Taylor and Laurent Series

Theorem 7.1. If $f$ is analytic on $|z - z_0| \lt R$Then

$f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z - z_0)^n$

And the series converges uniformly on compact subsets of $|z - z_0| \lt R$.

Proof. For $|z - z_0| \lt r \lt R$Apply Cauchy”s integral formula on $|\zeta - z_0| = r$:

$f(z) = \frac{1}{2\pi i}\int_{|\zeta - z_0| = r} \frac{f(\zeta)}{\zeta - z}\, d\zeta$

Write $\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \cdot \frac{1}{1 - (z - z_0)/(\zeta - z_0)}$ $= \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}$ (geometric series, convergent since $|z - z_0|/|\zeta - z_0| \lt 1$).

Substituting and integrating term by term gives the Taylor series. $\blacksquare$

Remark. The radius of convergence $R$ is the distance from $z_0$ to the nearest singularity of $f$.

$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \cdots$

$\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}$

$\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$

$\frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n, \quad |z| \lt 1$

$\ln(1 + z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n}, \quad |z| \lt 1$

Theorem 7.2 (Laurent Series). If $f$ is analytic on the annulus $r \lt |z - z_0| \lt R$ Then

$f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n = \cdots + \frac{a_{-2}}{(z - z_0)^2} + \frac{a_{-1}}{z - z_0} + a_0 + a_1(z - z_0) + \cdots$

Where

$a_n = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz$

For any simple closed contour $\gamma$ in the annulus encircling $z_0$.

The principal part is $\sum_{n=-\infty}^{-1} a_n(z - z_0)^n$ (negative powers). The analytic Part is $\sum_{n=0}^{\infty} a_n(z - z_0)^n$ (non-negative powers).

The Laurent series expansion depends on the annulus of convergence. A function may have different Laurent expansions in different annuli.

Proposition 7.3. The Laurent series expansion of $f$ in an annulus is unique.

• Wrong annulus choice. A function has different Laurent expansions in different annuli (e.g.\ $1/[z(z-1)]$ has distinct series in $0<|z|<1$ vs.\ $|z|>1$). Always identify all singularities and the correct annulus before expanding.

Definition. The residue at infinity of $f$ is defined as

$\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f(z)\, dz$

For sufficiently large $R$ (enclosing all finite singularities).

Proposition 7.4. For a function $f$ with finitely many singularities in $\mathbb{C}$:

$\sum_{\mathrm{all\ finite\ } z_k} \mathrm{Res}(f, z_k) + \mathrm{Res}(f, \infty) = 0$

Proof. By the residue theorem applied to $|z| = R$ enclosing all finite singularities:

$\int_{|z|=R} f\, dz = 2\pi i \sum_{\mathrm{finite} \mathrm{Res}(f, z_k)}$.

But $\mathrm{Res}(f, \infty) = -\frac{1}{2\pi i}\int_{|z|=R} f\, dz$So the sum is zero. $\blacksquare$