📂Complex Anaylsis

What is the Laurent Series?

Buildup

Taylor’s theorem generalizes the mean value theorem regarding the number of differentiations. It expands from dealing with something differentiated $1$ times to $n \in \mathbb{N}$ times. But if it was possible to generalize it to natural numbers, could it not be generalized to all integers? Of course, it’s not possible to differentiate $-n$ times, but what about considering integration, which is the inverse operation of differentiation? Here we introduce the Laurent’s Theorem without proof.

Assuming $f$ is analytic on two concentric circles $\mathscr{C}_{1}: |z-\alpha| = r_{1}$ and $\mathscr{C}_{2}: |z-\alpha| = r_{2}$ $(r_{2} < r_{1})$ centered at the singularity $\alpha$ of $f: A \subset \mathbb{C} \to \mathbb{C}$. Then, for all points between the two concentric circles, $f$ can be represented by $\displaystyle f(z) = \sum_{n = 0 }^{\infty} a_{n} (z-\alpha) ^{n} + \sum_{n = 1 }^{\infty} { {b_{n} } \over{ (z-\alpha) ^{n} } }$.

• $\displaystyle a_{n} = {{1} \over {2 \pi i}} \int_{\mathscr{C}_{1}} {{f(z)} \over {(z - \alpha)^{ 1 + n} }} dz \qquad , n = 0,1,2, \cdots$
• $\displaystyle b_{n} = {{1} \over {2 \pi i}} \int_{\mathscr{C}_{2}} {{f(z)} \over {(z - \alpha)^{ 1 - n} }} dz \qquad , n=1,2,3,\cdots$

Definition

The following series is called the Laurent series. $$ f(z) = \sum_{n = 0 }^{\infty} a_{n} (z-\alpha) ^{n} + \sum_{n = 1 }^{\infty} { {b_{n} } \over{ (z-\alpha) ^{n} } } $$

Explanation

A Generalization of Cauchy’s Integral Formula for Differentiation: Let $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic in a simply connected region $\mathscr{R}$.

If a simple closed path $\mathscr{C}$ in $\mathscr{R}$ encloses a point $\alpha$, then for a natural number $n$:

$$ {{f^{(n)} (\alpha) } \over {n!}} = {{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { (z - \alpha)^{1+n} }} dz $$

Using Cauchy’s Integral Formula will make it more evident that it is a generalization of Taylor’s theorem.

$$ f(z) = \sum_{n = 0 }^{\infty} {{f^{(n)} (\alpha) } \over {n!}} (z-\alpha) ^{n} + \sum_{n = 1 }^{\infty} { {b_{n} } \over{ (z-\alpha) ^{n} } } $$ In such a series form, $\displaystyle \sum_{n = 1 }^{\infty} { {b_{n} } \over{ (z-\alpha) ^{n} } }$ is referred to as the Principal Part. Notably, the coefficient of $\displaystyle {{1} \over {z-\alpha}}$, i.e., $b_{1}$, is defined as the Residue of $f$ at $\alpha$ and is expressed as $b_{1} = \text{Res}_{\alpha} f(z)$¹.