# Notes - Complex Analysis MT23, Residue theorem

> Source: https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/residue-theorem/ · Updated: 2023-12-15 · Tags: uni, notes

- [Course - Complex Analysis MT23](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/)
	- [Notes - Complex Analysis MT23, Complex integration](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/complex-integration/)
	- [Notes - Complex Analysis MT23, Jordan's lemma](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/jordans-lemma/)
	- [Notes - Complex Analysis MT23, Integral examples](https://ollybritton.com/notes/uni/part-a/mt23/complex-analysis/notes/integral-examples/)

### Flashcards
@State the residue theorem.::

Suppose

- $f : U \setminus S \to \mathbb C$ holomorphic
- $\gamma$ path contained in $U$
- $U$ is an open set
- $S \subset U$ finite set
- $S \cap \gamma^\star = \emptyset$

Then

$$
\int_\gamma f(z) \text d z = 2\pi i \sum_{a \in S} I(\gamma, a) \mathrm{Res}_a(f)
$$

@Prove the residue theorem, i.e. that if

- $f : U \setminus S \to \mathbb C$ holomorphic
- $\gamma$ path contained in $U$
- $U$ is an open set
- $S \subset U$ finite set
- $S \cap \gamma^\star = \emptyset$

then

$$
\int_
\gamma f(z) \text d z = 2\pi i \sum_
{a \in S} I(\gamma, a) \mathrm{Res}_
a(f)
$$

::

For each $a \in S$, let
$$
P_
a(f)(z) = \sum^{-\infty}_
{n=-1} c_n(a) (z - a)^n
$$
This is the "principal part" of $f$ at $a$, and is holomorphic on $\mathbb C \setminus \{a\}$. Then
$$
f - P_a(f)
$$
is holomorphic at $a \in S$, so
$$
g(z) = f(z) - \sum_
{a \in S} P_
a(f)
$$
is holomorphic on all of $U$. But then
$$
\int_
\gamma f(z) \text dz = \sum_
{a \in S} \int_
\gamma P_
a(f)(z) \text dz
$$
Since $P_a(f)$ converge uniformly on $\gamma^\star$,
$$
\begin{aligned}
\int_
\gamma P_a(f) \text dz &= \int_
\gamma \sum^{-\infty}_
{n=-1} c_n(a) (z-a)^n \text dz \\
&= \sum^\infty_
{n = 1} \int_
\gamma \frac{c_
{-n}(a)}{(z-a)^n} \text dz \\
&= \int_
\gamma \frac{c_{-1}(a)}{ z-a} \text dz \\
&= 2\pi i I(\gamma, a) \cdot \mathrm{Res}_
a(f)
\end{aligned}
$$
so we have the required result.

Suppose we are trying to calculate
$$
\int^\infty_{-\infty} \frac{\sin x}{x} \text d x
$$
via a semicircular contour placed at the real axis. This doesn't quite work since the integrand is not defined at $x = 0$. Hence we use a contour with a small circular arc around the singularity, explicitly
$$
\eta_R = (\nu_R^- \star \gamma_\epsilon \star \nu_R^+ ) \star \gamma_R
$$
where

- $\gamma_R : [0, \pi] \to \mathbb C$, $t \mapsto Re^{it}$
- $\gamma_\epsilon : [0, \pi] \to \mathbb C$, $\epsilon e^{i(\pi - t)}$
- $\nu_R^- : [-R, -\epsilon] \to \mathbb C$, $t \mapsto t$
- $\nu_R^+ : [\epsilon, R] \to \mathbb C$, $t \mapsto t$

@State a lemma which is useful in such situations, i.e. calculating the contribution from the indent.::
Suppose:

- $f : U \to \mathbb C$ is a meromorphic function
- $f$ has a **simple** pole $a \in U$
- $\gamma_\epsilon : [\alpha, \beta] \to \mathbb C$, $t \mapsto a + \epsilon e^{it}$

Then:

$$
\lim_{\varepsilon \to 0} \int_{\gamma_\epsilon} f(z) \text d z = i(\beta - \alpha)\mathrm{Res}_a (f)
$$

@Prove that if

- $f : U \to \mathbb C$ is a meromorphic function
- $f$ has a **simple** pole $a \in U$
- $\gamma_\epsilon : [\alpha, \beta] \to \mathbb C$, $t \mapsto a + \epsilon e^{it}$

then:

$$
\lim_{\varepsilon \to 0} \int_{\gamma_\epsilon} f(z) \text d z = i(\beta - \alpha)\mathrm{Res}_a (f)
$$

::

As $f$ has a simple pole at $a$, we can write

$$
f(z) = \frac{\mathrm{Res}_a(f)}{z - a} + g(z)
$$
So

$$
\begin{aligned}
\lim_
{\varepsilon \to 0} \int_
{\gamma_\epsilon} f(z) \text d z &= \lim_
{\varepsilon \to 0} \left( \int_
{\gamma_\varepsilon} \frac{\mathrm{Res}_
a(f)}{z - a} \text d z + \int_
{\gamma_\varepsilon} g(z) \text dz  \right) \\
&= \lim_
{\varepsilon \to 0} \int^\beta_
\alpha \frac{\mathrm{Res}_
a(f)}{\varepsilon e^{it}\,} i\varepsilon e^{it} \text dt \\
&= \int^\beta_
\alpha (i\mathrm{Res}_
a(f)) \text d t \\
&= i(\beta - \alpha)\mathrm{Res}_
a(f)
\end{aligned}
$$
where we can discard the integral involving $g(z)$ as it is holomorphic so bounded on $\gamma_\varepsilon$.

What's a useful strategy for finding the Laurent series of the ratio of two holomorphic functions
$$
\frac{f(z)}{g(z)}
$$
?::

Rewriting the Taylor series of $g(z)$ as $g(z) = c_k z^k \left( 1 + zh(z) \right)$ and then using the geometric series formula.

Suppose:

- $f : \mathbb C \to \mathbb C$ is a continuous function
- $\gamma_{\varepsilon, R} : [0, R] \to \mathbb C$, $t \mapsto t + i \varepsilon$ 

If $\gamma$ is the path $\gamma_{0, R}$, what can you deduce about
$$
\lim_{\varepsilon \to 0} \int_{\gamma
_{\varepsilon, R}
} f(z) \text dz
$$
and when is this result useful?::

$$
\lim_{\varepsilon \to 0} \int_{\gamma
_{\varepsilon, R}
} f(z) \text dz = \int_\gamma f(z) \text dz
$$
This result is useful when calculating keyhole contours.

### Proofs
@Prove that the function $\cot(\pi z)$ has simple poles at every integer with residue $\frac 1 \pi$, and that $\cot(\pi z)$ is uniformly bounded on the sequence of contours
$$
\gamma_N = \text{square with vertices } (N + 1/2)(\pm 1 \pm i)
$$
::

Todo, page 73.

@todo~

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