Example: Consider the Maclaurin series of f(z) = 1/(1-z). Note that f is analytic for |z| < 1 and indeed, on \mathbb C \setminus \{1\}. Moreover, f^{(n)}(z) = \frac{n!}{(1-z)^{n+1}}, \quad z \ne 1. In particular, f^{(n)}(0) = n!. This tells us that the Taylor series of f at 0 is given by T_{f,0}(z) = \sum_{n=1}^\infty z^n. This has \Lambda = \lim_{n \to \infty}1/1=1 which implies R = 1. Taylor’s theorem implies that T_{f,0} converges to f for |z| < 1.
Some things to note here. This is exactly the geometric series formula, which says \frac 1 {1-z}=\sum_{n=0}^\infty z^n for |z|<1. The series converges “out to the first singularity”, here 1.
Example: Find the Maclaurin series of 1/(2+4z) on \mathbb C \setminus \{-1/2\}. Note that we can manipulate it into the familiar form above. \frac 1 {2+4z} = \frac{1/2}{1+2z} = \frac {1/2}{1-(-2z)}=\frac 1 2 \sum_{n=0}^\infty(-2z)^n = \sum_{n=0}^\infty (-1)^n 2^{n-1}z^n, \quad \text{for }|-2z|<1. Example: f(z) = (1+2z^2)/(z^3+z^5). Some clever algebra tricks lead to f(z) = \frac 1{z^3}\left(\frac{2+2z^2}{1+z^2}-\frac 1 {1+z^2}\right) = \frac 1 {z^3}\left(2 - \frac 1 {1+z^2}\right) which is analytic on \mathbb C \setminus \{0, \pm i\}. Note that this wont converge around 0. For |z|<1, 1/(1+z^2) = \sum_{n=0}^\infty (-1)^n z^{2n}. \begin{aligned} f(z) &= \frac 1 {z^3} (2 - (1 - z^2 + z^4 - \cdots)) = \frac 1 {z^3} + \frac 1 z +- z + z^3 + \cdots. \end{aligned} Although this is not defined at 0, it is still useful. It is almost a power series but has some terms involving negative exponents of z.
By Weierstrass (1841) / Laurent (1843).
Theorem. Let f be analytic on the open annulus (donut-like shape) A = \{z : r_1 < |z-z_0| < r_2\}, centred at z_0. Let C be a positively oriented, simple, closed curve in A, and let z \in \operatorname{Int}C. Then, f has a series representation, the Laurent series, f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n + \sum_{n=1}^\infty \frac{b_n}{(z-z_0)^n} on A, where \begin{aligned} a_n &= \frac 1 {2\pi i}\int_C \frac{f(\xi)}{(\xi-z_0)^{n+1}}\,d\xi, \qquad b_n = \frac 1 {2\pi i}\int_C \frac{f(\xi)}{(\xi-z_0)^{-n+1}}\,d\xi. \end{aligned} The b_i’s are essentially coefficients of terms of the series with negative exponents (notice the negative power in the integrand the division by (z-z_0)^n).
Alternatively, this can be written as f(z) = \sum_{n=-\infty}^\infty c_n(z-z_0)^n, \quad \text{where }c_n = \frac 1 {2\pi i}\int_C \frac{f(\xi)}{(\xi-z_0)^{n+1}}\,d\xi. In particular with a Laurent series, b_1 = \frac 1 {2\pi i} \int_C \frac{f(\xi)}{(\xi-z_0)^{-1+1}}\,d\xi = \frac 1 {2\pi i}\int_C f(\xi)\,d\xi. This means that if we know the Laurent series, we know the value of this contour integral. This is so important that is has a name, the residue of f at z_0, denoted \operatorname{res}_{z=z_0}f(z). The residue of an analytic function is zero. For example, 1/z has residue 1 at the origin but 0 elsewhere. Here, we are also concerned about the very particular behaviour of the coefficient of z^{-1}.
Notes:
Example: Find the Laurent series of e^{1/z}. We can just use change of variables with the exponential Taylor series to get e^{1/z}= \sum_{n=0}^\infty \frac 1 {n! z^n} = 1 + \frac 1 z + \frac 1 {2!z^2} for all |z| > 0. Here, there is only one term in the Taylor series part, which is 1. Looking at the coefficient of z^{-1}, we know that \frac 1 {2\pi i}\int_C e^{1/\xi}d\xi = b_1 = 1 for all circles about the origin.
Example 7: Compute I = \int_C \frac{5z-2}{z(z-1)}\,dz where C = 2e^{i\theta} for \theta \in [0, 2\pi]. Note that the integrand f(z) = \frac{5z-2}{z(z-1)} is analytic everywhere except for 0 and 1.
Definition. We say f : \Omega \to \mathbb C has a singularity at z_0 if
In particular, we say f has an isolated singularity at z_0 if f is analytic on B_\epsilon(z_0)\setminus \{z_0\} for some \epsilon > 0.
Examples:
The function f(z) from Example 7 above has isolated singularities at 0 and 1.
\operatorname{Log}z has non-isolated singularities on the negative real axis including the origin.
\sin z / z has an isolated singularity at 0.
1/\sin(\pi/z) has isolated singularities at z = 1/k for k \in \mathbb Z and a non-isolated singularity at 0.