mathbb C infinitely often, with the possible exception of one point.
Theorem. If f has a pole of order N at z_0, then in B_r(z_0)\setminus \{z_0\} we can write f(z) = \frac{\phi(z)}{(z-z_0)^N} with \phi analytic in B_r and \phi(z_0)\ne 0. Moreover, \operatorname{res}_{z=z_0} f(z) = \frac{\phi^{(N-1)}(z_0)}{(N-1)!}.