Notes - Complex Analysis MT23, Laurent series
Flashcards
Can you define an open annulus $A(r, R, z _ 0)$?
Can you state the “Laurent series” theorem, about when you can define a function as a power series on an open annulus, and how you can determine the coefficients?
Suppose:
- $0 < r < R$
- $A = A(r, R, z _ 0)$, annulus centered at $z _ 0$
- $\overline A \subseteq U$
- $f : U \to \mathbb C$, holomorphic
Then $\exists ! c _ n \in \mathbb C$ such that $\forall z \in A$
\[f(z) = \sum^\infty_{n = -\infty} c_n(z-z_0)^n\]and
\[c_n = \frac{1}{2\pi i} \int_{\gamma_s} \frac{f(z)}{(z-z_0)^{n+1}\\,} \text dz\]where $\gamma _ s (t) = z _ 0 + se^{2\pi i t}$ for any $s \in [r, R]$.
Suppose:
- $f : U \to \mathbb C$, holomorphic
- $A = A(r, R, z _ 0)$, annulus centered at $z _ 0$
- $0 < r < R$
- $\overline A \subseteq U$
Then $\exists ! c _ n \in \mathbb C$ such that $\forall z \in A$
\[f(z) = \sum^\infty_{n = -\infty} c_n(z-z_0)^n\]
Can you give an explicit formula for $c _ n$?
Suppose:
- $f : U \setminus S \to \mathbb C$
- $f$ holomorphic
- $S$ is a discrete set
How (any why) can you give $f$ as a power series at each $a \in S$ and how can you define the principal of $f$ at each $a \in S$, and in this context, what is the residue of $f$ at $a$?
Since $f$ is a holomorphic on the punctured disc $B(a, r) \setminus \{z _ 0\}$ for sufficiently small $r > 0$, it has a Laurent expansion
\[f(z) = \sum^\infty_{n = -\infty} c_n (z-a)^n\]Then the principal part is given by
\[P_a(f) = \sum^{-\infty}_{n = -1} c_n (z- a)^n\]and the residue is the coefficient $c _ {-1}$.
The Laurent series theorem states that if:
- $0 < r < R$
- $A = A(r, R, z _ 0)$, annulus centered at $z _ 0$
- $\overline A \subseteq U$
- $f : U \to \mathbb C$, holomorphic
then $\exists ! c _ n \in \mathbb C$ such that
\[f(z) = \sum^\infty_{n = -\infty} c_n(z-z_0)^n\]
and
\[c_n = \frac{1}{2\pi i} \int_{\gamma_s} \frac{f(z)}{(z-z_0)^{n+1}
} \text dz\]
where $\gamma _ s (t) = z _ 0 + se^{2\pi i t}$ for any $s \in [r, R]$.
Suppose we have shown that the result holds for the path $\gamma _ r$ in particular. How can we then quickly show that we can actually use any circular contour $\gamma _ s$?
Note that if $r \le s _ 1 \le s _ 2 \le R$, then $f/(z - z _ 0)^{n+1}$ is holomorphic on the inside of $\Gamma := \gamma _ {s _ 2} - \gamma _ {s _ 1}$. Hence, by the homology form of Cauchy’s theorem,
\[0 = \int_\Gamma \frac{f(z)}{(z-z_0)^{n+1}\\,} \text dz = \int_{\gamma_{s_2}\\,} \frac{f(z)}{(z-z_0)^{n+1}\\,} \text dz - \int_{\gamma_{s_1}\\,} \frac{f(z)}{(z-z_0)^{n+1}\\,} \text{d} z\]The geometric series for $ \vert z \vert < 1$ is given by
\[\frac{1}{1-z} = \sum^\infty_{k = 0} z^k\]
Quickly prove that the following formula holds for $ \vert z \vert > 1$:
\[\frac{1}{1-z} = -\sum^\infty_{k=1} \frac{1}{z^k}\]
Quickly prove the “Laurent series” theorem, about when you can define a function as a power series on an open annulus, i.e. suppose:
- $0 < r < R$
- $A = A(r, R, z _ 0)$, annulus centered at $z _ 0$
- $\overline A \subseteq U$
- $f : U \to \mathbb C$, holomorphic
then $\exists ! c _ n \in \mathbb C$ such that
\[f(z) = \sum^\infty_{n = -\infty} c_n(z-z_0)^n\]
and
\[c_n = \frac{1}{2\pi i} \int_{\gamma_s} \frac{f(z)}{(z-z_0)^{n+1}
} \text dz\]
where $\gamma _ s (t) = z _ 0 + se^{2\pi i t}$ for any $s \in [r, R]$.
First we may assume by translation that $z _ 0 = 0$. By the cycles version of Cauchy’s integral formula applied to $f$ and $\Gamma = \gamma _ R - \gamma _ r$, we have that $\forall w \in A$,
\[f(w) = \frac{1}{2\pi i} \int_ {\gamma_ R} \frac{f(z)}{z-w} \text dz - \frac{1}{2\pi i} \int_{\gamma_r} \frac{f(z)}{z-w} \text dz\](Why? Note that the winding number of $\Gamma$ at $w$ is $1$ iff $w \in A$, and $0$ otherwise).
Fix some $w$ for which we wish to show the identity holds. Then for $ \vert z \vert > \vert w \vert $ we have $ \vert w/z \vert < 1$ and so can use the geometric series expansion,
\[\begin{aligned} \frac{1}{z-w} &= \frac{1}{z\left(1 - \frac w z\right)} \\\\ &= \sum^\infty_{n = 0} \frac{w^n}{z^{n+1} } \end{aligned}\]which converges uniformly in $z$ for $ \vert z \vert > \vert w \vert + \varepsilon$ for $\varepsilon > 0$. Then
\[\begin{aligned} \int_ {\gamma_ R} \frac{f(z)}{z-w} \text dz &= \int_ {\gamma_ R} \sum^\infty_ {n = 0} \frac{f(z) w^n}{z^{n+1} } \text dz \\\\ &= \sum_ {n = 0}^\infty \left( \int_ {\gamma_ R} \frac{f(z)}{z^{n+1} } \text dz\right)w^n \end{aligned}\]for each $w \in A$.
Similarly, if $ \vert z \vert < \vert w \vert $ then we can instead expand in terms of , then
\[\begin{aligned} -\frac{1}{z -w} &= \frac{1}{w-z} \\\\ &= \frac{1}{w\left(1 - \frac z w\right)} \\\\ &= \sum_{n = 0}^\infty \frac{z^n}{w^{n+1} } \\\\ &= \sum^{-\infty}_{n = -1} \frac{w^n}{z^{n+1} } \end{aligned}\]and this converges uniformly in $z$ for $ \vert z \vert < \vert w \vert - \varepsilon$ for $\varepsilon > 0$. Then
\[\begin{aligned} -\int_ {\gamma_ r} \frac{f(w)}{z - w} \text dz &= \int_{\gamma _r} \frac{f(w)}{w - z} \text dz \\\\ &= \int_ {\gamma_ r}\sum^{-\infty}_ {n = -1} \frac{f(z) w^n}{z^{n+1} } \text{d}z \\\\ &= \sum^{-\infty}_ {n=-1} \left( \int_ {\gamma_ r} \frac{f(z)}{z^{n+1} } \text dz \right) w^n \end{aligned}\]Taking $c _ n$ as defined above then gives the result.
The $c _ n$ can be computed using any circular contour as for any $r \le s _ 1 < s _ 2 \le R$, the fact that $f/(z-z _ 0)^{n+1}$ is holomoirphic on the inside of $\Gamma = \gamma _ {s _ 2} - \gamma _ {s _ 1}$ implies that
\[\int_ {\gamma _ {s_ 2} } \frac{f(z)}{(z-z_0)^{n+1} } \text dz = \int_{\gamma_{s_1} } \frac{f(z)}{(z-z_0)^{n+1} } \text dz\]Uniqueness can be seen by setting two expansions equal to one another, multiplying both sides by some $(z-z _ 0)^{-k-1}$ for some integer, and then integrating about a path $\gamma$ inside the annulus.