What is an example of an essential singularity? Consider f(z) = 1/(1-z^{-1}) for z \ne 0. What is the singularity at 0? Observe, f(z) = \frac {1}{1-\frac 1 z}=\frac {z}{z-1} = \frac{-z}{1-z}=-z(1+z+z^2+\cdots) \quad \text{for}\quad0 < |z| < 1. Thus, the singularity is removable because the Laurent series has only non-negative powers. Taking f(0) = 0 extends f to an analytic function on \mathbb C \setminus \{1\}. Moreover, this means the Laurent series is only value out to the singularity at z=1.
Recall that if f has a pole of order n at z_0 and we can write f(z) = \phi(z)/(z-z_0)^n for \phi analytic in B_r(z_0) and \phi(z_0)\ne 0, then \operatorname{Res}_{z=z_0}f= \frac {\phi^{(n-1)}(z_0)}{(n-1)!}. In particular, if z_0 is a simple pole, then \operatorname{Res}_{z=z_0}f=\phi(z_0).
Example: Consider f(z) = (z+i)/(z^2+9). f is analytic on \mathbb C \setminus\{\pm 3i\}, and \pm 3i are isolated singularities. Near z=3i, we can write f(z) = \frac {\phi(z)}{z-3i}\quad \text{where}\quad \phi(z) = \frac {z+i}{z+3i} noting that \phi is analytic and non-zero near 3i. The theorem tells us that \operatorname{Res}_{z=3i}f=\phi(3i)=4i/6i=2/3. We can do the same thing at -3i as well.
Example: f(z) = (z^3+2z)/(z-i)^3. Observe that this is analytic except at i. Near i, f(z) = \frac {\phi(z)}{(z-i)^3} \quad \text{and}\quad \phi(z) = z^3+2z with \phi analytic and \phi(i) \ne 0. Using the same theorem, \operatorname{Res}_{z=i}f={\phi''(i)}/{2!}=3i.
§82 (8 Ed §75)
We can generalise the theorem to talk about zeros of functions, because poles are essentially zeros of the denominator.
Lemma. If f is analytic at z_0, then f has a zero of order m at z_0 if \begin{cases} f^{(j)}(z_0)=0& \text{for }j=0, \ldots, m-1, \text{ and} \\ f^{(m)}(z_0)\ne 0. \end{cases} Example: f(z) = (z-i)^4(z-4) has a zero of order 4 at i and a simple zero (i.e. zero of order 1) at 4.
Theorem. f is analytic at z_0 and has a zero of order m at z_0 if and only if f(z) = (z-z_0)^m g(z) where g is analytic and g(z_0)\ne 0.
§83 (8 Ed §76)
Theorem. Suppose p and q are analytic at z_0, p(z_0)\ne 0, and q has a zero of order m at z_0. Then, p/q has a pole of order m at z_0.
Example: p(z)=1 and q(z)=z(e^z-1). We know that p/q has an isolated singularity at 0. p is analytic and non-zero everywhere (obviously). We can check that q(0)=0, q'(0)=0, q''(0)=1\ne 0. Thus, p/q has a pole of order 2 at 0.
Theorem. Let p, q be analytic at z_0. If p(z_0)\ne 0 and q(z_0)= 0, then p/q has a simple pole at z_0 and \operatorname{Res}_{z=z_0}\frac p q = \frac {p(z_0)}{q'(z_0)}. Note that there exist higher-order analogues, but they become messy.
The main application of this is contour integrals. In \mathbb R, recall that \begin{aligned} \int_{-\infty}^\infty f(x)\,dx=\lim_{m_1\to\infty}\int_{-m_1}^0 f(x)\,dx+\lim_{m_2\to\infty}\int_0^{m_2}f(x)\,dx. \end{aligned} Note we can replace the split 0 with any fixed c\in \mathbb R. If the limits on the right exist, then we say the integral exists with the given value.
We cannot in general replace the right-hand side with \lim_{m\to\infty}\int_{-m}^mf(x)\,dx. If we do this anyway, it defines the Cauchy principal value (PV) integral.
Example: Let’s look at this in practice. The below improper integral is undefined, because \begin{aligned} \int_{-\infty}^\infty x\,dx &= \lim_{m_1\to\infty}\int_{-m_1}^0x\,dx+\lim_{m_2\to\infty}\int_0^{m_2}x\,dx \\ &= \lim_{m_1\to-\infty}-m_1^2/2+\lim_{m_2\to\infty}m_2^2/2. \end{aligned} However, the principal value is \begin{aligned} \operatorname{PV}\int_{-\infty}^\infty x\,dx = \lim_{m\to\infty}\int_{-m}^m x\,dx=\lim_{m\to\infty}\left[\frac{m^2}2-\frac{m^2}2\right]=0. \end{aligned} Question: When does \operatorname{PV}\int_{-\infty}^\infty f=\int_{-\infty}^\infty f? One case is for even functions or non-negative functions. Specifically, if f is even (so f(x)=f(-x) for all x \in \mathbb R), then \int_0^\infty f(x)\,dx=\frac 1 2 \int_{-\infty}^\infty f(x)\,dx = \frac 1 2 \operatorname{PV} \int_{-\infty}^\infty f(x)\,dx and these integrals converge or diverge together.