Existence of Laurent Series

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Theorem

Let $z_0 \in \C$ be a complex number.

Let $R \in \R_{>0}$ be a real number.

Let $\map {B'} {z_0, R}$ be the open punctured disk at $z_0$ of radius $R$.

Let $f: \map {B'} {z_0, R} \to \C$ be holomorphic.


Then there exists a sequence $\sequence {a_n}_{n \mathop \in \Z}$ such that:

$\map f z = \displaystyle \sum_{n = -\infty}^\infty a_n \paren {z - z_0}^n$

for all $z \in B'(z_0, R)$.


Proof

Choose circles $C_1$ and $C_3$ centered at $z_0$ and connect them by path $C_2$ such that $z$ is inside $C_1 + C_2 - C_3 - C_2$ as shown below:

Path figure.png

This curve and its interior are contained in $B'$, so by Cauchy's Integral Formula:

\(\displaystyle \map f z\) \(=\) \(\displaystyle \dfrac 1 {2 \pi i} \int_{C_1 + C_2 - C_3 - C_2} \dfrac {\map f w} {\paren {w - z} } \rd w\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {2 \pi i} \int_{C_1 - C_3} \dfrac {\map f w} {\paren {w - z} } \rd w\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {2 \pi i} \int_{C_1} \dfrac {\map f w} {\paren {w - z} } \rd w - \dfrac 1 {2 \pi i} \int_{C_3} \dfrac {\map f w} {\paren {w - z} } \rd w\)



Since $C_1$ contains $z$ we have for all $w$ on $C_1$:

$\dfrac {\cmod {z - z_0} } {\cmod {w - z_0} } < 1$

Therefore:

\(\displaystyle \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z} } \rd w\) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z + z_0 - z_0} } \rd w\) adding and subtracting $z_0$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0} \paren {1 - \frac {z - z_0} {w - z_0} } } \rd w\) Factoring out $w - z_0$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0} } \cdot \frac 1 {\paren {1 - \frac {z - z_0} {w - z_0} } } \rd w\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0} } \cdot \sum_{n \mathop = 0}^\infty \paren {\frac {z - z_0} {w - z_0} }^n \rd w\) Sum of Infinite Geometric Sequence
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_1} \sum_{n \mathop = 0}^\infty \frac {\map f w} {\paren {w - z_0}^{n + 1} } \cdot \paren {z - z_0}^n \rd w\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0}^{n + 1} } \rd w} \paren {z - z_0}^n\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - z_0}^n\) letting $a_n$ equal the value in the large parentheses above, for all $n \ge 0$


Similarly, since $z$ lies outside $C_3$ we have for all $w$ on $C_3$:

$\dfrac {\cmod {w - z_0} } {\cmod {z - z_0} } < 1$

Therefore:

\(\displaystyle \frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {w - z} } \rd w\) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {w - z + z_0 - z_0} } \rd w\) Adding and subtracting $z_0$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \int_{C_3} -\frac {\map f w} {\paren {z - z_0} \paren {1 - \frac {w - z_0} {z - z_0} } } \rd w\) factoring out $z - z_0$
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {z - z_0} } \cdot \frac 1 {\paren {1 - \frac {w - z_0} {z - z_0} } } \rd w\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {z - z_0} } \cdot \sum_{n \mathop = 0}^\infty \paren {\frac {w - z_0} {z - z_0} }^n \rd w\) Sum of Infinite Geometric Sequence
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{C_3} \map f w \paren {w - z_0}^n \rd w} \paren {z - z_0}^ {-n - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{k \mathop = -1}^ {-\infty} \paren {\frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {w - z_0}^{k + 1} } \rd w} \paren {z - z_0}^k\) Re-indexing with $k = -n - 1$
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{k \mathop = -1}^ {-\infty} a_k \paren {z - z_0}^k\) Letting $a_k$ equal the value in the large parentheses above, for all $k < 0$


Combining what has been shown above yields:

\(\displaystyle \map f z\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - z_0}^n + \sum_{n \mathop = -1}^ {-\infty} a_n \paren {z - z_0}^n\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = -\infty}^\infty a_n \paren {z - z_0}^n\)

$\blacksquare$