Notes - Complex Analysis MT23, Residue theorem

Course - Complex Analysis MT23

Flashcards

@State the residue theorem.

residue-theorem-state

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

\[\int _ \gamma f(z) \text d z = 2\pi i \sum _ {a \in S} I(\gamma, a) \mathrm{Res} _ a(f)\]

Builds on ∆winding-number-definition, ∆principal-part-and-residue-definition

@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

\[\int _ \gamma f(z) \text d z = 2\pi i \sum _ {a \in S} I(\gamma, a) \mathrm{Res} _ a(f)\]

residue-theorem-proof

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.

Builds on ∆residue-theorem-state, ∆laurent-series-theorem-state, ∆cauchy-theorem-homotopy-form-state, ∆winding-number-integral-formula

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.

indentation-lemma-state

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:

\[\lim _ {\varepsilon \to 0} \int _ {\gamma _ \epsilon} f(z) \text d z = i(\beta - \alpha)\mathrm{Res} _ a (f)\]

Builds on ∆indented-contour-for-boundary-singularity, ∆simple-pole-definition, ∆principal-part-and-residue-definition

@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:

\[\lim _ {\varepsilon \to 0} \int _ {\gamma _ \epsilon} f(z) \text d z = i(\beta - \alpha)\mathrm{Res} _ a (f)\]

indentation-lemma-proof

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

\[f(z) = \frac{\mathrm{Res} _ a(f)}{z - a} + g(z)\]

\[\begin{aligned} \lim _ {\varepsilon \to 0} \int _ {\gamma _ \epsilon} f(z) \text d z &= \lim _ {\varepsilon \to 0} \left( \int _ {\gamma _ \varepsilon} \frac{\mathrm{Res} _ a(f)}{z - a} \text d z + \int _ {\gamma _ \varepsilon} g(z) \text dz \right) \\ &= \lim _ {\varepsilon \to 0} \int^\beta _ \alpha \frac{\mathrm{Res} _ a(f)}{\varepsilon e^{it}\,} i\varepsilon e^{it} \text dt \\ &= \int^\beta _ \alpha (i\mathrm{Res} _ a(f)) \text d t \\ &= i(\beta - \alpha)\mathrm{Res} _ a(f) \end{aligned}\]

where we can discard the integral involving $g(z)$ as it is holomorphic so bounded on $\gamma _ \varepsilon$.

Builds on ∆indentation-lemma-state, ∆laurent-series-at-pole, ∆estimation-lemma-and-proof

What’s a useful strategy for finding the Laurent series of the ratio of two holomorphic functions

\[\frac{f(z)}{g(z)}\]

laurent-series-of-ratio-strategy

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.

Builds on ∆laurent-series-theorem-state

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?

shifted-contour-limit

\[\lim _ {\varepsilon \to 0} \int _ {\gamma _ {\varepsilon, R} } f(z) \text dz = \int _ \gamma f(z) \text dz\]

This result is useful when calculating keyhole contours.

Builds on ∆path-integral-definition

Flashcards

Proofs

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