Laurent Series | Sahithyan's Notes

For $0 \le r < R \le \infty$ , the annulus $A(r, R, \alpha)$ is

A(r, R, \alpha) = \{z \in \mathbb{C} : r < |z-\alpha| < R\}

Let $c_n \in \mathbb{C}$ for each $n \in \mathbb{Z}$ . The doubly infinite series

\sum_{n=-\infty}^{\infty} c_n (z-\alpha)^n = \sum_{n=1}^{\infty} c_{-n}(z-\alpha)^{-n} + \sum_{n=0}^{\infty} c_n (z-\alpha)^n

is called a Laurent series, provided both series on the right-hand side converge.

Laurent’s Theorem

Let $f$ be analytic in the annulus $A(r, R, \alpha)$ . Then $f$ has a Laurent series representation on $A(r, R, \alpha)$ .

f(z) = \sum_{n=-\infty}^{\infty} c_n (z-\alpha)^n, \quad z \in A(r, R, \alpha)

The coefficients are

c_n = \frac{1}{2\pi i}\int_{|\zeta-\alpha|=s} \frac{f(\zeta)}{(\zeta-\alpha)^{n+1}}\,\text{d}\zeta

Here:

The Laurent series representation of $f$ on $A(r, R, \alpha)$ is unique.
A Laurent series can be differentiated term by term inside its annulus of convergence.

Examples:

$\frac{\cos z - 1}{z^4} = -\frac{1}{2z^2} + \frac{1}{24} - \frac{z^2}{720} + \cdots$ centered at $0$
$\exp\left(-\frac{1}{z^2}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!\, z^{2n}}$ centered at $0$

The same function has a different Laurent series on each annulus bounded by its singular points.

Consider

f(z) = \frac{3}{2+z-z^2} = \frac{1}{2-z} + \frac{1}{1+z}

The singular points are $z = 2$ and $z = -1$ with $|z| = 2$ and $|z| = 1$ . The centered series at $0$ are as follows.

$|z| < 1$
$\sum_{n=0}^{\infty} \left(\frac{1}{2^{n+1}} + (-1)^n\right) z^n$
$1 < |z| < 2$
$\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} + \sum_{n=0}^{\infty} (-1)^n z^{-n-1}$
$|z| > 2$
$\sum_{n=0}^{\infty} \left((-1)^n - 2^n\right) z^{-n-1}$

Find $c_{-2}$ of $f(z) = \frac{1}{z^2}$ centered at $0$ .

c_{-2} = \frac{1}{2\pi i}\int_{|\zeta|=s} \frac{\zeta^{-2}}{\zeta^{-2+1}}\,\text{d}\zeta = \frac{1}{2\pi i}\int_{|\zeta|=s} \zeta^{-1}\,\text{d}\zeta = \frac{1}{2\pi i} \cdot 2\pi i = 1

For general $n$ , the integrand is $\zeta^{-n-3}$ .

The other coefficients vanish since $\int_{|\zeta|=s} \zeta^{k}\,\text{d}\zeta = 0$ for $k \neq -1$ .

Find the Laurent series of $\frac{\cos z - 1}{z^4}$ centered at $0$ .

\cos z - 1 = -\frac{z^2}{2} + \frac{z^4}{24} - \frac{z^6}{720} + \cdots

Divide by $z^4$ .

\frac{\cos z - 1}{z^4} = -\frac{1}{2z^2} + \frac{1}{24} - \frac{z^2}{720} + \cdots

The series is valid for $0 < |z| < \infty$ .

Find the Laurent series of $\exp\left(-\frac{1}{z^2}\right)$ centered at $0$ .

Substitute $w = -\frac{1}{z^2}$ in $e^w = \sum \frac{w^n}{n!}$ .

\exp\left(-\frac{1}{z^2}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!\, z^{2n}} = 1 - \frac{1}{z^2} + \frac{1}{2z^4} - \cdots

The series is valid for $0 < |z| < \infty$ .

Find the Laurent series of $\frac{1}{z(1-z)}$ centered at $0$ in $0 < |z| < 1$ .

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

Find the Laurent series of $\frac{1}{(z-1)(z-3)}$ centered at $0$ in $1 < |z| < 3$ .

\frac{1}{(z-1)(z-3)} = \frac{1}{2}\left(\frac{1}{z-3} - \frac{1}{z-1}\right)

For $|z| < 3$ ,

\frac{1}{z-3} = -\frac{1}{3} \cdot \frac{1}{1 - \frac{z}{3}} = -\sum_{n=0}^{\infty} \frac{z^n}{3^{n+1}}

For $|z| > 1$ ,

\frac{1}{z-1} = \frac{1}{z} \cdot \frac{1}{1 - \frac{1}{z}} = \sum_{n=0}^{\infty} z^{-n-1}

\frac{1}{(z-1)(z-3)} = -\frac{1}{2}\sum_{n=0}^{\infty} \frac{z^n}{3^{n+1}} - \frac{1}{2}\sum_{n=0}^{\infty} z^{-n-1}

Find all Laurent series of $f(z) = \frac{3}{2+z-z^2}$ centered at $0$ .

Factor $2 + z - z^2 = (2-z)(1+z)$ . Partial fractions give

f(z) = \frac{1}{2-z} + \frac{1}{1+z}

The singular points $z = -1$ and $z = 2$ split the plane into $|z| < 1$ , $1 < |z| < 2$ , and $|z| > 2$ .

For $|z| < 2$ ,

\frac{1}{2-z} = \frac{1}{2} \cdot \frac{1}{1 - \frac{z}{2}} = \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}

For $|z| > 2$ ,

\frac{1}{2-z} = -\frac{1}{z} \cdot \frac{1}{1 - \frac{2}{z}} = -\sum_{n=0}^{\infty} 2^n z^{-n-1}

For $|z| < 1$ ,

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

For $|z| > 1$ ,

\frac{1}{1+z} = \frac{1}{z} \cdot \frac{1}{1 + \frac{1}{z}} = \sum_{n=0}^{\infty} (-1)^n z^{-n-1}

Combine the matching pieces.

For $|z| < 1$ , the series is $\sum_{n=0}^{\infty} \left(\frac{1}{2^{n+1}} + (-1)^n\right) z^n$ .
For $1 < |z| < 2$ , the series is $\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} + \sum_{n=0}^{\infty} (-1)^n z^{-n-1}$ .
For $|z| > 2$ , the series is $\sum_{n=0}^{\infty} \left((-1)^n - 2^n\right) z^{-n-1}$ .