---
title: "Laurent Series and the Residue Theorem"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 17
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem
---

# Chapter 17 — Laurent Series and the Residue Theorem

What happens to a [holomorphic function](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) near a point where it is *not* defined? The answer is a [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) trichotomy — removable point, pole, or essential singularity — read off a two-sided power series, the Laurent expansion. One coefficient of that expansion, the *[residue](#def-b3-residues-singularities)*, controls every [contour integral](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-contour) around the singularity: the [residue](#def-b3-residues-singularities) theorem converts hard definite integrals into finite algebra, counts zeros of functions (argument principle, Rouché), and proves the open mapping theorem. We first upgrade Cauchy’s theorem from star-shaped domains to its definitive, homology-free form — Dixon’s elegant argument — so that all contours with winding number zero around the complement become available.

## 17.1 The global Cauchy theorem

A *cycle* $\Gamma$ is a finite formal sum of closed [paths](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) $\gamma_1, \dots, \gamma_m$; integrals and indices along $\Gamma$ are the corresponding sums, and $\operatorname{im}\Gamma = \bigcup\operatorname{im}\gamma_j$.

**Theorem 17.1 (Cauchy, global form).**

Let $\Omega \subseteq \C$ be open, $f \in \mathcal H(\Omega)$, and $\Gamma$ a cycle in $\Omega$ such that

$$
\operatorname{Ind}_\Gamma(w) = 0
\qquad\text{for every } w \notin \Omega .
$$

Then, for all $z \in \Omega\setminus\operatorname{im}\Gamma$,

$$
\frac1{2\iu\pi}\int_\Gamma\frac{f(w)}{w - z}\,\dd w =
\operatorname{Ind}_\Gamma(z)\,f(z),
\qquad\text{and}\qquad
\int_\Gamma f(w)\,\dd w = 0 .
$$

**Proof (Dixon).** Define $g \colon \Omega\times\Omega \to \C$ by

$$
g(z, w) = \begin{cases}
\dfrac{f(w) - f(z)}{w - z} & w \neq z,\\[4pt]
f'(z) & w = z .
\end{cases}
$$

*$g$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)*: off the diagonal, clear. Near a diagonal point $(a, a)$, expand $f$ in a power series at $a$ ([Theorem 16.10](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-analytic)): $f(w) - f(z) =
\sum_{n\geq1}c_n\bigl((w-a)^n - (z-a)^n\bigr)$, and dividing each term by $w - z$ (factorization of $u^n - v^n$) gives, for $z, w \in D(a, r)$,

$$
g(z, w) =
\sum_{n\geq1}c_n\sum_{j=0}^{n-1}(w-a)^{\,j}(z-a)^{\,n-1-j},
$$

valid also on the diagonal (each inner sum becomes $n(z-a)^{n-1}$, summing to $f'(z)$). For $r$ small the series converges uniformly on $D(a,r)^2$ ($\abs{\text{term}} \leq n\abs{c_n}r^{n-1}$, summable inside the radius): the sum is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity).

Set $h(z) = \frac1{2\iu\pi}\int_\Gamma g(z, w)\,\dd w$ on $\Omega$: [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $g$ on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)), and [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) — by Morera ([Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv)’s criterion): for a triangle $T \subseteq \Omega$, Fubini gives $\int_{\partial
T}h = \frac1{2\iu\pi}\int_\Gamma\bigl(\int_{\partial T}g(z,
w)\dd z\bigr)\dd w = 0$, the inner integral vanishing because $z \mapsto g(z, w)$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $\Omega$ (at $z = w$ the singularity is removable: $g$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) there and [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) elsewhere — the extension argument of [Theorem 16.9](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-formula)’s proof).

On the [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $\Omega' = \{z \notin \operatorname{im}\Gamma
: \operatorname{Ind}_\Gamma(z) = 0\}$, define $h_1(z) =
\frac1{2\iu\pi}\int_\Gamma\frac{f(w)}{w - z}\dd w$: [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $\Omega'$ (Morera or differentiation under the integral). For $z \in \Omega\cap\Omega'$:

$$
h(z) = \frac1{2\iu\pi}\int_\Gamma\frac{f(w)}{w-z}\dd w
- f(z)\operatorname{Ind}_\Gamma(z) = h_1(z) .
$$

By hypothesis $\Omega\cup\Omega' = \C$ ($w \notin \Omega
\Rightarrow \operatorname{Ind}_\Gamma(w) = 0$), so $h$ and $h_1$ glue to an entire function $H$. Since the unbounded component of the complement of $\operatorname{im}\Gamma$ lies in $\Omega'$ and $h_1(z) \to 0$ as $\abs z \to \infty$ (ML bound), $H$ is bounded and tends to $0$: Liouville ([Corollary 16.12](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#cor-b3-holomorphic-liouville)) gives $H \equiv 0$. Thus $h \equiv 0$ on $\Omega$, which is the integral formula. Applying it, for fixed $a \in
\Omega\setminus\operatorname{im}\Gamma$, to $\tilde f(w) =
(w - a)f(w)$ at $z = a$:

$$
\frac1{2\iu\pi}\int_\Gamma f(w)\dd w =
\frac1{2\iu\pi}\int_\Gamma \frac{\tilde f(w)}{w - a}\dd w =
\operatorname{Ind}_\Gamma(a)\,\tilde f(a) = 0 .
$$

∎

## 17.2 Laurent series and isolated singularities

**Theorem 17.2 (Laurent expansion).**

Let $f$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on the annulus $A = \{r < \abs{z - a} <
R\}$ ($0 \leq r < R \leq \infty$). Then

$$
f(z) = \sum_{n\in\Z}c_n\,(z - a)^n
\qquad\text{on } A,
\qquad
c_n = \frac1{2\iu\pi}\int_{C_\rho}\frac{f(w)}{(w -
a)^{n+1}}\,\dd w
$$

for any $r < \rho < R$ (independent of $\rho$), the two half-series converging normally on [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) subannuli. The expansion is unique.

**Proof.** Fix $r < \rho_1 < \abs{z - a} < \rho_2 < R$ and let $\Gamma =
C_{\rho_2} - C_{\rho_1}$ (outer counterclockwise, inner clockwise): a cycle in $A$ with $\operatorname{Ind}_\Gamma(w) = 0$ for all $w \notin A$ (points inside the small disc: $1 - 1 = 0$; outside the big one: $0 - 0$). By [Theorem 17.1](#thm-b3-residues-globalcauchy), $\operatorname{Ind}_\Gamma(z) = 1 - 0 = 1$ gives

$$
f(z) = \frac1{2\iu\pi}\int_{C_{\rho_2}}\frac{f(w)}{w - z}\dd
w - \frac1{2\iu\pi}\int_{C_{\rho_1}}\frac{f(w)}{w - z}\dd w .
$$

Expand the first kernel as in [Theorem 16.10](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-analytic) (powers of $\frac{z -
a}{w - a}$, modulus $< 1$): the nonnegative part $\sum_{n\geq0}c_n(z-a)^n$. In the second, expand the other way: $\frac{-1}{w - z} = \frac1{(z-a)(1 - \frac{w - a}{z -
a})} = \sum_{m\geq0}\frac{(w-a)^m}{(z - a)^{m+1}}$, normally convergent on $C_{\rho_1}$: the negative part $\sum_{n\leq-1}c_n(z-a)^n$ with the stated coefficients (index $n = -m-1$). Independence of $\rho$: the coefficient integrals over $C_{\rho}$ and $C_{\rho'}$ differ by $\int_\Gamma$ over a null-index cycle of the [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $\frac{f(w)}{(w-a)^{n+1}}$ in $A$: zero, by [Theorem 17.1](#thm-b3-residues-globalcauchy) again. Uniqueness: integrate $\sum c_n(z-a)^n$ against $(z - a)^{-m-1}$ over $C_\rho$ term by term (normal convergence): only $n = m$ survives. ∎

**Definition 17.3.**

If $f$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on a punctured disc $D(a, R)\setminus
\{a\}$, expand by Laurent ($r = 0$). Three exclusive cases:

- all $c_n = 0$ for $n < 0$ : *[removable singularity](#thm-b3-residues-riemanncw)* (the nonnegative series extends $f$ [holomorphically](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) to $a$ );
- $c_n \neq 0$ for finitely many, at least one, $n < 0$ : a *pole* of order $m = -\min\{n : c_n \neq 0\}$ ; equivalently $f = g/(z-a)^m$ , $g$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) , $g(a)  \neq 0$ ; equivalently $\abs{f(z)} \to \infty$ as $z\to  a$ ;
- infinitely many negative $c_n \neq 0$ : *essential singularity* .

The *residue* is $\operatorname{Res}(f, a) = c_{-1}$. A function [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $\Omega$ minus a set of poles is *meromorphic* on $\Omega$.

**Theorem 17.4 (Riemann; Casorati–Weierstrass).**

Let $f$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $D(a,R)\setminus\{a\}$.

1. (Riemann) If $f$ is *bounded* near $a$ , the singularity is removable.
2. (Casorati–Weierstrass) If $a$ is essential, then $f\bigl(D(a,\varepsilon)\setminus\{a\}\bigr)$ is dense in $\C$ for every $\varepsilon$ .

**Proof.** (1) For $n < 0$ and $\rho \to 0$: $\abs{c_n} \leq
M\rho^{-n-1}\cdot\rho\cdot\rho^{-\,n}\dots$ by ML on $C_\rho$: $\abs{c_n} \leq \frac{1}{2\pi}\,2\pi\rho\cdot
M\rho^{-(n+1)} = M\rho^{-n} \to 0$ (as $-n > 0$): all negative coefficients vanish. (2) If some value $b$ were not approached: $\abs{f - b} \geq
\delta$ near $a$, so $g = 1/(f - b)$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and bounded near $a$: removable (1), $g$ extends with value $c$. If $c \neq 0$, $f = b + 1/g$ is bounded near $a$: removable — excluded. If $c = 0$, $g$ has a zero of finite order $m$ at $a$ ([Theorem 16.13](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-identity); $g \not\equiv 0$), and $f = b + 1/g$ has a pole of order $m$: excluded again. ∎

## 17.3 The residue theorem

**Theorem 17.5 (Residue theorem).**

Let $\Omega$ be open, $S \subseteq \Omega$ finite, $f \in
\mathcal H(\Omega\setminus S)$, and $\Gamma$ a cycle in $\Omega\setminus S$ with $\operatorname{Ind}_\Gamma(w) = 0$ for every $w \notin \Omega$. Then

$$
\frac{1}{2\iu\pi}\int_\Gamma f(z)\,\dd z
= \sum_{a\in S}\operatorname{Ind}_\Gamma(a)\,
\operatorname{Res}(f, a) .
$$

**Proof.** For each $a \in S$, let $P_a(z) = \sum_{n\leq-1}c_n^{(a)}(z -
a)^n$ be the principal part of $f$ at $a$: a series converging on $\C\setminus\{a\}$ (its radius in $1/(z-a)$ is infinite: the Laurent tail converges for all small $\abs{z-a}$ hence, being a power series in $(z-a)^{-1}$, everywhere), and [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) there. Then $g = f - \sum_{a\in S}P_a$ has removable singularities at each point of $S$ (its Laurent expansion at $a$ has no negative part: the other $P_{a'}$ are [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) at $a$), so $g$ extends [holomorphically](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) to $\Omega$, and [Theorem 17.1](#thm-b3-residues-globalcauchy) gives $\int_\Gamma g = 0$. It remains to integrate each $P_a$: term-by-term (normal convergence on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\operatorname{im}\Gamma$, which avoids $a$),

$$
\frac1{2\iu\pi}\int_\Gamma(z - a)^n\,\dd z = 0 \ (n \leq -2:
\text{primitive } \tfrac{(z-a)^{n+1}}{n+1}),
\qquad
\frac1{2\iu\pi}\int_\Gamma\frac{\dd z}{z - a} =
\operatorname{Ind}_\Gamma(a),
$$

so $\frac1{2\iu\pi}\int_\Gamma P_a =
c_{-1}^{(a)}\operatorname{Ind}_\Gamma(a)$. Sum over $a$. ∎

**Method 17.6 (Computing residues).**

Simple pole: $\operatorname{Res}(f, a) = \lim_{z\to a}(z -
a)f(z)$; for $f = g/h$ with $g(a) \neq 0$, $h(a) = 0$, $h'(a)
\neq 0$: $\operatorname{Res} = g(a)/h'(a)$. Pole of order $m$: $\operatorname{Res}(f, a) = \frac1{(m-1)!}\lim_{z\to
a}\bigl((z-a)^mf(z)\bigr)^{(m-1)}$. Essential singularities: expand and read $c_{-1}$ (e.g. from known series). Always verify which poles the contour actually encircles, and with which index.

**Example 17.7 (The four classical integral types).**

(a) *Rational over $\R$*: for $\int_\R\frac{\dd x}{1 +
x^4}$, close with a large semicircle $S_R$ in the upper half-plane: the integrand is $O(R^{-4})$ there, so $\int_{S_R} \to 0$ (ML), and the [residue](#def-b3-residues-singularities) theorem with the poles $\eu^{\iu\pi/4}, \eu^{3\iu\pi/4}$ (simple, [residues](#def-b3-residues-singularities) $\frac1{4z^3} = \frac{z}{4z^4} = -\frac z4$ at a pole) gives

$$
\int_\R\frac{\dd x}{1 + x^4}
= 2\iu\pi\Bigl(-\frac{\eu^{\iu\pi/4}}4 -
\frac{\eu^{3\iu\pi/4}}4\Bigr)
= \frac{\pi}{\sqrt2} .
$$

(b) *Fourier type*: for $t \geq 0$, $\int_\R\frac{\eu^{\iu tx}}{1 + x^2}\dd x =
2\iu\pi\operatorname{Res}\bigl(\tfrac{\eu^{\iu tz}}{1+z^2},
\iu\bigr) = 2\iu\pi\frac{\eu^{-t}}{2\iu} = \pi\eu^{-t}$ — the upper semicircle works because $\abs{\eu^{\iu tz}} =
\eu^{-t\operatorname{Im}z} \leq 1$ there; taking real parts: $\int_\R\frac{\cos(tx)}{1+x^2}\dd x = \pi\eu^{-\abs t}$, settling [Exercise 10.10](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#exo-b3-lebesgue-10)’s admitted formula. (c) *Trigonometric over a period*: substitute $z =
\eu^{\iu t}$, $\cos t = \frac{z + z^{-1}}2$, $\dd t =
\frac{\dd z}{\iu z}$: $\int_0^{2\pi}\frac{\dd t}{a + \cos t}$ ($a > 1$) becomes a [residue](#def-b3-residues-singularities) count inside the unit circle ([Exercise 17.2](#exo-b3-residues-2)). (d) *Series*: pair $f$ with $\pi\cot(\pi z)$, whose poles are the integers with [residue](#def-b3-residues-singularities) $1$: the weekend problem sums $\sum n^{-2}$ and $\sum n^{-4}$ this way.

![The semicircle contour for ∈t_ℝ x/1 + x4: as R ∈fty the arc contributes O(R-3), and the residue theorem counts the two enclosed poles (blue). The two lower poles (gray) are outside: index 0.](https://one-course.com/images/onecourse/chapters/math-5/b3-residues/fig-5a0f9191769e.svg)

*The semicircle contour for $\int_\R\frac{\dd x}{1 +
x^4}$: as $R \to \infty$ the arc contributes $O(R^{-3})$, and the [residue](#def-b3-residues-singularities) theorem counts the two enclosed poles (blue). The two lower poles (gray) are outside: index $0$.*

## 17.4 The argument principle and Rouché’s theorem

**Theorem 17.8 (Argument principle).**

Let $f$ be [meromorphic](#def-b3-residues-singularities) on $\Omega$, with zeros $z_j$ (orders $m_j$) and poles $p_k$ (orders $\mu_k$), and $\gamma$ a closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) in $\Omega$ avoiding them all, with $\operatorname{Ind}_\gamma = 0$ off $\Omega$. Then

$$
\frac1{2\iu\pi}\int_\gamma\frac{f'(z)}{f(z)}\,\dd z
= \sum_j m_j\operatorname{Ind}_\gamma(z_j)
- \sum_k \mu_k\operatorname{Ind}_\gamma(p_k)
$$

(finitely many terms are nonzero). For a simple counterclockwise contour, the integral counts zeros minus poles inside, with multiplicity — and equals the winding number of the image [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) $f\circ\gamma$ around $0$.

**Proof.** Near a zero of order $m$: $f = (z-a)^mg$, $g(a) \neq 0$, so $\frac{f'}f = \frac m{z - a} + \frac{g'}g$ with the second term [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) near $a$: a simple pole of [residue](#def-b3-residues-singularities) $m$. Near a pole of order $\mu$: $f = (z-a)^{-\mu}g$ gives [residue](#def-b3-residues-singularities) $-\mu$. Elsewhere $\frac{f'}f$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo). (The zeros and poles with nonzero index lie in a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) region encircled by $\gamma$; by the identity theorem they are finite in number there, $f \not\equiv 0$.) Apply [Theorem 17.5](#thm-b3-residues-residue). The last remark: $\frac1{2\iu\pi}\int_\gamma\frac{f'}f =
\frac1{2\iu\pi}\int_{f\circ\gamma}\frac{\dd w}w =
\operatorname{Ind}_{f\circ\gamma}(0)$ (substitute $w =
f(\gamma(t))$). ∎

**Theorem 17.9 (Rouché).**

Let $f, g$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $\Omega$, and $\gamma$ a closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) with $\operatorname{Ind}_\gamma \in \{0,1\}$, zero off $\Omega$ (a simple contour). If

$$
\abs{g(z)} < \abs{f(z)} \qquad \text{on }
\operatorname{im}\gamma,
$$

then $f$ and $f + g$ have the same number of zeros (with multiplicity) in the region $\{\operatorname{Ind}_\gamma =
1\}$.

**Proof.** For $t \in \intcc01$, $f_t = f + tg$ has no zero on $\operatorname{im}\gamma$ ($\abs{f_t} \geq \abs f - \abs g >
0$), so

$$
N(t) = \frac1{2\iu\pi}\int_\gamma
\frac{f_t'(z)}{f_t(z)}\,\dd z
$$

is well defined; it counts the zeros in the enclosed region ([Theorem 17.8](#thm-b3-residues-argument); no poles). $N$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) in $t$ (the integrand is jointly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), the denominators uniformly bounded below — dominated convergence) and integer-valued: constant. $N(0) = N(1)$. ∎

**Corollary 17.10 (Open mapping theorem).**

A nonconstant [holomorphic function](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on a [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) is an open map. In particular (again) the maximum principle holds, and a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) bijection has [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) inverse.

**Proof.** Let $f(a) = b$; $f - b$ has a zero of some finite order $m
\geq 1$ at $a$ (identity theorem: $f \not\equiv b$). Choose $r$ with $f - b$ zero-free on $\bar D(a, r)\setminus\{a\}$ ([isolated zeros](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-identity)) and let $\delta = \min_{\abs{z - a} =
r}\abs{f(z) - b} > 0$. For $\abs{w - b} < \delta$: on the circle, $\abs{(b - w)} < \delta \leq \abs{f - b}$, so Rouché ($f - b$ versus constant $b - w$) says $f - w$ has exactly $m$ zeros in $D(a, r)$: every such $w$ is attained — $f(D(a,r)) \supseteq D(b, \delta)$: open. Maximum principle: an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum of $\abs f$ is impossible for a nonconstant $f$, its image around $f(a)$ containing points of larger modulus. Inverse: a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) bijection $f$ is open, so $f^{-1}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity); the zero of $f - f(a)$ at $a$ is simple ($m \geq 2$ would give $m$ preimages of nearby values — distinct ones, since $f'$ vanishes only at isolated points, so near $a$ the $m$ zeros of $f - w$ are simple and distinct for generic small $w$: contradiction with injectivity); then $f'(a) \neq 0$ and the difference quotient of $f^{-1}$ converges: $\bigl(f^{-1}\bigr)'(b) =
1/f'(a)$. ∎

## 17.5 Exercises

**Exercise 17.1 ★.**

Classify the singularity at $0$ and compute the [residue](#def-b3-residues-singularities):

$$
\frac{\sin z}{z},\qquad
\frac{\eu^z - 1}{z^2},\qquad
\frac{1}{z(z-1)^2},\qquad
\frac{\cos z}{z^3},\qquad
\eu^{1/z},\qquad
\frac1{\sin z} .
$$

Also give the [residue](#def-b3-residues-singularities) of the third at $z = 1$ and of the last at $z = \pi$.

**Solution of Exercise 17.1.**

$\frac{\sin z}z = 1 - \frac{z^2}6 + \cdots$: removable, [residue](#def-b3-residues-singularities) $0$. $\frac{\eu^z - 1}{z^2} = \frac1z + \frac12 +
\frac z6 + \cdots$: simple pole, [residue](#def-b3-residues-singularities) $1$. $\frac1{z(z-1)^2}$: simple pole at $0$ with [residue](#def-b3-residues-singularities) $\frac1{(0-1)^2} = 1$; double pole at $1$ with [residue](#def-b3-residues-singularities) $\frac{\dd}{\dd z}\bigl(\frac1z\bigr)\big|_{z=1} = -1$. $\frac{\cos z}{z^3} = \frac1{z^3} - \frac1{2z} + \cdots$: pole of order $3$, [residue](#def-b3-residues-singularities) $-\frac12$. $\eu^{1/z} = \sum_{n\geq0}
\frac{z^{-n}}{n!}$: essential, [residue](#def-b3-residues-singularities) $1$. $\frac1{\sin z}$: simple poles at $n\pi$; [residue](#def-b3-residues-singularities) $\frac1{\cos 0} = 1$ at $0$, $\frac1{\cos\pi} = -1$ at $\pi$ ([Method 17.6](#met-b3-residues-compute), $g/h'$).

**Exercise 17.2 ★.**

For $a > 1$ compute, via $z = \eu^{\iu t}$:

$$
\int_0^{2\pi}\frac{\dd t}{a + \cos t}
= \frac{2\pi}{\sqrt{a^2 - 1}} .
$$

Check the limit behaviors $a \to 1^+$ and $a \to \infty$.

**Solution of Exercise 17.2.**

With $z = \eu^{\iu t}$, $\cos t = \frac{z + z^{-1}}2$, $\dd t
= \frac{\dd z}{\iu z}$:

$$
\int_0^{2\pi}\frac{\dd t}{a + \cos t}
= \oint_{\abs z = 1}\frac{2\,\dd z}{\iu\,(z^2 + 2az + 1)} .
$$

The roots $z_\pm = -a \pm \sqrt{a^2 - 1}$ satisfy $z_+z_- =
1$ with $\abs{z_+} < 1 < \abs{z_-}$; the [residue](#def-b3-residues-singularities) at $z_+$ is $\frac{1}{z_+ - z_-} = \frac1{2\sqrt{a^2-1}}$, so the integral is $\frac2\iu\cdot2\iu\pi\cdot\frac1{2\sqrt{a^2-1}} =
\frac{2\pi}{\sqrt{a^2-1}}$. As $a \to 1^+$ it blows up (the integrand peaks at $t = \pi$); as $a \to \infty$ it behaves like $\frac{2\pi}a$, matching $\int\frac{\dd t}a$.

**Exercise 17.3 ★★.**

Compute with semicircle contours, justifying the arc estimates:

$$
\int_\R\frac{x^2}{1 + x^6}\,\dd x = \frac\pi3,
\qquad
\int_\R\frac{\dd x}{(1 + x^2)^{2}} = \frac\pi2
\quad\text{(re-deriving \text{Exercise 14.3})}.
$$

**Solution of Exercise 17.3.**

First integral: upper poles of $\frac{z^2}{1+z^6}$ at $p =
\eu^{\iu\pi/6}, \iu, \eu^{5\iu\pi/6}$; at each, $\operatorname{Res} = \frac{p^2}{6p^5} = \frac{p^3}{6p^6} =
-\frac{p^3}6$, and $p^3$ takes the values $\iu, -\iu, \iu$: sum of [residues](#def-b3-residues-singularities) $-\frac{\iu}{6}$. The arc is $O(R^{-4})\cdot O(R) \to 0$:

$$
\int_\R\frac{x^2\,\dd x}{1 + x^6} =
2\iu\pi\Bigl(-\frac \iu6\Bigr) = \frac\pi3 .
$$

Second: double pole at $\iu$ of $\frac1{(1+z^2)^2} =
\frac1{(z-\iu)^2(z+\iu)^2}$:

$$
\operatorname{Res} = \frac{\dd}{\dd z}\,(z +
\iu)^{-2}\Big|_{z=\iu} = \frac{-2}{(2\iu)^3} =
\frac{-2}{-8\iu} = -\frac\iu4,
\qquad
\int_\R\frac{\dd x}{(1+x^2)^2} =
2\iu\pi\cdot\Bigl(-\frac\iu4\Bigr) = \frac\pi2 ,
$$

consistent with [Exercise 14.3](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#exo-b3-fouriertransform-3)(b).

**Exercise 17.4 ★★.**

Prove, for $t \geq 0$ and $a > 0$:

$$
\int_\R\frac{\cos(tx)}{x^2 + a^2}\,\dd x =
\frac{\pi}{a}\,\eu^{-at},
$$

and deduce the Fourier transform of $x \mapsto \eu^{-a\abs
x}$ by inversion — comparing with [Exercise 14.1](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#exo-b3-fouriertransform-1).

**Solution of Exercise 17.4.**

Close $\frac{\eu^{\iu tz}}{z^2 + a^2}$ in the upper half-plane ($t \geq 0$): there $\abs{\eu^{\iu tz}} =
\eu^{-t\operatorname{Im}z} \leq 1$, so the arc contributes $O(R^{-2})\cdot O(R) \to 0$. The single enclosed pole $\iu a$ is simple with [residue](#def-b3-residues-singularities) $\frac{\eu^{-at}}{2\iu a}$:

$$
\int_\R\frac{\eu^{\iu tx}}{x^2 + a^2}\dd x =
\frac{\pi}{a}\,\eu^{-at},
\qquad\text{hence}\qquad
\int_\R\frac{\cos(tx)}{x^2+a^2}\dd x = \frac\pi
a\,\eu^{-a\abs t}
$$

(real part; even in $t$). This is the inversion counterpart of $\widehat{\eu^{-a\abs x}} = \frac{2a}{a^2+\xi^2}$ ([Exercise 14.1](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#exo-b3-fouriertransform-1)): the two computations confirm each other through [Theorem 14.5](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#thm-b3-fouriertransform-inversion).

**Exercise 17.5 ★★.**

Expand $f(z) = \dfrac1{(z-1)(z-2)}$ in [Laurent series](#thm-b3-residues-laurent) in each of the three regions $\abs z < 1$, $1 < \abs z < 2$, $\abs z >
2$. Why do the three expansions differ? Explain why the coefficient of $z^{-1}$ in the second and third expansions is *not* a [residue](#def-b3-residues-singularities) of $f$ at $0$ ($f$ has no singularity there), and compute the actual [residues](#def-b3-residues-singularities) of $f$, at $1$ and at $2$.

**Solution of Exercise 17.5.**

Partial fractions: $f = \frac1{z-2} - \frac1{z-1}$. On $\abs z < 1$ (Taylor): $f = \sum_{n\geq0}\bigl(1 - 2^{-n-1}\bigr)z^n$. On $1 < \abs z < 2$: $\frac1{z-2} =
-\sum_{n\geq0}\frac{z^n}{2^{n+1}}$ and $-\frac1{z-1} =
-\sum_{n\geq1}z^{-n}$: a genuine two-sided series. On $\abs z > 2$: $f = \sum_{n\geq1}\bigl(2^{n-1} -
1\bigr)z^{-n}$. The three differ because Laurent expansions are attached to *annuli*, not points: each region has its own geometric expansions. The $z^{-1}$-coefficients ($-1$ and $0$ respectively) are integrals over circles encircling *the singularities inside*, not [residues](#def-b3-residues-singularities) at $0$ ($f$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) at $0$): for $1 < \abs z < 2$ the coefficient $-1$ is $\operatorname{Res}(f, 1)$; for $\abs z > 2$ the coefficient $0$ is $\operatorname{Res}(f,1) +
\operatorname{Res}(f,2) = -1 + 1$. The [residues](#def-b3-residues-singularities) of $f$: $-1$ at $1$ and $+1$ at $2$.

**Exercise 17.6 ★★.**

(a) Show that $\eu^{1/z}$ has an essential singularity at $0$ and verify Casorati–Weierstrass by hand: solve $\eu^{1/z} =
w$ explicitly for any $w \neq 0$, exhibiting solutions arbitrarily close to $0$. (b) Show that $\abs{\eu^{1/z}}$ is unbounded on every punctured [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $0$ yet $\eu^{1/z}$ has no pole: which limit fails?

**Solution of Exercise 17.6.**

(a) The [Laurent series](#thm-b3-residues-laurent) $\sum_nz^{-n}/n!$ has infinitely many negative terms: essential. Solving $\eu^{1/z} = w$ ($w \neq
0$): $\frac1z = \log\abs w + \iu\arg w + 2\iu\pi k$, so

$$
z_k = \frac1{\log\abs w + \iu\arg w + 2\iu\pi k}
\xrightarrow[k\to\infty]{} 0 :
$$

every nonzero value is attained infinitely often near $0$ — stronger than [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma). (b) Along $z = 1/x$, $x \to +\infty$: $\eu^x \to \infty$; along $z = \iu/y$: modulus $1$. A pole requires $\abs{f(z)} \to \infty$ *along every approach*: here the limit simply does not exist, even in $[0, +\infty]$.

**Exercise 17.7 ★★.**

Count with Rouché: (a) the zeros of $z^7 - 4z^3 + z - 1$ in $\abs z < 1$; (b) the zeros of $z^4 + 5z + 1$ in $\abs z < 1$ and in $1 <
\abs z < 2$; (c) re-prove d’Alembert–Gauss: a monic degree-$n$ polynomial has $n$ zeros in some large disc *(compare with $z^n$)*.

**Solution of Exercise 17.7.**

(a) On $\abs z = 1$: $\abs{z^7 + z - 1} \leq 3 < 4 =
\abs{-4z^3}$. Rouché with $f = -4z^3$, $g = z^7 + z - 1$: three zeros in the disc. (b) On $\abs z = 1$: $\abs{z^4 + 1} \leq 2 < 5 = \abs{5z}$: one zero in $\abs z < 1$. On $\abs z = 2$: $\abs{5z + 1} \leq
11 < 16 = \abs{z^4}$: four zeros in $\abs z < 2$. Hence three zeros in the annulus. (c) For $P = z^n + a_{n-1}z^{n-1} + \dots$: on $\abs z = R >
1 + \sum\abs{a_k}$, $\abs{P - z^n} \leq
\bigl(\sum\abs{a_k}\bigr)R^{n-1} < R^n = \abs{z^n}$: $P$ has exactly $n$ zeros in $D(0, R)$ — d’Alembert–Gauss with multiplicity, by pure counting.

**Exercise 17.8 ★★★.**

(Hurwitz) Let $f_n \to f$ uniformly on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), $f_n \in
\mathcal H(\Omega)$, $\Omega$ [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected), $f \not\equiv 0$. (a) Show that if all $f_n$ are zero-free, so is $f$. *(If $f(a) = 0$: argument principle on a small circle around $a$, and [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv) to pass to the limit in $\int f_n'/f_n$.)* (b) Show that if all $f_n$ are injective, $f$ is injective or constant. *(Apply (a) to $z \mapsto f_n(z) - f_n(w)$ on $\Omega\setminus\{w\}$.)*

**Solution of Exercise 17.8.**

(a) Suppose $f(a) = 0$, $f \not\equiv 0$: choose $r$ with $f$ zero-free on the circle $C = \partial D(a, r)$ ([isolated zeros](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-identity)) and $m = \min_C\abs f > 0$. By [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv), $f_n \to f$ and $f_n' \to f'$ uniformly on $C$; for large $n$, $\abs{f_n}
\geq m/2$ on $C$, so

$$
\frac1{2\iu\pi}\int_C\frac{f_n'}{f_n}
\longrightarrow \frac1{2\iu\pi}\int_C\frac{f'}{f} \geq 1
$$

(the limit counts the zero $a$; convergence because numerators converge uniformly and denominators are uniformly bounded below). The left side is an integer counting zeros of $f_n$ in the disc: it must be $\geq 1$ eventually — contradicting zero-freeness. So $f$ is zero-free.

(b) Fix $w \in \Omega$ and apply (a) on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $\Omega\setminus\{w\}$ (removing a point of an open [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) subset of $\C$ preserves [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)) to $g_n(z) = f_n(z) -
f_n(w)$, zero-free there by injectivity, converging to $g = f
- f(w)$. If $f$ is nonconstant, $g \not\equiv 0$ on $\Omega\setminus\{w\}$, so $g$ is zero-free there: $f(z) \neq
f(w)$ for all $z \neq w$. As $w$ was arbitrary, $f$ is injective.

**Exercise 17.9 ★★★.**

For $n \geq 2$, integrate $\frac1{1 + z^n}$ over the boundary of the sector $\{0 \leq \arg z \leq \frac{2\pi}n,\ \abs z
\leq R\}$ and deduce

$$
\int_0^{+\infty}\frac{\dd x}{1 + x^n} =
\frac{\pi}{n\,\sin(\pi/n)} .
$$

Verify $n = 2$ against $\arctan$, and the limit $n \to
\infty$.

**Solution of Exercise 17.9.**

The sector boundary consists of $[0, R]$, the arc $A_R$, and the ray $\eu^{2\iu\pi/n}[0, R]$ reversed. Inside lies the single pole $p = \eu^{\iu\pi/n}$ of $\frac1{1+z^n}$, with [residue](#def-b3-residues-singularities) $\frac1{np^{n-1}} = \frac{p}{np^n} = -\frac pn$. On the return ray, $z = \eu^{2\iu\pi/n}x$ gives $z^n = x^n$ and $\dd z = \eu^{2\iu\pi/n}\dd x$; the arc is $O(R^{-n})\cdot O(R) \to 0$. Hence

$$
\bigl(1 - \eu^{2\iu\pi/n}\bigr)
\int_0^\infty\frac{\dd x}{1 + x^n}
= 2\iu\pi\Bigl(-\frac{\eu^{\iu\pi/n}}n\Bigr),
\quad\text{so}\quad
\int_0^\infty\frac{\dd x}{1+x^n}
= \frac{2\iu\pi}{n\,\bigl(\eu^{\iu\pi/n} -
\eu^{-\iu\pi/n}\bigr)} = \frac{\pi}{n\sin(\pi/n)} .
$$

$n = 2$: $\frac\pi{2\sin(\pi/2)} = \frac\pi2 =
[\arctan]_0^\infty$. As $n \to \infty$: the value tends to $1$, and indeed the integrand tends to $\mathbf 1_{\intco01}$ (with DCT domination $\min(1, x^{-2})$ for $n \geq 2$).

**Exercise 17.10 ★★.**

Let $f$ be a rational function with $\deg(\text{denominator})
\geq \deg(\text{numerator}) + 2$. Show that the sum of *all* [residues](#def-b3-residues-singularities) of $f$ is zero *(integrate over larger and larger circles)*. Use this to recompute the partial fraction decomposition of $\frac1{z(z-1)(z-2)}$ with no linear algebra.

**Solution of Exercise 17.10.**

On $\abs z = R$ large, $\abs f \leq C R^{-2}$: $\abs{\oint_{C_R}
f} \leq 2\pi R\cdot CR^{-2} \to 0$. But for $R$ beyond all poles, the [residue](#def-b3-residues-singularities) theorem gives $\oint_{C_R}f =
2\iu\pi\sum_{\text{all }p}\operatorname{Res}(f, p)$: the total sum vanishes. For $f = \frac1{z(z-1)(z-2)}$: [residues](#def-b3-residues-singularities) $\frac1{(-1)(-2)} = \frac12$ at $0$, $\frac1{1\cdot(-1)} = -1$ at $1$, $\frac1{2\cdot1} = \frac12$ at $2$ — summing to $0$ as predicted, and

$$
\frac1{z(z-1)(z-2)} = \frac{1/2}{z} - \frac1{z - 1} +
\frac{1/2}{z-2} :
$$

the [residues](#def-b3-residues-singularities) *are* the partial fraction coefficients, and the zero-sum identity supplies a free consistency check (or determines the last coefficient from the others).

**Exercise 17.11 ★★★.**

(The keyhole: Euler’s reflection integral) For $0 < a < 1$, compute

$$
I(a) = \int_0^{\infty}\frac{x^{a-1}}{1 + x}\,\dd x =
\frac{\pi}{\sin(\pi a)}
$$

by integrating $f(z) = \frac{z^{a-1}}{1+z} =
\frac{\eu^{(a-1)\log z}}{1 + z}$ (logarithm cut along $\R_+$, $\arg z \in \intoo0{2\pi}$) over the keyhole contour: out along the top of the cut from $\varepsilon$ to $R$, around $C_R$, back under the cut, around $C_\varepsilon$. Justify: the two straight stretches differ by the factor $\eu^{2\iu\pi(a-1)}$, the circle contributions vanish ($R^{a-1}\cdot R \to 0$ and $\varepsilon^{a-1}\cdot\varepsilon \to 0$), and the unique pole $z = -1$ has [residue](#def-b3-residues-singularities) $\eu^{\iu\pi(a - 1)}$. Deduce also $\Gamma(a)\Gamma(1 - a) = \frac\pi{\sin\pi a}$ *(write $\Gamma(a)\Gamma(1-a) = B(a, 1-a)$ by [Problem 10.1](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#pb-b3-lebesgue-1) and substitute $t =
\frac{x}{1+x}$)*.

**Solution of Exercise 17.11.**

On the keyhole, with the chosen determination: just above the cut, $\log z = \ln x$; just below, $\log z =
\ln x + 2\iu\pi$. The four pieces give

$$
\Bigl(1 - \eu^{2\iu\pi(a-1)}\Bigr)\int_\varepsilon^R
\frac{x^{a-1}}{1+x}\dd x + \int_{C_R} + \int_{C_\varepsilon}
= 2\iu\pi\operatorname{Res}(f, -1) .
$$

Arcs: $\abs{f} \leq \frac{R^{a-1}}{R - 1}$ on $C_R$, length $2\pi R$: contribution $O(R^{a-1}) \to 0$ ($a < 1$); $\abs f \leq \frac{\varepsilon^{a-1}}{1 - \varepsilon}$ on $C_\varepsilon$, length $2\pi\varepsilon$: $O(\varepsilon^a)
\to 0$ ($a > 0$). [Residue](#def-b3-residues-singularities): at $z = -1 = \eu^{\iu\pi}$, $\operatorname{Res} = \eu^{(a-1)\iu\pi}$. Hence

$$
I(a) = \frac{2\iu\pi\,\eu^{\iu\pi(a-1)}}{1 -
\eu^{2\iu\pi(a-1)}}
= \frac{2\iu\pi}{\eu^{-\iu\pi(a-1)} - \eu^{\iu\pi(a-1)}}
= \frac{\pi}{-\sin(\pi(a-1))} = \frac{\pi}{\sin\pi a} .
$$

Gamma reflection: $B(a, 1-a) =
\int_0^1t^{a-1}(1-t)^{-a}\dd t$; the substitution $t =
\frac x{1+x}$, $1 - t = \frac1{1+x}$, $\dd t =
\frac{\dd x}{(1+x)^2}$ turns it into $\int_0^\infty
\frac{x^{a-1}}{1+x}\dd x = I(a)$, and Euler’s formula $B(a, 1-a) = \Gamma(a)\Gamma(1-a)/\Gamma(1)$ ([Problem 10.1](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#pb-b3-lebesgue-1)) gives $\Gamma(a)\Gamma(1-a) = \frac\pi{\sin\pi a}$ — in particular $\Gamma(\tfrac12) = \sqrt\pi$ once more.

**Exercise 17.12 ★★.**

(Counting zeros with the argument principle, numerically) Let $P(z) = z^4 + 8z + 1$. (a) How many zeros in the unit disc? *(Rouché against $8z + 1$.)* (b) How many in the annulus $1 < \abs z < 3$? *(Rouché against $z^4$ on $\abs z = 3$.)* Sharpen: show every zero has modulus $< 2.1$. (c) How many in the right half-plane? *(Count on $\abs z = 2$ first; then track the image of the imaginary axis: $P(\iu t) = t^4 + 1 + 8\iu t$ has positive real part throughout, so no zeros on the axis, and the argument variation along it is computable — conclude with a large half-disc.)*

**Solution of Exercise 17.12.**

(a) On $\abs z = 1$: $\abs{z^4} = 1 < 7 \leq \abs{8z + 1}$ ($\abs{8z} - 1 = 7$): $P$ has as many zeros in $\mathbb D$ as $8z + 1$, namely *one* (at $-\frac18$).

(b) On $\abs z = 3$: $\abs{8z + 1} \leq 25 < 81 =
\abs{z^4}$: Rouché against $z^4$ gives all four zeros in $\abs z < 3$, hence $4 - 1 = 3$ zeros in the annulus $1 <
\abs z < 3$. Sharpening: a zero with $\abs z = r \geq 2.1$ would satisfy $r^4 = \abs{8z + 1} \leq 8r + 1$, but $r^4 -
8r - 1$ is increasing for $r \geq 2$ and equals $19.45 -
16.8 - 1 = 1.65 > 0$ at $r = 2.1$: impossible. So the three outer zeros lie in $1 < \abs z < 2.1$. (Numerically: a real zero near $-1.95$ and a conjugate pair near $1.04 \pm
1.73\iu$, of modulus $2.02$ — which is why a Rouché attempt at radius exactly $2$ must fail: the theorem demands strict domination, and the zeros sit just outside.)

(c) No zeros on $\iu\R$: $\operatorname{Re}P(\iu t) = t^4 +
1 \geq 1$. Zeros in the right half-plane: use the argument principle on the boundary of the half-disc $\{\abs z \leq
R,\ \operatorname{Re}z \geq 0\}$. On the large arc, $\arg P
\approx \arg z^4$ turns by $4\cdot\pi = 2\pi\cdot2$ (the arc spans angle $\pi$). Along the imaginary axis from $\iu R$ down to $-\iu R$: $P(\iu t) = (t^4 + 1) + 8\iu t$ stays in the right half-plane ($\operatorname{Re} > 0$), so $\arg P$ varies within $\intoo{-\pi/2}{\pi/2}$ and returns with net change $\to 0$ as $R \to \infty$ (endpoints both $\approx \arg t^4 = 0$). Total winding: $\frac{4\pi + 0}
{2\pi} = 2$: *two* zeros in the right half-plane — consistent with the numerics: the conjugate pair $\approx
1.04 \pm 1.73\iu$ has positive real part, the real zeros $\approx -0.125$ and $\approx -1.96$ negative.

## 17.6 Problem: $\zeta(2k)$ by the cotangent

**Problem 17.1.**

Weekend problem — summing $\sum n^{-2k}$ with [residues](#def-b3-residues-singularities)

The [residue](#def-b3-residues-singularities) theorem sums series: pairing a rational function with $\pi\cot(\pi z)$, whose poles sit at the integers, turns $\sum_{n}f(n)$ into a [residue](#def-b3-residues-singularities) count. We prove the method and compute $\zeta(2) = \frac{\pi^2}{6}$ and $\zeta(4) =
\frac{\pi^4}{90}$ — the values found by Fourier series in Year 2 and by operator traces in [Chapter 15](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#ch-b3-spectral), now by contour integration.

**Part I — The cotangent kernel.**

1. Show that $\pi\cot(\pi z)$ is [meromorphic](#def-b3-residues-singularities) on $\C$ with simple poles exactly at $z = n \in \Z$ , each of [residue](#def-b3-residues-singularities) $1$ *(compute $\lim_{z\to n}(z-n)\pi\cot\pi z$)* .
2. Compute the beginning of the Laurent expansion at $0$: $$\pi\cot(\pi z) = \frac1z - \frac{\pi^2}{3}\,z -  \frac{\pi^4}{45}\,z^3 + O(z^5) ,$$ by dividing the power series of $\cos$ by that of $\sin$ (justify the division: $\frac{\sin\pi z}{\pi z}$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and nonzero near $0$, so its reciprocal is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo); identify coefficients through order $3$).
3. Let $C_N$ be the boundary of the square with vertices $(\pm1\pm\iu)(N + \frac12)$ . Show that $\abs{\cot(\pi z)} \leq 2$ on $C_N$ for every $N \geq  1$ . *(On vertical sides, $\cot(\pi(\pm(N +  \frac12) + \iu y)) = \mp\tan(\iu\pi y)$, of modulus $\abs{\tanh(\pi y)} \leq 1$; on horizontal sides $\abs y = N + \frac12$, bound $\abs{\cot(\pi(x\pm\iu y))} \leq \coth(\pi y) \leq  \coth(\pi/2) < 1.1$.)*

**Part II — The summation theorem.**

4. Let $f$ be rational, [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) at the integers, with $\deg(\text{denom}) \geq \deg(\text{num}) + 2$. Using the [residue](#def-b3-residues-singularities) theorem on $C_N$ and the bound of question 3, prove: $$\lim_{N\to\infty}\ \sum_{n = -N}^{N} f(n)  = -\sum_{p\ \text{pole of}\ f}  \operatorname{Res}\bigl(\pi\cot(\pi z)f(z),\,p\bigr).$$
5. Where does the argument need the degree condition? Show by example (take $f(z) = 1/(z + \frac12)$ ) that for slower decay the symmetric limit may still exist while the two-sided series diverges — and that the formula then computes the *principal value* .

**Part III — The values.**

6. Apply the method to $f(z) = 1/z^2$: here $f$ has its pole *at* an integer, so run the argument directly — integrate $g(z) = \frac{\pi\cot(\pi  z)}{z^2}$ over $C_N$, show the integral $\to 0$, and compute $\operatorname{Res}(g, 0)$ from question 2. Conclude: $$2\sum_{n\geq1}\frac1{n^2} = \frac{\pi^2}{3},  \qquad \zeta(2) = \frac{\pi^2}6 .$$
7. Same with $g(z) = \frac{\pi\cot(\pi z)}{z^4}$ : compute $\operatorname{Res}(g, 0)$ and deduce $\zeta(4) = \frac{\pi^4}{90}$ .
8. Explain the general pattern: for every $k \geq 1$ , $\zeta(2k)$ is $-\frac12$ times the coefficient of $z^{2k-1}$ in the Laurent expansion of $\pi\cot(\pi  z)$ at $0$ — a rational multiple of $\pi^{2k}$ . Compute $\zeta(6)$ by pushing question 2’s division one step further. What does the method say about $\zeta(3)$ — and why does it say nothing?

**Part IV — The partial fraction expansion of the cotangent.**

9. Fix $w \in \C\setminus\Z$ and apply the method of Part II to $f(z) = \dfrac{1}{(z - w)(z + w)}$ — noting that $\pi\cot(\pi z)f(z)$ now has additional simple poles at $\pm w$, whose [residues](#def-b3-residues-singularities) must join the count. Deduce the partial fraction expansion $$\pi\cot(\pi w) = \frac1w +  \sum_{n\geq1}\frac{2w}{w^2 - n^2},$$ the series converging normally on [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) subsets of $\C\setminus\Z$.
10. Recover from this expansion, by expanding each term in powers of $w$ (justify the interchange), the *same* Laurent coefficients as in question 2 — the circle closes: Euler’s formula $\sum\frac1{n^2} = \frac{\pi^2}6$ is the coefficient of $w$ in the cotangent’s two faces. Compare with the Fourier-series proof (Year 2) and the trace proof ( [Problem 15.1](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#pb-b3-spectral-1) ): three theories, one number.

**Part V — Euler’s product for the sine.** The expansion of question 9 is the logarithmic derivative of an infinite product; we now prove Euler’s 1734 factorization honestly.

11. For $N \geq 1$ set $P_N(z) =  z\prod_{n=1}^{N}\bigl(1 - \frac{z^2}{n^2}\bigr)$ . Show that $P_N$ converges, uniformly on every disc $\bar D(0, R)$ , to an entire function $P$ whose zeros are exactly the integers, all simple. *(For $n  \geq 2R$ write the factor as $\exp\log(1 -  z^2/n^2)$ with the principal logarithm of [Exercise 16.3](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#exo-b3-holomorphic-3), bound $\abs{\log(1+u)}  \leq 2\abs u$ for $\abs u \leq \frac12$ via the series, and exponentiate the normally convergent sum of logarithms; the finitely many remaining factors are a polynomial. Conclude with [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv).)*
12. Show that on $\C\setminus\Z$, $$\frac{P'(z)}{P(z)} = \frac1z +  \sum_{n\geq1}\frac{2z}{z^2 - n^2} = \pi\cot(\pi z)$$ *(differentiate the finite products, pass to the limit using [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv) and the zero-freeness of $P$ off $\Z$, and quote question 9)*.
13. Show that $Q = \sin(\pi z)/P(z)$ extends to a zero-free entire function with $Q' = 0$, and conclude *Euler’s product*: $$\sin(\pi z) = \pi z\prod_{n\geq1}  \Bigl(1 - \frac{z^2}{n^2}\Bigr)  \qquad (z \in \C) .$$
14. (Wallis, 1655) Evaluate at $z = \frac12$: $$\frac\pi2 = \prod_{n\geq1}\frac{4n^2}{4n^2 - 1}  = \lim_{N\to\infty}  \frac{2\cdot2\cdot4\cdot4\cdots(2N)(2N)}  {1\cdot3\cdot3\cdot5\cdots(2N-1)(2N+1)} .$$
15. For $\abs z < 1$ , expand the logarithm of the product as a double series (justify the rearrangement) and recover $\zeta(2) =  \frac{\pi^2}6$ by matching the coefficient of $z^3$ in $\sin(\pi z) = \pi z - \frac{\pi^3}6z^3 + \cdots$ — the product’s face of Euler’s number.

**Part VI — Sister kernels.** The cotangent has siblings; each prices its own family of series.

16. Differentiate question 9’s expansion term by term (justified by [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv)) to obtain, normally on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) of $\C\setminus\Z$, $$\frac{\pi^2}{\sin^2(\pi z)} =  \sum_{n\in\Z}\frac1{(z - n)^2} .$$
17. Evaluate at $z = \frac12$ : $\sum_{m\geq0}\frac1{(2m+1)^2} = \frac{\pi^2}8$ ; recover $\zeta(2)$ once more by splitting the integers by parity.
18. Verify the duplication identity $\tan\theta =  \cot\theta - 2\cot(2\theta)$ and deduce $$\pi\tan(\pi z) =  \sum_{m\geq0}\frac{8z}{(2m+1)^2 - 4z^2} ,$$ normally on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) avoiding $\frac12 + \Z$.
19. Expand around $0$ ($\abs z < \frac12$; Fubini again): with $\lambda(s) =  \sum_{m\geq0}(2m+1)^{-s}$, $$\pi\tan(\pi z) =  \sum_{k\geq0}8\cdot4^k\,\lambda(2k+2)\,z^{2k+1} ;$$ compare with $\tan u = u + \frac{u^3}3 + O(u^5)$ to recover $\lambda(2) = \frac{\pi^2}8$ and to get $\lambda(4) = \frac{\pi^4}{96}$, then cross-check $\zeta(4) = \frac{\pi^4}{90}$ via $\lambda(4) = (1 -  2^{-4})\,\zeta(4)$.
20. Verify $\frac1{\sin\theta} = \cot\frac\theta2 -  \cot\theta$ and deduce $$\frac{\pi}{\sin(\pi z)} = \frac1z +  \sum_{n\geq1}(-1)^n\,\frac{2z}{z^2 - n^2} .$$ Check the signs against the [residues](#def-b3-residues-singularities) of $\pi/\sin(\pi z)$ at the integers.
21. Read off the coefficient of $z$ : $\eta(2) =  \sum_{n\geq1}\frac{(-1)^{n-1}}{n^2} =  \frac{\pi^2}{12}$ , and confirm the consistency $\eta(2) = (1 - 2^{1-2})\,\zeta(2)$ .
22. (Finale) Evaluate question 20’s expansion at $z =  \frac12$ and deduce *Leibniz’s formula* $$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17 +  \cdots$$ Close with a short paragraph: one kernel per arithmetic — which kernel prices which family of series, and why all of them are structurally blind to $\zeta(3)$.

**Part VII — The full price list: Bernoulli numbers.**

23. Combine the partial fraction expansion of $\pi  z\cot(\pi z)$ with the generating function of the Bernoulli numbers ($\frac{w}{\eu^w - 1} =  \sum_n\frac{B_n}{n!}w^n$, [Problem 16.1](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#pb-b3-holomorphic-1), Part VI): from $$\pi z\cot(\pi z) = \iu\pi z +  \frac{2\iu\pi z}{\eu^{2\iu\pi z} - 1}$$ (prove this identity first), deduce the closed form $$\zeta(2k) = (-1)^{k+1}\,  \frac{(2\pi)^{2k}\,B_{2k}}{2\,(2k)!}  \qquad (k \geq 1).$$
24. Verify the formula against $B_2 = \frac16$ , $B_4 =  -\frac1{30}$ , $B_6 = \frac1{42}$ : recover $\zeta(2) =  \frac{\pi^2}6$ , $\zeta(4) = \frac{\pi^4}{90}$ , and compute $\zeta(6) = \frac{\pi^6}{945}$ .
25. (Euler’s recursion) Expand both sides of $\bigl(z\cot z\bigr)' = \cot z - z(1 + \cot^2z)$ — or square the cotangent series directly — to prove $$\Bigl(k + \frac12\Bigr)\zeta(2k) =  \sum_{j=1}^{k-1}\zeta(2j)\,\zeta(2k - 2j)  \qquad (k \geq 2),$$ and check it computes $\zeta(4)$ from $\zeta(2)$ and $\zeta(6)$ from $\zeta(2), \zeta(4)$ — all even zeta values from the single seed $\frac{\pi^2}6$, with no new integration.

**Solution of Problem 17.1.**

**1.** $\sin(\pi z)$ has simple zeros exactly at $\Z$ ($\sin\pi z = 0$ iff $z \in \Z$, and $(\sin\pi z)' = \pi\cos\pi
z \neq 0$ there), and $\cos(\pi n) \neq 0$: $\pi\cot(\pi z) =
\pi\cos(\pi z)/\sin(\pi z)$ has simple poles at $\Z$ with

$$
\operatorname{Res}(\pi\cot\pi z,\ n)
= \frac{\pi\cos(\pi n)}{\pi\cos(\pi n)} = 1
$$

($g/h'$ rule, [Method 17.6](#met-b3-residues-compute)).

**2.** $\frac{\sin(\pi z)}{\pi z} = 1 - \frac{(\pi z)^2}6
+ \frac{(\pi z)^4}{120} - \cdots$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and nonzero near $0$: its reciprocal is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) ([Definition 16.1](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo): quotient), with series $1 +
\frac{(\pi z)^2}{6} + \frac{7(\pi z)^4}{360} + \cdots$ (identify: $(1 - u)^{-1}$-style coefficients from $u =
\frac{(\pi z)^2}6 - \frac{(\pi z)^4}{120}$: the $z^4$ coefficient is $\frac1{36} - \frac1{120} = \frac{7}{360}$). Multiply by $\cos(\pi z) = 1 - \frac{(\pi z)^2}2 + \frac{(\pi
z)^4}{24} - \cdots$ and divide by $z$:

$$
\pi\cot(\pi z) = \frac1z\Bigl[1 + \pi^2z^2\Bigl(\frac16 -
\frac12\Bigr) + \pi^4z^4\Bigl(\frac7{360} - \frac1{12} +
\frac1{24}\Bigr)\Bigr] + \cdots
= \frac1z - \frac{\pi^2}3\,z - \frac{\pi^4}{45}\,z^3 -
\cdots
$$

($\frac7{360} - \frac{30}{360} + \frac{15}{360} =
-\frac8{360} = -\frac1{45}$).

**3.** Vertical sides $z = \pm(N + \frac12) + \iu y$: by $\pi$-periodicity of $\cot$, $\cot(\pi z) = \cot(\pm\frac\pi2
+ \iu\pi y) = -\tan(\iu\pi y) = -\iu\tanh(\pi y)$, of modulus $\leq 1$. Horizontal sides $z = x \pm \iu(N + \frac12)$: from $\abs{\cot(a + \iu b)}^2 = \frac{\cos^2a +
\sinh^2b}{\sin^2a + \sinh^2b} \leq \frac{1 +
\sinh^2b}{\sinh^2b} = \coth^2 b$,

$$
\abs{\cot(\pi z)} \leq \coth\bigl(\pi(N + \tfrac12)\bigr)
\leq \coth(\pi/2) < 1.1 .
$$

Both bounds are $\leq 2$.

**4.** Apply [Theorem 17.5](#thm-b3-residues-residue) to $F(z) =
\pi\cot(\pi z)f(z)$ on $C_N$ ($N$ beyond all poles of $f$):

$$
\frac1{2\iu\pi}\oint_{C_N}F = \sum_{n=-N}^{N}f(n) +
\sum_p\operatorname{Res}(F, p),
$$

the integer poles contributing $f(n)$ (question 1; $f$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) there). On $C_N$: $\abs F \leq 2\pi\cdot
C\abs z^{-2} \leq 2\pi C N^{-2}$, and the perimeter is $8(N +
\frac12)$: the integral is $O(1/N) \to 0$. Let $N \to
\infty$: the displayed summation formula.

**5.** The decay $\abs f = O(\abs z^{-2})$ killed the [contour integral](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-contour) *and* made $\sum\abs{f(n)}$ converge. For $f(z) = \frac1{z + \frac12}$: the symmetric sums $\sum_{-N}^N\frac1{n + \frac12}$ telescope to $0$ (the terms $n$ and $-n - 1$ cancel), and the right side is $-\operatorname{Res}\bigl(\frac{\pi\cot\pi z}{z + \frac12},
-\frac12\bigr) = -\pi\cot(-\frac\pi2) = 0$: consistent — but $\sum\abs{f(n)}$ diverges; the method computes the symmetric (principal value) limit only.

**6.** $g(z) = \frac{\pi\cot(\pi z)}{z^2}$: poles at the nonzero integers with [residues](#def-b3-residues-singularities) $\frac1{n^2}$, and at $0$ where, by question 2,

$$
g(z) = \frac1{z^3} - \frac{\pi^2}{3z} - \frac{\pi^4}{45}z -
\cdots :
\qquad \operatorname{Res}(g, 0) = -\frac{\pi^2}3 .
$$

The [contour integral](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-contour) over $C_N$ tends to $0$ as in question 4 ($\abs{g} = O(N^{-2})$ on $C_N$). Hence $0 =
\sum_{n\neq0}\frac1{n^2} - \frac{\pi^2}3$: $2\zeta(2) =
\frac{\pi^2}3$, $\zeta(2) = \frac{\pi^2}6$.

**7.** $g(z) = \frac{\pi\cot(\pi z)}{z^4} = \frac1{z^5}
- \frac{\pi^2}{3z^3} - \frac{\pi^4}{45z} - \cdots$: [residue](#def-b3-residues-singularities) at $0$ equal to $-\frac{\pi^4}{45}$, and $0 =
2\zeta(4) - \frac{\pi^4}{45}$: $\zeta(4) =
\frac{\pi^4}{90}$.

**8.** With $g = \pi\cot(\pi z)/z^{2k}$: the [residue](#def-b3-residues-singularities) at $0$ is the coefficient $a_{2k-1}$ of $z^{2k-1}$ in the expansion of $\pi\cot(\pi z)$, and the vanishing contour gives $2\zeta(2k) + a_{2k-1} = 0$: $\zeta(2k) = -a_{2k-1}/2$, a rational multiple of $\pi^{2k}$ since the cotangent’s coefficients are. One more division step yields $a_5 =
-\frac{2\pi^6}{945}$, whence $\zeta(6) = \frac{\pi^6}{945}$. For $\zeta(3)$: the natural kernel $g = \pi\cot(\pi z)/z^3$ produces $\sum_{n\neq0}\frac1{n^3} = 0$ by oddness — the method proves $0 = 0$ and is structurally blind to odd zeta values (no closed form for $\zeta(3)$ is known; its irrationality, Apéry 1978, needed entirely different ideas).

**9.** $F(z) = \frac{\pi\cot(\pi z)}{(z - w)(z + w)}$ has poles at the integers ([residues](#def-b3-residues-singularities) $\frac1{n^2 - w^2}$, noting the sign: $f(n) = \frac1{(n-w)(n+w)} = \frac1{n^2 -
w^2}$) and simple poles at $\pm w$ with [residues](#def-b3-residues-singularities) $\frac{\pi\cot(\pm\pi w)}{\pm2w} = \frac{\pi\cot(\pi
w)}{2w}$ each ($\cot$ is odd). Question 4’s argument ($\abs f = O(\abs z^{-2})$) gives

$$
\sum_{n\in\Z}\frac1{n^2 - w^2} + \frac{\pi\cot(\pi w)}{w} =
0,
\qquad\text{i.e.}\qquad
\pi\cot(\pi w) = \frac1w + \sum_{n\geq1}\frac{2w}{w^2 -
n^2},
$$

(the $n = 0$ term is $-\frac1{w^2}$; regroup $\pm n$). Normal convergence on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) of $\C\setminus\Z$: for $\abs w \leq
R$ and $n \geq 2R$, $\abs{\frac{2w}{w^2 - n^2}} \leq
\frac{2R}{n^2 - R^2} \leq \frac{8R}{3n^2}$.

**10.** For $\abs w \leq r < 1$: $\frac{2w}{w^2 - n^2} =
-\frac{2w}{n^2}\cdot\frac1{1 - w^2/n^2} =
-2\sum_{k\geq0}\frac{w^{2k+1}}{n^{2k+2}}$, with $\abs{\text{terms}} \leq 2r^{2k+1}/n^{2k+2}$, summable over $(n, k)$: Fubini for series rearranges

$$
\pi\cot(\pi w) = \frac1w -
2\sum_{k\geq0}\zeta(2k+2)\,w^{2k+1} .
$$

Matching with question 2: $-2\zeta(2) = -\frac{\pi^2}3$ and $-2\zeta(4) = -\frac{\pi^4}{45}$ — the same values. Three roads to $\frac{\pi^2}6$: [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) (Fourier series), the trace of the string’s Green operator, and the cotangent’s two expansions; that a sum over frequencies, an operator trace, and a [contour integral](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-contour) agree is no accident — each is a face of the same spectral identity.

**11.** Fix $R \geq 1$ and let $n_0$ be the smallest integer $\geq 2R$. For $\abs z \leq R$ and $n \geq n_0$: $\abs{z^2/n^2} \leq \frac14$, so $1 - z^2/n^2 \in
\bar D(1, \frac14)$, where the principal logarithm is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo), and

$$
\Bigl|\log\Bigl(1 - \frac{z^2}{n^2}\Bigr)\Bigr|
\leq \sum_{k\geq1}\frac1k\,\Bigl|\frac{z^2}{n^2}\Bigr|^k
\leq \frac{\abs{z^2/n^2}}{1 - \abs{z^2/n^2}}
\leq \frac{2R^2}{n^2} :
$$

the sum $S(z) = \sum_{n\geq n_0}\log(1 - z^2/n^2)$ converges normally on $\bar D(0, R)$, with [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) partial sums $S_N$ and $\abs{S_N} \leq 2R^2\zeta(2)$ uniformly. Since $\abs{\eu^a - \eu^b} \leq
\eu^{\max(\abs a,\abs b)}\abs{a - b}$ (mean value bound on the segment), the tail products $\prod_{n_0\leq n\leq N} =
\eu^{S_N}$ converge uniformly on $\bar D(0, R)$ to the zero-free $\eu^S$. Multiplying by the fixed polynomial $z\prod_{n<n_0}(1 - z^2/n^2)$: $P_N \to P$ uniformly on $\bar D(0, R)$, and [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv) makes $P$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) there; $R$ being arbitrary, $P$ is entire. On $\bar D(0, R)$ the zeros of $P$ are those of the polynomial prefactor — the integers of modulus $\leq R$, each simple ($(1 - z/n)(1 + z/n)$ has distinct simple zeros, $\eu^S$ none): the zero set of $P$ is $\Z$, all zeros simple.

**12.** Logarithmic differentiation of the finite product, away from its zeros:

$$
\frac{P_N'(z)}{P_N(z)} = \frac1z +
\sum_{n=1}^{N}\frac{-2z/n^2}{1 - z^2/n^2}
= \frac1z + \sum_{n=1}^{N}\frac{2z}{z^2 - n^2} .
$$

On a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq \C\setminus\Z$: $P_N \to P$ and $P_N' \to P'$ uniformly ([Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv)), and $\min_K
\abs P > 0$ ($P$ vanishes only on $\Z$), so eventually $\abs{P_N} \geq \frac12\min_K\abs P$ and $P_N'/P_N \to
P'/P$ uniformly on $K$. The middle member converges to $\frac1z + \sum_{n\geq1}\frac{2z}{z^2-n^2} = \pi\cot(\pi z)$ by question 9: hence $P'/P = \pi\cot(\pi z)$ on $\C\setminus\Z$.

**13.** $\sin(\pi z)$ and $P$ are entire with the same zero set $\Z$, all zeros simple (questions 1 and 11). Near $m \in \Z$ write $\sin(\pi z) = (z - m)\,\sigma(z)$ and $P(z) = (z - m)\,\psi(z)$ with $\sigma, \psi$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and nonvanishing at $m$ (factor the power series): $Q =
\sin(\pi z)/P = \sigma/\psi$ extends [holomorphically](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and zero-free across each integer, and is zero-free on $\C\setminus\Z$ as a quotient of zero-free functions. There,

$$
\frac{Q'}{Q} = \frac{(\sin\pi z)'}{\sin\pi z} -
\frac{P'}{P} = \pi\cot(\pi z) - \pi\cot(\pi z) = 0 ,
$$

so the entire function $Q'$ vanishes on $\C\setminus\Z$, hence everywhere by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $Q$ is constant. As $z \to
0$: $\sin(\pi z)/z \to \pi$ and $P(z)/z \to 1$, so $Q =
\pi$:

$$
\sin(\pi z) = \pi z\prod_{n\geq1}
\Bigl(1 - \frac{z^2}{n^2}\Bigr) .
$$

**14.** At $z = \frac12$: $1 = \sin\frac\pi2 =
\frac\pi2\prod_{n\geq1}\bigl(1 - \frac1{4n^2}\bigr) =
\frac\pi2\prod\frac{4n^2-1}{4n^2}$, so

$$
\frac\pi2 = \prod_{n\geq1}\frac{4n^2}{4n^2 - 1}
= \lim_{N\to\infty}\prod_{n=1}^N
\frac{(2n)(2n)}{(2n-1)(2n+1)}
= \lim_{N\to\infty}
\frac{2\cdot2\cdot4\cdot4\cdots(2N)(2N)}
{1\cdot3\cdot3\cdot5\cdots(2N-1)(2N+1)} :
$$

Wallis’s product, a one-line corollary of Euler’s factorization.

**15.** For $\abs z \leq r < 1$ every factor lies in $D(1, r^2) \subseteq D(1, 1)$, so $P(z)/z = \exp\bigl(
\sum_{n\geq1}\log(1 - z^2/n^2)\bigr)$: each partial product is the exponential of a partial sum, and both sides pass to the limit by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $\exp$. The double series

$$
\sum_{n\geq1}\log\Bigl(1 - \frac{z^2}{n^2}\Bigr)
= -\sum_{n\geq1}\sum_{k\geq1}\frac{z^{2k}}{k\,n^{2k}}
= -\sum_{k\geq1}\frac{\zeta(2k)}k\,z^{2k}
= -\zeta(2)\,z^2 + O(z^4)
$$

rearranges by Fubini for series: $\sum_{n,k}
\frac{r^{2k}}{kn^{2k}} \leq \sum_k\zeta(2k)r^{2k} \leq
\zeta(2)\frac{r^2}{1-r^2} < \infty$. Hence

$$
P(z) = z\,\exp\bigl(-\zeta(2)z^2 + O(z^4)\bigr)
= z - \zeta(2)\,z^3 + O(z^5) ,
$$

and question 13 compares this with $\sin(\pi z) = \pi z -
\frac{\pi^3}6z^3 + O(z^5)$: $\pi\zeta(2) = \frac{\pi^3}6$, i.e. $\zeta(2) = \frac{\pi^2}6$. The additive face (question 10) and the multiplicative face compute the same number.

**16.** On a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq \C\setminus\Z$ the partial sums $S_N = \frac1z + \sum_{n\leq N}\bigl(
\frac1{z-n} + \frac1{z+n}\bigr)$ (question 9, terms regrouped as $\frac{2z}{z^2-n^2} = \frac1{z-n} +
\frac1{z+n}$) converge uniformly to $\pi\cot(\pi z)$, so [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv) gives $S_N' \to
(\pi\cot\pi z)' = -\pi^2/\sin^2(\pi z)$ uniformly on $K$. Since $S_N' = -\sum_{\abs n\leq N}(z - n)^{-2}$:

$$
\frac{\pi^2}{\sin^2(\pi z)} =
\sum_{n\in\Z}\frac1{(z - n)^2} ,
$$

the convergence normal on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) of $\C\setminus\Z$ (terms $O(n^{-2})$).

**17.** At $z = \frac12$ the left side is $\pi^2$; on the right, $(\frac12 - n)^2 = \frac{(2n-1)^2}4$ with $2n -
1$ running over all odd integers exactly once as $n$ runs over $\Z$:

$$
\pi^2 = \sum_{n\in\Z}\frac{4}{(2n-1)^2}
= 8\sum_{m\geq0}\frac1{(2m+1)^2},
\qquad
\sum_{m\geq0}\frac1{(2m+1)^2} = \frac{\pi^2}8 .
$$

Parity split: $\zeta(2) = \frac{\pi^2}8 +
\sum_{n\geq1}\frac1{(2n)^2} = \frac{\pi^2}8 +
\frac{\zeta(2)}4$, so $\frac34\zeta(2) = \frac{\pi^2}8$ and $\zeta(2) = \frac{\pi^2}6$ once more.

**18.** With $c = \cot\theta$ and $\cot(2\theta) =
\frac{c^2-1}{2c}$: $\cot\theta - 2\cot(2\theta) = c -
\frac{c^2-1}c = \frac1c = \tan\theta$. Hence $\pi\tan(\pi z)
= \pi\cot(\pi z) - 2\pi\cot(2\pi z)$, and question 9 at $z$ and at $2z$ gives

$$
\pi\cot(\pi z) = \frac1z +
\sum_{n\geq1}\frac{2z}{z^2 - n^2},
\qquad
2\pi\cot(2\pi z) = \frac1z +
\sum_{n\geq1}\frac{8z}{4z^2 - n^2} .
$$

Both series converge absolutely at each fixed $z$ off the poles, so the difference may be regrouped at will: in the second series the even terms $n = 2m$ give $\frac{8z}{4z^2 -
4m^2} = \frac{2z}{z^2 - m^2}$ and cancel the first series entirely, leaving

$$
\pi\tan(\pi z) = -\sum_{m\geq0}\frac{8z}{4z^2 - (2m+1)^2}
= \sum_{m\geq0}\frac{8z}{(2m+1)^2 - 4z^2} ,
$$

normally on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) avoiding $\frac12 + \Z$ (terms $O(m^{-2})$).

**19.** For $\abs z \leq r < \frac12$:

$$
\frac{8z}{(2m+1)^2 - 4z^2} = \frac{8z}{(2m+1)^2}
\sum_{k\geq0}\Bigl(\frac{4z^2}{(2m+1)^2}\Bigr)^{k},
$$

with $\sum_{m,k}8r\,(4r^2)^k(2m+1)^{-2k-2} < \infty$ since $4r^2 < 1$: Fubini rearranges the double sum into

$$
\pi\tan(\pi z) =
\sum_{k\geq0}8\cdot4^k\,\lambda(2k+2)\,z^{2k+1} .
$$

Against $\pi\tan(\pi z) = \pi^2z + \frac{\pi^4}3z^3 +
O(z^5)$: the coefficient of $z$ gives $8\lambda(2) = \pi^2$ — question 17 again — and that of $z^3$ gives $32\lambda(4) = \frac{\pi^4}3$, i.e. $\lambda(4) =
\frac{\pi^4}{96}$. Removing the even denominators, $\lambda(4) = \zeta(4) - 2^{-4}\zeta(4) =
\frac{15}{16}\zeta(4)$: $\zeta(4) = \frac{16}{15}\cdot
\frac{\pi^4}{96} = \frac{\pi^4}{90}$, matching question 7.

**20.** $\cot\frac\theta2 - \cot\theta =
\frac{\cos\frac\theta2\,\sin\theta -
\cos\theta\,\sin\frac\theta2}{\sin\frac\theta2\,\sin\theta}
= \frac{\sin\frac\theta2}{\sin\frac\theta2\,\sin\theta} =
\frac1{\sin\theta}$, the numerator being $\sin(\theta -
\frac\theta2)$. With $\theta = \pi z$, question 9 at $\frac z2$ reads $\pi\cot\frac{\pi z}2 = \frac2z +
\sum_{n\geq1}\frac{4z}{z^2 - 4n^2}$, so

$$
\frac{\pi}{\sin(\pi z)}
= \pi\cot\frac{\pi z}2 - \pi\cot(\pi z)
= \frac1z + \sum_{n\geq1}\frac{4z}{z^2 - 4n^2}
- \sum_{n\geq1}\frac{2z}{z^2 - n^2} .
$$

Absolute convergence permits the parity regrouping: even $n
= 2m$ in the subtracted series contributes $\frac{2z}{z^2-4m^2}$, leaving $+\frac{2z}{z^2-4m^2}$ from the first sum, while odd $n$ survives with sign $-$:

$$
\frac{\pi}{\sin(\pi z)} = \frac1z +
\sum_{n\geq1}(-1)^n\,\frac{2z}{z^2 - n^2} .
$$

Signs check: $\operatorname{Res}\bigl(\pi/\sin(\pi z),
n\bigr) = \pi/(\pi\cos\pi n) = (-1)^n$ ([Method 17.6](#met-b3-residues-compute)), and $(-1)^n\frac{2z}{z^2 -
n^2} = \frac{(-1)^n}{z-n} + \frac{(-1)^n}{z+n}$ carries exactly that [residue](#def-b3-residues-singularities) at $\pm n$.

**21.** For $\abs z \leq r < 1$, expanding each term as in question 19 ($(-1)^n\frac{2z}{z^2-n^2} =
(-1)^{n-1}\frac{2z}{n^2}\sum_k\frac{z^{2k}}{n^{2k}}$) and applying Fubini:

$$
\frac{\pi}{\sin(\pi z)} = \frac1z +
2\sum_{k\geq0}\eta(2k+2)\,z^{2k+1},
\qquad
\eta(s) = \sum_{n\geq1}\frac{(-1)^{n-1}}{n^s} .
$$

Taylor side: $\sin(\pi z) = \pi z(1 - \frac{(\pi z)^2}6 +
O(z^4))$ gives $\frac\pi{\sin\pi z} = \frac1z +
\frac{\pi^2}6z + O(z^3)$: $2\eta(2) = \frac{\pi^2}6$, so $\eta(2) = \frac{\pi^2}{12}$. Consistency: $\eta(2) =
\zeta(2) - 2\sum_n(2n)^{-2} = (1 - 2^{1-2})\zeta(2) =
\frac{\zeta(2)}2 = \frac{\pi^2}{12}$.

**22.** At $z = \frac12$, question 20 gives

$$
\pi = 2 + \sum_{n\geq1}(-1)^n\frac{1}{\frac14 - n^2}
= 2 + 4\sum_{n\geq1}\frac{(-1)^{n-1}}{(2n-1)(2n+1)} .
$$

With $\frac1{(2n-1)(2n+1)} = \frac12\bigl(\frac1{2n-1} -
\frac1{2n+1}\bigr)$, and both alternating series convergent (Leibniz criterion), the sum splits as $\frac12\bigl[L -
(1 - L)\bigr] = L - \frac12$, where $L = 1 - \frac13 +
\frac15 - \cdots$ and $\frac13 - \frac15 + \frac17 - \cdots
= 1 - L$. Hence $\pi = 2 + 4(L - \frac12) = 4L$:

$$
\frac\pi4 = 1 - \frac13 + \frac15 - \frac17 + \cdots
$$

The moral: each kernel is [meromorphic](#def-b3-residues-singularities) with poles on an arithmetic progression and prescribed [residues](#def-b3-residues-singularities). The cotangent puts [residue](#def-b3-residues-singularities) $1$ at every integer and sums $f(n)$; its derivative squares the poles and prices $\lambda(2)$; the tangent moves the poles to $\frac12 + \Z$ and prices the odd denominators; $\pi/\sin$ keeps the integer poles but alternates the [residues](#def-b3-residues-singularities) $(-1)^n$, hence the alternating series. The first-order kernels are all odd functions: pairing $n$ with $-n$ doubles the even-power coefficients and annihilates the odd ones, so $\zeta(2k)$ pours out mechanically while $\zeta(3)$ never appears. The blindness is parity, not lack of technique.

**23.** With $w = 2\iu\pi z$:

$$
\iu\pi z + \frac{2\iu\pi z}{\eu^{2\iu\pi z} - 1}
= \iu\pi z\,\frac{\eu^{2\iu\pi z} + 1}{\eu^{2\iu\pi z} - 1}
= \iu\pi z\,\frac{\eu^{\iu\pi z} +
\eu^{-\iu\pi z}}{\eu^{\iu\pi z} - \eu^{-\iu\pi z}}
= \pi z\,\frac{\cos\pi z}{\sin\pi z} = \pi z\cot(\pi z) .
$$

Hence, using the generating function on $w = 2\iu\pi z$ (and $B_1 = -\frac12$ cancelling the $\iu\pi z$ term, odd $B$’s vanishing beyond):

$$
\pi z\cot(\pi z) = \sum_{k\geq0}\frac{B_{2k}}{(2k)!}
(2\iu\pi z)^{2k}
= 1 + \sum_{k\geq1}(-1)^k\frac{(2\pi)^{2k}B_{2k}}{(2k)!}
z^{2k} .
$$

On the other hand the partial fraction expansion (Part IV) gives $\pi z\cot(\pi z) = 1 - 2\sum_{k\geq1}\zeta(2k)z^{2k}$ (expand each $\frac{2z^2}{z^2 - n^2} =
-2\sum_k\frac{z^{2k}}{n^{2k}}$ and sum over $n$, normal convergence justifying the interchange for $\abs z < 1$). Comparing coefficients: $-2\zeta(2k) =
(-1)^k\frac{(2\pi)^{2k}B_{2k}}{(2k)!}$, the stated formula.

**24.** $k = 1$: $\frac{(2\pi)^2}{2\cdot2}\cdot\frac16
= \frac{\pi^2}6$. $k = 2$: $-\frac{(2\pi)^4}{2\cdot24}
\cdot\bigl(-\frac1{30}\bigr) = \frac{16\pi^4}{48\cdot30} =
\frac{\pi^4}{90}$. $k = 3$: $\frac{(2\pi)^6}{2\cdot720}
\cdot\frac1{42} = \frac{64\pi^6}{1440\cdot42} =
\frac{\pi^6}{945}$.

**25.** Write $C(z) = \pi z\cot(\pi z) = 1 -
2\sum_{k\geq1}\zeta(2k)z^{2k}$ (question 23). Direct differentiation of $C = \pi z\cot(\pi z)$, using $(\cot u)'
= -1 - \cot^2u$:

$$
zC'(z) = \pi z\cot(\pi z) - \pi^2z^2\bigl(1 +
\cot^2(\pi z)\bigr) = C - \pi^2z^2 - C^2 .
$$

Now expand both sides in powers of $z^2$. Left side: $\sum_k(-4k)\,\zeta(2k)\,z^{2k}$. Right side: squaring the series,

$$
C - C^2 = 2\sum_{k\geq1}\zeta(2k)z^{2k}
- 4\sum_{k\geq2}\Bigl(\sum_{j=1}^{k-1}\zeta(2j)
\zeta(2k-2j)\Bigr)z^{2k},
$$

and the term $-\pi^2z^2 = -6\zeta(2)z^2$ only adjusts $k =
1$. Comparing the coefficients of $z^{2k}$ for $k \geq 2$:

$$
-4k\,\zeta(2k) = 2\,\zeta(2k) -
4\sum_{j=1}^{k-1}\zeta(2j)\,\zeta(2k-2j),
$$

i.e. $\bigl(k + \frac12\bigr)\zeta(2k) =
\sum_{j=1}^{k-1}\zeta(2j)\zeta(2k-2j)$. (At $k = 1$ the identity reads $-4\zeta(2) = 2\zeta(2) - 6\zeta(2)$: a consistency check, not new information.) Applications: $k =
2$: $\frac52\zeta(4) = \zeta(2)^2 = \frac{\pi^4}{36}$, so $\zeta(4) = \frac{\pi^4}{90}$; $k = 3$: $\frac72\zeta(6) =
2\zeta(2)\zeta(4) = \frac{\pi^6}{270}$, so $\zeta(6) =
\frac{2}{7}\cdot\frac{\pi^6}{270} = \frac{\pi^6}{945}$. One transcendental seed ($\zeta(2) = \frac{\pi^2}6$), and pure algebra generates every even zeta value.