Notes - Complex Analysis MT23, Residue theorem

Created: December 15, 2023

Flashcards

Can you 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

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

Quickly 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

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

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 $\gamma^\star$,

\[\begin{aligned} \int_ \gamma P_a(f) \text dz &= \int_ \gamma \sum^{-\infty}_ {n=-1} c_n(a) (z-a)^n \text dz \\\\ &= \sum^\infty_ {n = 1} \int_ \gamma \frac{c_ {-n}(a)}{(z-a)^n} \text dz \\\\ &= \int_ \gamma \frac{c_{-1}(a)}{ z-a} \text dz \\\\ &= I(\gamma, a) \cdot \text{Res}_ a(f) \end{aligned}\]

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

  • $\gamma _ R : [0, \pi] \to \mathbb C$, $t \mapsto Re^{it}$
  • $\gamma _ \epsilon : [0, \pi] \to \mathbb C$, $\epsilon e^{i(\pi - t)}$
  • $\nu _ R^+ : [-R, -\epsilon] \to \mathbb C$, $t \mapsto t$
  • $\nu _ R^+ : [\epsilon, R] \to \mathbb C$, $t \mapsto t$

Can you 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:

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

Quickly 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:

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

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

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

So

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

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

\[\gamma_N = \text{square with vertices } (N+1/2) \times {(\pm 1} {\pm i})\]

Todo, page 73.



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