--- title: "Holomorphic Functions" book: "University Mathematics — Year 3" subject: math language: en chapter: 16 exercises: 12 source: https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions --- # Chapter 16 — Holomorphic Functions Complex differentiability looks like a small variation on the real theory — one limit, one quotient. It is instead a different universe. A function differentiable *once* on an open subset of $\C$ is automatically infinitely differentiable, analytic, determined on a whole domain by its values near a single point, and constrained by rigid global principles (Liouville, maximum modulus). All of this flows from one miracle, Cauchy’s theorem: the integral of a [holomorphic function](#def-b3-holomorphic-holo) along a closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) in a star-shaped domain vanishes. This chapter proves the miracle (Goursat’s argument, with no regularity assumed beyond differentiability), harvests its classical consequences, and settles a theorem this book has used on credit since [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois): every nonconstant complex polynomial has a root. Throughout, $\Omega \subseteq \C$ is open, and $D(a, r)$ denotes the open disc. ## 16.1 Complex differentiability **Definition 16.1.** $f \colon \Omega \to \C$ is *holomorphic* on $\Omega$ if for every $z_0 \in \Omega$ $$ f'(z_0) = \lim_{h\to0}\frac{f(z_0 + h) - f(z_0)}{h} $$ exists ($h \in \C^*$). Sums, products, quotients (nonvanishing denominators), compositions of holomorphic functions are holomorphic, with the usual formulas (the Year 1–2 proofs are verbatim: they only use field operations and limits). $\mathcal H(\Omega)$ denotes the set of holomorphic functions on $\Omega$. **Proposition 16.2 (Cauchy–Riemann).** Write $f(x + \iu y) = P(x,y) + \iu Q(x,y)$. Then $f$ is [holomorphic](#def-b3-holomorphic-holo) at $z_0$ iff $f$ is $\R$-differentiable at $z_0$ (as a map of two real variables) and $$ \frac{\partial P}{\partial x} = \frac{\partial Q}{\partial y}, \qquad \frac{\partial P}{\partial y} = -\frac{\partial Q}{\partial x} \qquad \text{at } z_0 ; $$ equivalently, the real differential is multiplication by the complex number $f'(z_0)$. **Proof.** $\C$-differentiability says $f(z_0 + h) = f(z_0) + ch + o(\abs h)$ with $c = f'(z_0)$: an $\R$-linear differential which is multiplication by $c = a + \iu b$, i.e. has matrix $\bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr)$ in the basis $(1, \iu)$ — exactly the displayed relations for the partials. Conversely such a differential is $\C$-linear, and the $o(\abs h)$ definitions match. ∎ **Example 16.3.** Polynomials in $z$, rational functions off their poles, and — by Year 2’s term-by-term differentiation theorem for power series, whose proof works identically over $\C$ — every sum of a power series $\sum a_n(z - a)^n$ inside its disc of convergence: [holomorphic](#def-b3-holomorphic-holo), with derivative $\sum na_n(z - a)^{n-1}$ (same radius). In particular $\exp z = \sum z^n/n!$ is entire ([holomorphic](#def-b3-holomorphic-holo) on $\C$) with $\exp' = \exp$. On the other hand $z \mapsto \bar z$, $\abs z$, $\operatorname{Re}z$ are nowhere [holomorphic](#def-b3-holomorphic-holo) (Cauchy–Riemann fails everywhere): [holomorphy](#def-b3-holomorphic-holo) is orientation-and-angle-preserving rigidity, not smoothness. ## 16.2 Contour integrals **Definition 16.4.** A *[path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected)* is a piecewise $\mathcal C^1$ map $\gamma \colon \intcc ab \to \C$; it is *closed* if $\gamma(a) = \gamma(b)$. For [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ on the image of $\gamma$: $$ \int_\gamma f(z)\,\dd z = \int_a^b f(\gamma(t))\,\gamma'(t)\,\dd t, \qquad \Bigl|\int_\gamma f\,\dd z\Bigr| \leq \sup_{\gamma}\abs f\cdot\operatorname{length}(\gamma) $$ (the *ML inequality*; length $= \int_a^b\abs{\gamma'}$). The integral is invariant under increasing $\mathcal C^1$ reparametrization and changes sign under orientation reversal. **Proposition 16.5 (Primitives).** For [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ on $\Omega$, the following are equivalent: (i) $f$ has a *primitive* $F \in \mathcal H(\Omega)$ ($F' = f$); (ii) $\int_\gamma f\,\dd z = 0$ for every closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) $\gamma$ in $\Omega$. In that case $\int_\gamma f\,\dd z = F(\gamma(b)) - F(\gamma(a))$ for every [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected). **Proof.** (i)$\Rightarrow$: $\frac{\dd}{\dd t}F(\gamma(t)) = F'(\gamma(t))\gamma'(t)$ (chain rule, valid piecewise), so the integral telescopes to the endpoint difference; closed [paths](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) give $0$. (ii)$\Rightarrow$(i): fix $z_*$ in a [connected component](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components), define $F(z) = \int_{\gamma_z}f\,\dd z$ along any [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) from $z_*$ to $z$ (well defined: two [paths](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) differ by a closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected)); for $h$ small, taking the segment from $z$ to $z + h$, $$ \frac{F(z + h) - F(z)}{h} - f(z) = \frac1h\int_{[z, z+h]}\bigl(f(w) - f(z)\bigr)\dd w \longrightarrow 0 $$ by the ML inequality and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$ at $z$. ∎ **Definition 16.6 (Winding number).** For a closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) $\gamma$ and $z \notin \operatorname{im}\gamma$, the *index* is $$ \operatorname{Ind}_\gamma(z) = \frac1{2\iu\pi} \int_\gamma\frac{\dd w}{w - z} . $$ It is an *integer*: setting $\varphi(t) = \int_a^t\frac{\gamma'(s)}{\gamma(s) - z}\dd s$, the function $(\gamma(t) - z)\eu^{-\varphi(t)}$ has zero derivative (piecewise), hence is constant; at $t = b$, $\eu^{\varphi(b)} = \frac{\gamma(b) - z}{\gamma(a) - z} = 1$, so $\varphi(b) \in 2\iu\pi\Z$. As a function of $z$, the index is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\C\setminus\operatorname{im}\gamma$ (dominated convergence), hence constant on each [connected component](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components), and $0$ on the unbounded component (ML: the integral tends to $0$ as $z \to \infty$). For the circle $\gamma(t) = a + r\eu^{\iu t}$, $t \in \intcc0{2\pi}$: $\operatorname{Ind}_\gamma(z) = 1$ for $z \in D(a,r)$ (compute at $z = a$: $\frac1{2\iu\pi} \int_0^{2\pi}\frac{r\iu\eu^{\iu t}}{r\eu^{\iu t}}\dd t = 1$; constancy does the rest). ## 16.3 Cauchy’s theorem **Theorem 16.7 (Goursat).** Let $f \in \mathcal H(\Omega)$ and $T \subseteq \Omega$ a closed solid triangle. Then $\int_{\partial T}f\,\dd z = 0$ (boundary run once, any orientation). **Proof.** Let $I(T) = \int_{\partial T}f\,\dd z$. Joining the midpoints of the sides splits $T$ into four half-size triangles $T^{(1)}, \dots, T^{(4)}$, and the inner edges cancel in pairs: $I(T) = \sum_iI(T^{(i)})$. Choose $T_1$ among them with $\abs{I(T_1)} \geq \frac14\abs{I(T)}$, and iterate: a nested sequence $T \supseteq T_1 \supseteq T_2 \supseteq\cdots$ with $$ \abs{I(T_n)} \geq 4^{-n}\abs{I(T)}, \qquad \operatorname{diam}T_n = 2^{-n}\operatorname{diam}T, \quad \operatorname{length}(\partial T_n) = 2^{-n}\operatorname{length}(\partial T). $$ The intersection $\bigcap T_n$ is a single point $z_0$ (nested [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) with vanishing diameters, [Theorem 6.13](https://one-course.com/books/math/5/en/chapter/6-general-topology#thm-b3-topology-compactprops)(3)). Differentiability at $z_0$: given $\varepsilon$, for $n$ large, on $T_n$, $$ f(z) = f(z_0) + f'(z_0)(z - z_0) + R(z), \qquad \abs{R(z)} \leq \varepsilon\abs{z - z_0} \leq \varepsilon\operatorname{diam}T_n . $$ The affine part has a primitive: its integral on the closed $\partial T_n$ vanishes ([Proposition 16.5](#prop-b3-holomorphic-primitive)), leaving $$ \abs{I(T_n)} = \Bigl|\int_{\partial T_n}R\Bigr| \leq \varepsilon\operatorname{diam}(T_n)\, \operatorname{length}(\partial T_n) = \varepsilon\,4^{-n}\operatorname{diam}(T) \operatorname{length}(\partial T) . $$ Comparing with $\abs{I(T_n)} \geq 4^{-n}\abs{I(T)}$: $\abs{I(T)} \leq \varepsilon\cdot\text{const}$ for every $\varepsilon$: $I(T) = 0$. ∎ **Theorem 16.8 (Cauchy’s theorem, star-shaped version).** Let $\Omega$ be *star-shaped* about $c$ (every segment $[c, z]$, $z \in \Omega$, lies in $\Omega$) — e.g. convex. Every $f \in \mathcal H(\Omega)$ has a primitive on $\Omega$; consequently $\int_\gamma f\,\dd z = 0$ for *every* closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) $\gamma$ in $\Omega$. **Proof.** Define $F(z) = \int_{[c,z]}f\,\dd w$. For $z, z + h \in \Omega$ with $[z, z+h] \subseteq \Omega$ (true for small $h$), the triangle with vertices $c, z, z+h$ lies in $\Omega$ (star-shapedness: each of its points lies on a segment $[c, w]$ with $w \in [z, z+h] \subseteq \Omega$): Goursat gives $$ F(z + h) - F(z) = \int_{[z, z+h]}f\,\dd w, $$ and the difference-quotient computation of [Proposition 16.5](#prop-b3-holomorphic-primitive) yields $F' = f$. The vanishing of [closed-path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) integrals follows from the same proposition. ∎ **Theorem 16.9 (Cauchy’s integral formula).** Let $f \in \mathcal H(\Omega)$, $\bar D(a, r) \subseteq \Omega$, and $C_r$ the circle $\partial D(a,r)$ run once counterclockwise. Then for every $z \in D(a, r)$: $$ f(z) = \frac1{2\iu\pi}\int_{C_r}\frac{f(w)}{w - z}\,\dd w . $$ **Proof.** Fix $z$ and define on $\Omega$ $$ g(w) = \begin{cases} \dfrac{f(w) - f(z)}{w - z} & w \neq z,\\[4pt] f'(z) & w = z : \end{cases} $$ $g$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\Omega$ and [holomorphic](#def-b3-holomorphic-holo) off $z$. Goursat holds for $g$ on every triangle $T \subseteq \Omega'$, where $\Omega'$ is a disc slightly larger than $\bar D(a,r)$ inside $\Omega$: if $z \notin T$, directly; if $z \in T$, split $T$ into small triangles having $z$ as a vertex plus triangles avoiding $z$; on a triangle with vertex $z$ the ML bound gives $\abs{\int_{\partial T'}g} \leq \sup_{T'}\abs g\cdot\operatorname{length} \to 0$ as the triangle shrinks, and the remaining pieces vanish by Goursat — hence $\int_{\partial T}g = 0$ in all cases. The proof of [Theorem 16.8](#thm-b3-holomorphic-cauchy) used only this triangle property: $g$ has a primitive on the convex $\Omega'$, so $\int_{C_r}g = 0$, i.e. $$ \frac1{2\iu\pi}\int_{C_r}\frac{f(w)}{w - z}\dd w = f(z)\,\frac{1}{2\iu\pi}\int_{C_r}\frac{\dd w}{w - z} = f(z)\operatorname{Ind}_{C_r}(z) = f(z) . $$ ∎ ## 16.4 Analyticity and its cascade **Theorem 16.10 (Holomorphic === analytic).** Let $f \in \mathcal H(\Omega)$ and $D(a, R) \subseteq \Omega$. Then $$ f(z) = \sum_{n\geq0}c_n\,(z - a)^n \quad \text{on } D(a, R), \qquad c_n = \frac1{2\iu\pi}\int_{C_r}\frac{f(w)}{(w - a)^{n+1}}\,\dd w \ \ (0 < r < R), $$ the coefficients being independent of $r$. Consequently $f$ is infinitely $\C$-differentiable, $c_n = f^{(n)}(a)/n!$, and the *Cauchy estimates* hold: $$ \abs{c_n} \;\leq\; \frac{\sup_{\abs{w - a} = r}\abs{f(w)}}{r^{n}} . $$ **Proof.** For $\abs{z - a} < r$: expand the Cauchy kernel in the geometric series $$ \frac1{w - z} = \frac1{(w - a)\bigl(1 - \frac{z - a}{w - a}\bigr)} = \sum_{n\geq0}\frac{(z - a)^n}{(w - a)^{n+1}}, $$ normally convergent in $w$ on $C_r$ ($\abs{\frac{z-a}{w-a}} = \frac{\abs{z-a}}r < 1$): integrate term by term against $\frac{f(w)}{2\iu\pi}$ (uniform convergence justifies the interchange) and apply [Theorem 16.9](#thm-b3-holomorphic-formula). A power series is infinitely differentiable with $c_n = f^{(n)}(a)/n!$ (Year 2), which also shows the $c_n$ do not depend on $r$. The estimates: bound the coefficient integral by ML. ∎ **Example 16.11 (Singularities dictate radii).** Why does the innocent real function $\frac1{1 + x^2}$ have a Taylor series at $x = 3$ converging only for $\abs{x - 3} < \sqrt{10}$, when nothing goes wrong on the real line? Because the theorem above makes the radius of convergence at $a$ equal to the distance from $a$ to the nearest point where [holomorphy](#def-b3-holomorphic-holo) fails. Here $f(z) = \frac1{1 + z^2}$ is [holomorphic](#def-b3-holomorphic-holo) exactly on $\C\setminus\{\pm\iu\}$, so the expansion at $a = 3$ converges on the largest disc avoiding $\pm\iu$, of radius $\abs{3 - \iu} = \sqrt{10}$ — and cannot converge on a larger one, since the sum would extend $f$ [holomorphically](#def-b3-holomorphic-holo) to a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\pm\iu$, where $\abs f \to \infty$. The real theory sees the mysterious radius $\sqrt{10}$; the complex plane sees two poles. This is the practical rule: *to find a radius of convergence, locate the singularities* — e.g. the Taylor series of $\tan$ at $0$ has radius $\frac\pi2$ (nearest zeros of $\cos$), and the Bernoulli generating function $\frac z{\eu^z - 1}$ ([Problem 16.1](#pb-b3-holomorphic-1), Part VI) has radius $2\pi$ (nearest nonzero zeros of $\eu^z - 1$: $\pm2\iu\pi$). **Corollary 16.12 (Liouville; d’Alembert–Gauss).** A bounded entire function is constant. Consequently every nonconstant polynomial over $\C$ has a root: *$\C$ is [algebraically closed](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-closure)*. **Proof.** If $\abs f \leq M$ on $\C$: for every $a$ and $r$, $\abs{c_1(a)} = \abs{f'(a)} \leq M/r \to 0$: $f' \equiv 0$, and $f$ is constant (on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\C$: zero derivative implies locally constant — integrate along segments). If $P$ had no root, $1/P$ would be entire and bounded ($\abs{P(z)} \to \infty$ as $\abs z \to \infty$: the leading term dominates, so $\abs{1/P}$ is small outside a large disc and [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) disc): constant — absurd for nonconstant $P$. (The weekend problem gives a second, elementary proof and the [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) consequences.) ∎ **Theorem 16.13 (Zeros are isolated; identity theorem).** Let $\Omega$ be *[connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)* and $f \in \mathcal H(\Omega)$, $f \not\equiv 0$. Then every zero $a$ of $f$ has finite *order*: $f(z) = (z - a)^m\,g(z)$ with $g \in \mathcal H(\Omega)$, $g(a) \neq 0$, and the zeros of $f$ have no accumulation point in $\Omega$. Consequently, if two [holomorphic functions](#def-b3-holomorphic-holo) on $\Omega$ agree on a set with an accumulation point in $\Omega$, they agree everywhere. **Proof.** Let $Z$ be the set of points where all derivatives of $f$ vanish. $Z$ is closed (intersection of closed sets) and open: if all $c_n = 0$ at $a$, the power series expansion makes $f \equiv 0$ on a disc around $a$. [Connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected): $Z = \varnothing$ or $Z = \Omega$; the latter is excluded by $f \not\equiv 0$. So at a zero $a$, some coefficient is nonzero: let $m$ be minimal with $c_m \neq 0$; then $f(z) = (z - a)^m\sum_{n\geq m}c_n(z-a)^{n-m}$ on a disc, and the sum defines $g$ [holomorphic](#def-b3-holomorphic-holo) near $a$ with $g(a) = c_m \neq 0$; extend $g = f/(z-a)^m$ off $a$ ([holomorphic](#def-b3-holomorphic-holo) there). Since $g(a) \neq 0$ and $g$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), $f$ has no other zero in a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $a$: zeros are isolated, and a set of isolated points has no accumulation point in $\Omega$ (an accumulation point of zeros is a zero — [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) — and would not be isolated). Identity: apply to the difference, whose zero set has an accumulation point, forcing it into the $Z = \Omega$ branch. ∎ **Theorem 16.14 (Mean value and maximum modulus).** Let $f \in \mathcal H(\Omega)$. 1. (Mean value) For $\bar D(a, r) \subseteq \Omega$ : $f(a) = \frac1{2\pi}\int_0^{2\pi}f(a + r\eu^{\iu t})\,\dd t$ . 2. (Maximum principle) If $\Omega$ is [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) and $\abs f$ attains a local maximum at some point of $\Omega$ , then $f$ is constant. Consequently, for $\Omega$ bounded and $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\bar\Omega$ : $\sup_{\bar\Omega}\abs f = \sup_{\partial\Omega}\abs f$ . **Proof.** (1) is Cauchy’s formula at the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions): parametrize $C_r$. (2) Say $\abs f \leq \abs{f(a)}$ on $\bar D(a, \rho) \subseteq \Omega$. If $f(a) = 0$, $f \equiv 0$ near $a$. Otherwise, for $0 < r \leq \rho$, the mean value gives $$ \abs{f(a)} \leq \frac1{2\pi}\int_0^{2\pi}\abs{f(a + r\eu^{\iu t})}\,\dd t \leq \abs{f(a)} : $$ the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) nonnegative integrand $\abs{f(a)} - \abs{f(a + r\eu^{\iu t})}$ has zero mean, hence vanishes: $\abs f$ is *constant* $= \abs{f(a)} \ne 0$ on the disc. A [holomorphic function](#def-b3-holomorphic-holo) of constant nonzero modulus on a disc is constant: differentiating $P^2 + Q^2 = \text{const}$ gives $PP_x + QQ_x = 0$ and $PP_y + QQ_y = 0$; substituting the Cauchy–Riemann relations $P_y = -Q_x$, $Q_y = P_x$ into the second equation yields the linear system $$ P\,P_x + Q\,Q_x = 0, \qquad -P\,Q_x + Q\,P_x = 0, $$ whose determinant is $P^2 + Q^2 \neq 0$: $P_x = Q_x = 0$, so $f' = P_x + \iu Q_x = 0$ on the disc: $f$ constant there. The identity theorem spreads constancy to all of $\Omega$. The boundary form: $\abs f$ attains its sup on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\bar\Omega$; an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum makes $f$ constant, and the sup is attained on the boundary in every case. ∎ **Theorem 16.15 (Weierstrass convergence theorem).** If $f_n \in \mathcal H(\Omega)$ converge to $f$ uniformly on every [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) subset of $\Omega$, then $f \in \mathcal H(\Omega)$ and $f_n^{(k)} \to f^{(k)}$ uniformly on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), for every $k$. **Proof.** $f$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). For any closed triangle $T \subseteq \Omega$: $\int_{\partial T}f = \lim\int_{\partial T}f_n = 0$ (uniform convergence on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\partial T$; Goursat for $f_n$). By the argument of [Theorem 16.8](#thm-b3-holomorphic-cauchy), $f$ has local primitives (discs are convex; only the triangle property was used), i.e. $f = F'$ locally with $F$ [holomorphic](#def-b3-holomorphic-holo); $F$ is analytic ([Theorem 16.10](#thm-b3-holomorphic-analytic)), hence so is $f = F'$: [holomorphic](#def-b3-holomorphic-holo). (This is *Morera’s theorem*: [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with vanishing triangle integrals implies [holomorphic](#def-b3-holomorphic-holo).) Derivatives: for $\bar D(a, 2r) \subseteq \Omega$ and $z \in \bar D(a, r)$, Cauchy’s formula for derivatives (differentiate [Theorem 16.9](#thm-b3-holomorphic-formula) under the integral, or use the coefficient formula) gives $$ \abs{f_n'(z) - f'(z)} = \Bigl|\frac{1}{2\iu\pi}\int_{C_{2r}}\frac{f_n(w) - f(w)}{(w - z)^2}\,\dd w\Bigr| \leq \frac{2r\,\sup_{C_{2r}}\abs{f_n - f}}{r^2} \to 0 $$ uniformly on $\bar D(a,r)$; cover a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) by finitely many such discs, and iterate for higher $k$. ∎ ![Cauchy’s formula: the values of a holomorphic function inside a disc are a weighted average of its values on the bounding circle. Everything rigid about holomorphy — analyticity, Liouville, the maximum principle — unfolds from this one identity.](https://one-course.com/images/onecourse/chapters/math-5/b3-holomorphic/fig-9bf653c7459a.svg) *Cauchy’s formula: the values of a [holomorphic function](#def-b3-holomorphic-holo) *inside* a disc are a weighted average of its values on the bounding circle. Everything rigid about [holomorphy](#def-b3-holomorphic-holo) — analyticity, Liouville, the maximum principle — unfolds from this one identity.* **Method 16.16.** The daily toolkit. To prove a function [holomorphic](#def-b3-holomorphic-holo): exhibit it as a power series, a composition, a locally uniform limit ([Theorem 16.15](#thm-b3-holomorphic-weierstrassconv)), or an integral with [holomorphic](#def-b3-holomorphic-holo) parameter ([Exercise 16.7](#exo-b3-holomorphic-7) — differentiate under $\int$ or apply Morera–Fubini). To prove identities: prove them on a segment or subdomain and invoke the identity theorem. To bound: [Cauchy estimates](#thm-b3-holomorphic-analytic) on the largest circle available. To prove constancy/nonexistence: Liouville or the maximum principle. Always know *where* your function is [holomorphic](#def-b3-holomorphic-holo) and *which* discs fit in $\Omega$. ## 16.5 Exercises **Exercise 16.1 ★.** (a) At which points are $z \mapsto \bar z$, $\abs z^2$, $\operatorname{Re}z$ complex-differentiable? [Holomorphic](#def-b3-holomorphic-holo) on an [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology)? (b) Show that $P(x, y) = x^2 - y^2$ is the real part of a [holomorphic function](#def-b3-holomorphic-holo) on $\C$, found explicitly, and determine *all* of them. **Solution of Exercise 16.1.** (a) $\bar z$: $P = x$, $Q = -y$, so $P_x = 1 \neq -1 = Q_y$: nowhere $\C$-differentiable. $\abs z^2$: $P = x^2 + y^2$, $Q = 0$: Cauchy–Riemann demands $2x = 0 = 2y$: differentiable at $0$ only — and [holomorphic](#def-b3-holomorphic-holo) nowhere (no [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology)). $\operatorname{Re}z$: $P_x = 1 \neq 0 = Q_y$: nowhere. (b) $x^2 - y^2 = \operatorname{Re}(z^2)$: $f(z) = z^2$ works. All solutions: if $\operatorname{Re}f = \operatorname{Re}g$ with $f, g$ [holomorphic](#def-b3-holomorphic-holo) on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\C$, then $h = f - g$ has $\operatorname{Re}h = 0$; Cauchy–Riemann gives $h' = P_x + \iu Q_x = 0 - \iu P_y = 0$: $h$ is an imaginary constant. Answer: $f(z) = z^2 + \iu c$, $c \in \R$. **Exercise 16.2 ★.** Compute from the definitions: $\int_{C}z^n\,\dd z$ for all $n \in \Z$, $C$ the unit circle; $\int_\gamma\bar z\,\dd z$ along the segment $[0, 1+\iu]$ and along the two-segment [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) through $1$: conclude that $\bar z$ has no primitive on any [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of these [paths](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected). **Solution of Exercise 16.2.** On the unit circle $\gamma(t) = \eu^{\iu t}$: $$ \int_C z^n\,\dd z = \int_0^{2\pi}\eu^{\iu nt}\,\iu\eu^{\iu t}\dd t = \iu\int_0^{2\pi}\eu^{\iu(n+1)t}\dd t = \begin{cases} 2\iu\pi & n = -1,\\ 0 & n \neq -1.\end{cases} $$ For $\bar z$: along $[0, 1+\iu]$, $\gamma(t) = t(1 + \iu)$: $\int_0^1 t(1 - \iu)(1 + \iu)\dd t = \int_0^12t\,\dd t = 1$. Along $0 \to 1 \to 1 + \iu$: $\int_0^1t\dd t + \int_0^1(1 - \iu t)\,\iu\,\dd t = \frac12 + \iu + \frac12 = 1 + \iu$. Different values between the same endpoints: by [Proposition 16.5](#prop-b3-holomorphic-primitive), $\bar z$ has no primitive on any [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) containing both [paths](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected). **Exercise 16.3 ★★.** (a) Show that the principal logarithm $\log z = \ln\abs z + \iu\arg z$ ($\arg \in \intoo{-\pi}\pi$) is [holomorphic](#def-b3-holomorphic-holo) on $\C\setminus\intoc{-\infty}0$ with derivative $\frac1z$ *(primitive of $\frac1z$ on the star-shaped cut plane: [Theorem 16.8](#thm-b3-holomorphic-cauchy); fix the constant)*. (b) Show that no [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) logarithm exists on $\C^*$ *(its derivative-free obstruction: the [index](#def-b3-holomorphic-index) of the unit circle)*. (c) Expand $\log(1 + z)$ in power series on $D(0,1)$. **Solution of Exercise 16.3.** (a) The cut plane $\Omega = \C\setminus\intoc{-\infty}0$ is star-shaped about $1$, and $\frac1z \in \mathcal H(\Omega)$: [Theorem 16.8](#thm-b3-holomorphic-cauchy) provides a primitive $L$ with $L(1) = 0$. Then $\bigl(z\eu^{-L(z)}\bigr)' = \eu^{-L}(1 - z\cdot\frac1z) = 0$: $z = c\,\eu^{L(z)}$ with $c = 1$ (at $z = 1$). Writing $L = u + \iu v$: $\abs z = \eu^u$ and $z = \abs z\eu^{\iu v}$ with $v$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), $v(1) = 0$, $v \in \intoo{-\pi}\pi$ ($v$ is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) argument of $z$ on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\Omega$, so its image avoids the odd multiples of $\pi$ — no point of $\Omega$ lies on $\R_-$ — and, containing $v(1) = 0$, stays in $\intoo{-\pi}\pi$: $v$ is the principal argument): $L = \log$. (b) If $g$ were a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) logarithm on $\C^*$: $h(t) = g(\eu^{\iu t})$ satisfies $\eu^{h(t)} = \eu^{\iu t}$, so $h(t) - \iu t \in 2\iu\pi\Z$, and by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $h(t) = \iu t + 2\iu\pi k$ for a fixed integer $k$. Then $g(1) = h(0) = 2\iu\pi k$ and $g(1) = h(2\pi) = 2\iu\pi(k + 1)$: contradiction. (c) On $D(0,1)$: $\log(1 + z) = \sum_{n\geq1}\frac{(-1)^{n+1}}{n}z^n$ — both sides vanish at $0$ and have derivative $\frac1{1+z} = \sum(-1)^nz^n$ ([Example 16.3](#ex-b3-holomorphic-examples)); a primitive is unique up to a constant on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) disc. **Exercise 16.4 ★★.** (a) Let $f$ be entire with $\abs{f(z)} \leq C(1 + \abs z)^{n}$. Show that $f$ is a polynomial of degree $\leq n$ *([Cauchy estimates](#thm-b3-holomorphic-analytic) on large circles)*. (b) Let $f$ be entire with $\operatorname{Re}f$ bounded above. Show that $f$ is constant *(consider $\eu^{f}$)*. (c) Deduce “little Picard for affine maps”: an entire function omitting a half-plane is constant. **Solution of Exercise 16.4.** (a) Expand at $0$ (radius $\infty$): by the [Cauchy estimates](#thm-b3-holomorphic-analytic) on $C_r$, $\abs{c_k} \leq C(1 + r)^n/r^k \to 0$ as $r \to \infty$ for $k > n$: $f = \sum_{k\leq n}c_kz^k$. (b) If $\operatorname{Re}f \leq M$: $g = \eu^f$ is entire with $\abs g = \eu^{\operatorname{Re}f} \leq \eu^M$: constant by Liouville. Then $g' = f'g = 0$ with $g$ nonvanishing: $f' = 0$, and $f$ is constant. (c) If $f$ omits the half-plane $H$, an affine map $w \mapsto \alpha w + \beta$ sends $\C\setminus H$ into $\{\operatorname{Re} \leq M\}$; apply (b) to $\alpha f + \beta$. **Exercise 16.5 ★★.** (a) Let $f$ be [holomorphic](#def-b3-holomorphic-holo) on a [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\Omega \ni 0$ with $f(\frac1n) = \frac1{n^2}$ for all large $n$. Determine $f$. (b) Does some [holomorphic](#def-b3-holomorphic-holo) $f$ on $\C^*$ satisfy $f(\frac1n) = \frac{(-1)^n}{n}$ for all $n \geq 1$? Justify. (c) Exhibit two distinct [holomorphic functions](#def-b3-holomorphic-holo) on $D(0,1)\sqcup D(3,1)$ agreeing on $D(0,1)$: where does the identity theorem use [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)? **Solution of Exercise 16.5.** (a) $g(z) = f(z) - z^2$ vanishes at the points $\frac1n$, which accumulate at $0 \in \Omega$: by the identity theorem ($\Omega$ [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)), $g \equiv 0$: $f(z) = z^2$. (b) Yes: $f(z) = z\cos(\pi/z)$ is [holomorphic](#def-b3-holomorphic-holo) on $\C^*$ (composition) and $f(\frac1n) = \frac1n\cos(n\pi) = \frac{(-1)^n}n$. No contradiction with (a): the accumulation point $0$ of the interpolation nodes does *not* belong to $\C^*$, so the identity theorem is silent — two distinct functions ($z\cos(\pi/z)$ and, say, the one from another interpolation) may share these values. (c) $f \equiv 0$ everywhere, versus $g = 0$ on $D(0,1)$ and $g = 1$ on $D(3,1)$: [holomorphic](#def-b3-holomorphic-holo) on the disconnected union, equal on $D(0,1)$, different. The identity theorem’s open-closed argument needs [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) to propagate from one component to the other — and cannot. **Exercise 16.6 ★★.** Let $f$ be [holomorphic](#def-b3-holomorphic-holo) on the open unit disc $\mathbb D$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\bar{\mathbb D}$, with $\abs f \equiv 1$ on the boundary circle. (a) If $f$ has no zero in $\mathbb D$, show $f$ is constant *(apply the maximum principle to $f$ and to $1/f$)*. (b) Give an example with a zero where $f$ is not constant. **Solution of Exercise 16.6.** (a) By the maximum principle applied on the bounded domain: $\sup_{\mathbb D}\abs f = \sup_{\partial\mathbb D}\abs f = 1$. Since $f$ has no zeros, $1/f$ is [holomorphic](#def-b3-holomorphic-holo) on $\mathbb D$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior), with boundary modulus $1$: likewise $\abs{1/f} \leq 1$, i.e. $\abs f \geq 1$. So $\abs f \equiv 1$: the modulus attains an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum, and [Theorem 16.14](#thm-b3-holomorphic-maximum)(2) forces $f$ constant. (b) $f(z) = z$: boundary modulus $1$, zero at the origin, nonconstant — the zero is exactly what blocks the $1/f$ argument. **Exercise 16.7 ★★.** ([Holomorphy](#def-b3-holomorphic-holo) under the integral) Let $\mu$ be a finite [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) on a space $X$ and $g \colon X\times\Omega \to \C$ with: $g(x, \cdot) \in \mathcal H(\Omega)$ for each $x$, $g$ [measurable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable) in $x$, and $\abs g \leq h(x)$ with $h$ [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1), locally uniformly in $z$. Show $G(z) = \int_Xg(x, z)\dd\mu(x)$ is [holomorphic](#def-b3-holomorphic-holo) on $\Omega$. *(Morera: triangle integrals vanish by Fubini and Goursat; [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) by dominated convergence. Then apply to $\Gamma(z) = \int_0^\infty t^{z-1}\eu^{-t}\dd t$ on $\{\operatorname{Re}z > 0\}$.)* **Solution of Exercise 16.7.** [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $G$: dominated convergence with dominator $h$ (locally uniform bound). [Holomorphy](#def-b3-holomorphic-holo) by Morera (established inside [Theorem 16.15](#thm-b3-holomorphic-weierstrassconv)): for a closed triangle $T$ in a disc where $\abs g \leq h$, $$ \int_{\partial T}G(z)\,\dd z = \int_X\Bigl(\int_{\partial T}g(x, z)\,\dd z\Bigr)\dd\mu(x) = 0, $$ the interchange by Fubini ($\int_X\int_{\partial T}\abs g \leq \operatorname{length}(\partial T)\int h < \infty$) and the inner vanishing by Goursat. For $\Gamma$: on the strip $a \leq \operatorname{Re}z \leq b$ ($0 < a \leq b$), $\abs{t^{z-1}\eu^{-t}} = t^{\operatorname{Re}z-1}\eu^{-t} \leq (t^{a-1} + t^{b-1})\eu^{-t}$, [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) on $\intoo0{+\infty}$: $\Gamma$ is [holomorphic](#def-b3-holomorphic-holo) on $\{\operatorname{Re} z > 0\}$ (the [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) is only $\sigma$-finite, but the argument needs only the [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) dominator). By the identity theorem, the functional equation $\Gamma(z + 1) = z\Gamma(z)$, proved on $\intoo0{+\infty}$ ([Example 10.16](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma)), holds on the whole half-plane. **Exercise 16.8 ★★★.** (Gauss–Lucas) Let $P \in \C[X]$ be nonconstant. Show that every root of $P'$ lies in the convex hull of the roots of $P$. *(Write $\frac{P'}{P} = \sum_k\frac{m_k}{z - a_k}$ at a root $z$ of $P'$ that is not a root of $P$, take conjugates, and read a convex combination.)* Illustrate on $P = z^3 - 1$. **Solution of Exercise 16.8.** Write $P = c\prod_k(X - a_k)^{m_k}$ ([Problem 16.1](#pb-b3-holomorphic-1)). Let $P'(z) = 0$. If $P(z) = 0$, then $z$ is one of the $a_k$: in the hull. Otherwise, the logarithmic derivative gives $$ 0 = \frac{P'(z)}{P(z)} = \sum_k\frac{m_k}{z - a_k} = \sum_k m_k\,\frac{\bar z - \bar a_k}{\abs{z - a_k}^2} ; $$ conjugating, $\sum_kw_k(z - a_k) = 0$ with $w_k = m_k/\abs{z - a_k}^2 > 0$: $z = \sum_k\frac{w_k}{\sum w}\,a_k$, a convex combination of the roots. For $P = z^3 - 1$: roots the cube roots of unity, $P' = 3z^2$ with double root $0$ — the centroid of the equilateral triangle. **Exercise 16.9 ★★★.** Let $f$ be entire and doubly periodic: $f(z + 1) = f(z + \iu) = f(z)$ for all $z$. Show that $f$ is constant. *(Bound $f$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) fundamental square, then everywhere; Liouville.)* Moral: nonconstant elliptic functions must have poles — the theme of [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues). **Solution of Exercise 16.9.** The closed unit square $K = \{x + \iu y : 0 \leq x, y \leq 1\}$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): $M = \sup_K\abs f < \infty$. Every $z \in \C$ differs from a point of $K$ by an element of $\Z + \iu\Z$ (subtract integer parts), and $f$ is invariant under those translations (iterate the two relations): $\abs f \leq M$ on $\C$. Liouville: $f$ is constant. Hence any nonconstant doubly periodic meromorphic function — the elliptic functions of the classical theory — must have poles. **Exercise 16.10 ★★.** (a) Show that $P = \operatorname{Re}f$ of a [holomorphic](#def-b3-holomorphic-holo) $f$ satisfies the mean value property $P(a) = \frac1{2\pi}\int_0^{2\pi}P(a + r\eu^{\iu t})\dd t$ and is *harmonic*: $\partial^2_{xx}P + \partial^2_{yy}P = 0$ (differentiate Cauchy–Riemann; use [Theorem 16.10](#thm-b3-holomorphic-analytic) for the needed smoothness). (b) Deduce the maximum principle for real parts of [holomorphic functions](#def-b3-holomorphic-holo) on bounded domains. **Solution of Exercise 16.10.** (a) Take real parts in the mean value formula ([Theorem 16.14](#thm-b3-holomorphic-maximum)(1)). Smoothness: $f$ is analytic, so $P, Q \in \mathcal C^\infty$; differentiating Cauchy–Riemann: $P_{xx} = (Q_y)_x = (Q_x)_y = (-P_y)_y = -P_{yy}$ (Schwarz symmetry of second derivatives): $\Delta P = 0$. (b) If $\operatorname{Re}f$ attained an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum on a [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\Omega$: $g = \eu^f$ has $\abs g = \eu^{\operatorname{Re}f}$ attaining an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum, so $g$, hence $\operatorname{Re}f = \ln\abs g$, is constant ([Theorem 16.14](#thm-b3-holomorphic-maximum)(2)). On a bounded domain with [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) up to the boundary, $\sup_{\bar\Omega}\operatorname{Re}f = \sup_{\partial\Omega}\operatorname{Re}f$. **Exercise 16.11 ★★★.** (Schwarz reflection) Let $\Omega^+ = \{z : \abs z < 1,\ \operatorname{Im}z > 0\}$, $I = \intoo{-1}1$, and $f$ [holomorphic](#def-b3-holomorphic-holo) on $\Omega^+$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\Omega^+\cup I$, *real-valued on $I$*. Define $$ F(z) = \begin{cases} f(z) & z \in \Omega^+\cup I,\\ \overline{f(\bar z)} & \bar z \in \Omega^+ . \end{cases} $$ (a) Show that $F$ is well defined and [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\Omega = \Omega^+\cup I\cup\Omega^-$, and [holomorphic](#def-b3-holomorphic-holo) on $\Omega^\pm$ *(for $\Omega^-$: check Cauchy–Riemann for $\overline{f(\bar z)}$, or expand $f$ in local power series and conjugate coefficients)*. (b) Show that $F$ is [holomorphic](#def-b3-holomorphic-holo) on all of $\Omega$ by Morera’s criterion: $\int_{\partial T}F = 0$ for every triangle $T \subseteq \Omega$ *(split triangles at $I$ and push their horizontal sides off the axis by $\varepsilon$, using uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity))*. (c) Deduce: a [holomorphic function](#def-b3-holomorphic-holo) on the disc, real on a diameter, satisfies $f(\bar z) = \overline{f(z)}$; and a nonconstant [holomorphic function](#def-b3-holomorphic-holo) cannot be real-valued on any nonempty open subset of its ([connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)) domain. **Solution of Exercise 16.11.** (a) The two formulas agree on $I$ ($z = \bar z$ and $f$ real there: $\overline{f(\bar z)} = \overline{f(z)} = f(z)$), and $z \mapsto \overline{f(\bar z)}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\Omega^-\cup I$ as a composition of [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $F$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\Omega$. [Holomorphy](#def-b3-holomorphic-holo) on $\Omega^-$: near $z_0 \in \Omega^-$, expand $f(w) = \sum c_n(w - \bar z_0)^n$ near $\bar z_0 \in \Omega^+$; then $$ \overline{f(\bar z)} = \sum_n\bar c_n\,(z - z_0)^n, $$ a convergent power series: [holomorphic](#def-b3-holomorphic-holo). (b) Triangles avoiding $I$ are handled by Goursat in $\Omega^\pm$. For a triangle meeting $I$, cut it by the real axis into at most three triangles/quadrilaterals, each with one side on $I$; for such a piece $P$ contained in, say, $\overline{\Omega^+}$, the [contour integral](#def-b3-holomorphic-contour) is the limit as $\varepsilon \to 0^+$ of the integral over $P + \iu \varepsilon$-type translates (uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $F$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) piece makes the boundary integrals converge, the side on $I$ being approached from above), and each translate lies in $\Omega^+$ where Goursat gives $0$. Summing the pieces: $\int_{\partial T}F = 0$. Morera (the criterion inside [Theorem 16.15](#thm-b3-holomorphic-weierstrassconv)): $F$ is [holomorphic](#def-b3-holomorphic-holo) on $\Omega$. (c) On the disc, $G(z) = \overline{f(\bar z)}$ is [holomorphic](#def-b3-holomorphic-holo) by (a)’s computation and agrees with $f$ on the diameter, a set with accumulation points: $G = f$ everywhere (identity theorem). If $f$ were real on a nonempty [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $U$: on $U$ both partials of $Q = \operatorname{Im}f$ vanish, and Cauchy–Riemann transfers this to $P = \operatorname{Re}f$ ($P_x = Q_y = 0$, $P_y = -Q_x = 0$), so $f' = P_x + \iu Q_x = 0$ on $U$: $f$ is constant on $U$, hence everywhere by the identity theorem ($\Omega$ [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)). **Exercise 16.12 ★★.** (The complex Pythagorean equation) Find all pairs of *entire* functions with $f^2 + g^2 = 1$. (a) Show that $h = f + \iu g$ is entire and zero-free, and that every zero-free entire function is $\eu^{\varphi}$ for some entire $\varphi$ *($h'/h$ is entire, hence has a primitive on the star-shaped $\C$; adjust the constant and show $h\eu^{-\varphi}$ is constant)*. (b) Conclude $f = \cos\varphi$, $g = \sin\varphi$ with $\varphi$ entire, and check the converse. What are the entire solutions of $f^2 + g^2 = 0$? **Solution of Exercise 16.12.** (a) $1 = f^2 + g^2 = (f + \iu g)(f - \iu g)$, so $h = f + \iu g$ never vanishes (its cofactor would have to blow up). For zero-free entire $h$: $h'/h$ is entire, and $\C$ is star-shaped, so it has a primitive $\varphi_0$ ([Theorem 16.8](#thm-b3-holomorphic-cauchy)); then $\bigl(h\eu^{-\varphi_0}\bigr)' = \eu^{-\varphi_0}(h' - h\varphi_0') = 0$: $h = c\,\eu^{\varphi_0}$ with $c \neq 0$, and absorbing a constant $\log c$ into $\varphi = \varphi_0 + \log c$ (any complex logarithm of $c$): $h = \eu^{\varphi}$. (b) With $h = \eu^{\varphi}$ and $h^{-1} = f - \iu g = \eu^{-\varphi}$: $$ f = \frac{\eu^{\varphi} + \eu^{-\varphi}}2, \qquad g = \frac{\eu^{\varphi} - \eu^{-\varphi}}{2\iu} . $$ Writing $\varphi = \iu\psi$ with $\psi = -\iu\varphi$ entire, these read $f = \cos\psi$, $g = \sin\psi$: the entire solutions are exactly the pairs $(\cos\psi, \sin\psi)$ with $\psi$ entire, and the converse is the identity $\cos^2 + \sin^2 = 1$. For $f^2 + g^2 = 0$: $(f + \iu g)(f - \iu g) = 0$ in the integral domain $\mathcal H(\C)$ ( $\C$ [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected): zero divisors would violate the identity theorem): $g = \pm\iu f$ with $f$ arbitrary entire. ## 16.6 Problem: the fundamental theorem of algebra, twice **Problem 16.1.** Weekend problem — $\C$ is algebraically closed: d’Alembert’s proof, Liouville’s proof, and the harvest Let $P(z) = z^n + a_{n-1}z^{n-1} + \dots + a_0$, $n \geq 1$. We prove twice that $P$ has a root, then collect what algebra has been waiting for since [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois). **Part I — Coercivity and the minimum.** 1. Show that $\abs{P(z)} \to +\infty$ as $\abs z \to \infty$ : precisely, $\abs{P(z)} \geq \frac12\abs z^n$ for $\abs z \geq R_0$ suitable. 2. Deduce that $\abs P$ attains a global minimum on $\C$ : there is $z_0$ with $\abs{P(z_0)} = \inf_\C\abs P$ *(compactness of a large closed disc, [Corollary 6.17](https://one-course.com/books/math/5/en/chapter/6-general-topology#cor-b3-topology-heineborel))* . **Part II — d’Alembert’s descent.** Suppose, for contradiction, $P(z_0) \neq 0$. 3. Expand $Q(h) = P(z_0 + h)/P(z_0)$ as a polynomial in $h$ : $Q(h) = 1 + c_kh^k + h^{k+1}S(h)$ with $c_k \neq 0$ , $k \geq 1$ , $S$ a polynomial. 4. Choose the direction of descent: for small $t > 0$, set $h = t\,\omega$ where $\omega^k = -1/c_k$ (why does such $\omega$ exist? — *prove* the existence of $k$-th roots of any complex number via polar form, independently of the theorem being proved). Show $$\abs{Q(t\omega)} \leq 1 - t^k + C\,t^{k+1}$$ for $t$ small, with an explicit constant $C$. 5. Conclude $\abs{Q(t\omega)} < 1$ for small $t$ — contradicting the minimality of $\abs{P(z_0)}$ . Hence $P(z_0) = 0$ : *every nonconstant complex polynomial has a root* (d’Alembert–Argand). **Part III — Liouville’s one-liner, in full.** 6. Write out carefully the proof of [Corollary 16.12](#cor-b3-holomorphic-liouville) : if $P$ has no root, verify that $1/P$ is entire, bounded (quantify, using question 1), hence constant, and conclude. Compare the two proofs: which ingredients does each use? (Compactness appears in both — where?) **Part IV — The harvest.** 7. Show that every $P \in \C[X]$ of degree $n$ splits: $P = c\prod_{i}(X - \alpha_i)^{m_i}$ with $\sum m_i = n$ *(induction, Euclidean division by $(X - \alpha)$)* . 8. Show that the [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) polynomials of $\R[X]$ are the linear ones and the quadratics with negative discriminant *(pair conjugate roots)* ; deduce that every real polynomial of odd degree has a real root, and give a second, order-theoretic proof of that last fact (intermediate value theorem) — checking they agree on $X^3 - X - 1$ . 9. Deduce the debts this book can now repay: (i) every endomorphism of a nonzero finite-dimensional $\C$ -vector space has an eigenvalue, so every complex matrix has a [Jordan form](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#thm-b3-modules-jordan) ( [Theorem 3.18](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#thm-b3-modules-jordan) ); (ii) the field $\bar\Q$ of [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) numbers used in [Remark 4.10](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#rem-b3-galois-closureexamples) is indeed an [algebraic closure](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-closure) of $\Q$ . 10. (Finale) Pinpoint where each proof would break over a field like $\Q(\iu)$ : which steps use the existence of $k$ -th roots (question 4), and which use compactness or [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) (questions 2 and 6)? Conclude in five lines: the theorem is genuinely *analytic* — every proof somewhere invokes the [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) or [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) of $\R$ — even though its statement is purely [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) . **Part V — The rigidity ladder of entire functions.** Liouville is the first rung of a ladder; we climb it. 11. ([Cauchy estimates](#thm-b3-holomorphic-analytic)) From the Cauchy formula on the circle of radius $r$ around $a$, prove $$\bigl|f^{(n)}(a)\bigr| \leq \frac{n!\,\sup_{\abs{z-a}=r}\abs f}{r^n} ,$$ and recover Liouville as the case $n = 1$, $r \to \infty$. 12. (Polynomial growth) Show that an entire $f$ with $\abs{f(z)} \leq A + B\abs z^m$ for all $z$ is a polynomial of degree $\leq m$ *(kill the Taylor coefficients beyond $m$ with question 11)* . 13. (Bounded real part) Show that an entire $f$ with $\operatorname{Re}f$ bounded above is constant *(apply Liouville to $\eu^{f}$)* . 14. (Double periodicity) Let $f$ be entire with $f(z + 1) = f(z)$ and $f(z + \iu) = f(z)$ for all $z$ . Show $f$ is constant. Conclude: a nonconstant “elliptic” function must have singularities — the historical reason poles enter complex analysis. 15. (Dense range) Show that the range of a nonconstant entire function is dense in $\C$ : if $f(\C)$ misses a disc $D(a, r)$ , then $\frac1{f - a}$ is entire and bounded. (Picard proved the range misses at most *one point* ; [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) is the level our tools reach.) 16. (Proper $\Rightarrow$ polynomial) Suppose $f$ is entire and $\abs{f(z)} \to \infty$ as $\abs z \to \infty$ . Show: the zeros of $f$ are finite in number ( $z_1, \dots, z_p$ , with multiplicities $m_i$ ); the quotient $g = f/\prod(z - z_i)^{m_i}$ is entire and zero-free; $1/g$ has polynomial growth, hence (question 12) is a polynomial, necessarily constant (zero-free); conclude that $f$ is a polynomial. So among entire functions, polynomials are exactly the proper ones — $\eu^z$ fails properness along $\R_-$ . **Part VI — Harmonic shadows and a mean value of Gauss.** 17. Let $f = u + \iu v$ be [holomorphic](#def-b3-holomorphic-holo) on an [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology). Verify that $u = \operatorname{Re}f$ satisfies the mean value property $$u(a) = \frac1{2\pi}\int_0^{2\pi} u\bigl(a + r\eu^{\iu\theta}\bigr)\,\dd\theta$$ (real part of the Cauchy formula), and deduce the maximum principle for $u$ on a bounded domain, with the same [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) proof as for $\abs f$. 18. (Gauss’s mean value) For $a \in \C$ and $r > 0$ with $\abs a \neq r$, prove $$\frac1{2\pi}\int_0^{2\pi} \log\bigl|a - r\eu^{\iu\theta}\bigr|\,\dd\theta = \log\max\bigl(\abs a, r\bigr)$$ *(if $\abs a > r$: $z \mapsto \log\abs{a - z}$ is the real part of a [holomorphic](#def-b3-holomorphic-holo) logarithm on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of the closed disc — why does one exist? — so question 17 applies; if $\abs a < r$: factor $\abs{a - r\eu^{\iu\theta}} = r\,\abs{1 - \frac ar\eu^{-\iu\theta}}$ and reuse the first case)*. 19. (Mahler [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure)) For $P = c\prod_{i=1}^n(X - \alpha_i) \in \C[X]$, deduce *Jensen’s formula for polynomials*: $$\frac1{2\pi}\int_0^{2\pi}\log\bigl|P(\eu^{\iu\theta}) \bigr|\,\dd\theta = \log\Bigl(\abs c\prod_{i}\max(1, \abs{\alpha_i})\Bigr) :$$ the geometric mean of $\abs P$ on the unit circle reads off the roots outside the disc. Verify on $P = X^2 - X$ and on $P = 2X - 1$. 20. (Bernoulli numbers) Define the coefficients $B_n$ by $\frac{z}{\eu^z - 1} = \sum_{n\geq0}\frac{B_n}{n!} z^n$ near $0$ (why is the left side analytic at $0$ ?). Derive the recurrence $\sum_{k=0}^{n}\binom{n+1}kB_k = 0$ ( $n \geq 1$ ) from $(\eu^z - 1)\cdot\frac z{\eu^z-1} = z$ , compute $B_0, \dots, B_6$ , and show $B_{2k+1} = 0$ for $k \geq 1$ *(the function $\frac z{\eu^z-1} + \frac z2$ is even)* . These numbers will price every $\zeta(2k)$ in [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues) . 21. (Reality) Show that an entire function taking real values on $\R$ satisfies $f(\bar z) = \overline{f(z)}$ everywhere *(compare Taylor coefficients at $0$, or apply the identity theorem to $z \mapsto \overline{f(\bar z)}$)* ; deduce again that nonreal roots of real polynomials come in conjugate pairs (question 8’s pairing, reproved analytically). 22. (The moral) Assemble the rigidity ladder: bounded $\Rightarrow$ constant; polynomially bounded $\Rightarrow$ polynomial; proper $\Rightarrow$ polynomial; missing a disc $\Rightarrow$ constant; doubly periodic $\Rightarrow$ constant. Contrast in a short paragraph with real $\mathcal C^\infty$ functions ( [bump functions](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#def-b3-lp-mollifier) , [Theorem 12.9](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-regularization) ): why does [holomorphy](#def-b3-holomorphic-holo) , a purely local condition, impose global law and order? **Part VII — Last harvest.** 23. (Landau’s inequality) For $P = \sum_{k=0}^na_kX^k$, prove the mean value $\frac1{2\pi}\int_0^{2\pi} \abs{P(\eu^{\iu\theta})}^2\dd\theta = \sum_k\abs{a_k}^2$ (orthogonality of the $\eu^{\iu k\theta}$), then, using the pointwise bound $\log t \leq t - 1$ to compare the means of $\log\abs P^2$ and $\abs P^2$, deduce from question 19 that $$\abs c\prod_{i}\max\bigl(1, \abs{\alpha_i}\bigr) \;\leq\; \Bigl(\sum_{k=0}^{n}\abs{a_k}^2\Bigr)^{1/2} :$$ the product of the roots outside the unit disc is controlled by the $\ell^2$ size of the coefficients (handle roots *on* the circle by applying the inequality to $P(rX)$ and letting $r \to 1$). Check it on $X^2 - X$. 24. (Bernoulli numbers grow factorially) Show that the radius of convergence of $\sum\frac{B_n}{n!}z^n$ is *exactly* $2\pi$: at least $2\pi$ because $z/(\eu^z - 1)$ extends [holomorphically](#def-b3-holomorphic-holo) to $D(0, 2\pi)$, at most $2\pi$ because the sum would otherwise stay bounded near $2\pi\iu$, where $\abs{z/(\eu^z-1)} \to \infty$. Deduce $$\limsup_{k\to\infty} \Bigl(\frac{\abs{B_{2k}}}{(2k)!}\Bigr)^{1/2k} = \frac1{2\pi} :$$ Bernoulli numbers grow factorially. Admitting $B_{12} = -\frac{691}{2730}$ (the recurrence of question 20, run further), compare $\abs{B_{12}} \approx 0.25311$ with the sharper prediction $2\,(2k)!/(2\pi)^{2k} \approx 0.25305$ at $k = 6$ — four matching digits of an asymptotic law that [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues) will prove exactly, via $\zeta(2k)$. 25. (Roots move [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ) Let $(P_j)$ be monic of degree $n$ with coefficients converging to those of (monic) $P$ . Prove the *Cauchy bound* : every root of a monic $Q = X^n + \sum_{k m$: $\abs{c_k} \leq (A + Br^m)/r^k \to 0$ as $r \to \infty$, so $c_k = 0$: $f = \sum_{k\leq m}c_kz^k$ is a polynomial of degree at most $m$. **13.** $g = \eu^f$ is entire with $\abs g = \eu^{\operatorname{Re}f} \leq \eu^M$: constant by question 11. Then $0 = g' = f'g$ with $g$ zero-free: $f' = 0$ and $f$ is constant. **14.** Let $M = \sup_K\abs f$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) closed unit square $K$. Every $z$ differs from a point of $K$ by an element of $\Z + \iu\Z$ (subtract integer parts), and iterating the two periodicity relations leaves $f$ unchanged: $\abs f \leq M$ on all of $\C$, and question 11 makes $f$ constant. So a nonconstant function invariant under the lattice cannot be entire: the elliptic functions of the classical theory must carry poles — the historical gateway to [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues). **15.** If $f(\C)$ misses the disc $D(a, r)$, then $\abs{f(z) - a} \geq r$ for all $z$, so $g = 1/(f - a)$ is entire with $\abs g \leq 1/r$: constant by Liouville, hence $f$ constant. Contrapositive: the range of a nonconstant entire function meets every disc — it is dense in $\C$. **16.** Choose $R$ with $\abs f \geq 1$ outside $D(0, R)$. The zeros of $f$ lie in the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\bar D(0, R)$; were they infinite, they would accumulate there, and [Theorem 16.13](#thm-b3-holomorphic-identity) would force $f \equiv 0$ — impossible. Call them $z_1, \dots, z_p$, with multiplicities $m_1, \dots, m_p$, put $M = \sum m_i$ and $\Pi(z) = \prod_i(z - z_i)^{m_i}$. Factoring each zero out of the power series, $g = f/\Pi$ is entire and zero-free. For $\abs z \geq \max(R, 2\max_i\abs{z_i})$: $\abs{z - z_i} \leq 2\abs z$ and $\abs f \geq 1$, so $\abs{1/g} = \abs\Pi/\abs f \leq 2^M\abs z^M$; on the remaining [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) disc $1/g$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), hence bounded: $\abs{1/g} \leq A + B\abs z^M$ everywhere. By question 12, $1/g$ is a polynomial; it is zero-free, so by question 7 it is a nonzero constant $c$: $f = \frac1c\Pi$ is a polynomial. Conversely, question 1 makes every nonconstant polynomial proper. And $\eu^z$ is honestly excluded: along $\R_-$, $\abs{\eu^z} = \eu^x \to 0$ while $\abs z \to \infty$. **17.** Parametrize [Theorem 16.9](#thm-b3-holomorphic-formula) at the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions): with $w = a + r\eu^{\iu\theta}$, $\dd w = \iu r\eu^{\iu\theta}\dd\theta$, $$ f(a) = \frac1{2\iu\pi}\int_{C_r}\frac{f(w)}{w - a}\,\dd w = \frac1{2\pi}\int_0^{2\pi} f\bigl(a + r\eu^{\iu\theta}\bigr)\,\dd\theta ; $$ taking real parts gives the mean value property of $u$. If $u$ attains a maximum at an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) point of the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\Omega$: $\abs{\eu^f} = \eu^u$ attains an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum, so $\eu^f$ is constant by [Theorem 16.14](#thm-b3-holomorphic-maximum)(2), and $u = \log\abs{\eu^f}$ is constant. On a bounded domain with [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) up to the boundary, $\sup_{\bar\Omega}u = \sup_{\partial\Omega}u$, exactly as for $\abs f$. **18.** *Case $\abs a > r$.* Pick $R$ with $r < R < \abs a$: on the convex disc $D(0, R)$ the function $a - z$ is [holomorphic](#def-b3-holomorphic-holo) and zero-free, and $z \mapsto -1/(a - z)$ has a primitive $L$ there ([Theorem 16.8](#thm-b3-holomorphic-cauchy)); after adjusting the constant, $\bigl(\eu^{-L}(a - z)\bigr)' = \eu^{-L}\bigl(-L'\,(a - z) - 1\bigr) = 0$ gives $\eu^L = a - z$: a [holomorphic](#def-b3-holomorphic-holo) logarithm exists, and $\log\abs{a - z} = \operatorname{Re}L(z)$. Question 17’s mean value property at $0$, radius $r$: $$ \frac1{2\pi}\int_0^{2\pi} \log\bigl|a - r\eu^{\iu\theta}\bigr|\,\dd\theta = \operatorname{Re}L(0) = \log\abs a . $$ *Case $\abs a < r$.* From $a - r\eu^{\iu\theta} = -r\eu^{\iu\theta}\bigl(1 - \frac ar\eu^{-\iu\theta}\bigr)$, the mean equals $\log r$ plus the mean of $\log\abs{1 - \frac ar\eu^{-\iu\theta}}$. The substitution $\theta \mapsto 2\pi - \theta$, then — writing $\frac ar = \rho\eu^{\iu\varphi}$, $\rho < 1$, the case $a = 0$ being trivial — the shift $\theta \mapsto \theta - \varphi$ (both preserve means over a period) turn this into the mean of $\log\abs{1 - \rho\eu^{\iu\theta}}$: the first case with $(a, r) = (1, \rho)$, which gives $\log 1 = 0$. Total: $\log r = \log\max(\abs a, r)$ in both cases. **19.** $\log\abs{P(\eu^{\iu\theta})} = \log\abs c + \sum_i\log\abs{\alpha_i - \eu^{\iu\theta}}$, each root repeated per its multiplicity; averaging in $\theta$ and applying question 18 with $r = 1$ to each root off the unit circle yields $\log\bigl(\abs c\prod_i\max(1, \abs{\alpha_i})\bigr)$. Checks. For $P = 2X - 1 = 2(X - \frac12)$ the formula predicts $\log 2$; directly, $\abs{2\eu^{\iu\theta} - 1} = 2\abs{\frac12 - \eu^{\iu\theta}}$ and the mean of $\log\abs{\frac12 - \eu^{\iu\theta}}$ is $\log\max(\frac12, 1) = 0$: mean $\log 2$. For $P = X^2 - X = X(X - 1)$ the root $1$ sits *on* the circle; the formula predicts $0$. Directly, the mean of $\log\abs{\eu^{\iu\theta}}$ is $0$, and with $\abs{\eu^{\iu\theta} - 1} = 2\abs{\sin\frac\theta2}$: $$ \frac1{2\pi}\int_0^{2\pi} \log\Bigl(2\sin\frac\theta2\Bigr)\dd\theta = \frac1\pi\int_0^\pi\log(2\sin u)\,\dd u = \log 2 + \frac J\pi, \qquad J = \int_0^\pi\log\sin u\,\dd u . $$ The substitution $u = 2v$ and $\sin 2v = 2\sin v\cos v$ give $J = \pi\log2 + 2\int_0^{\pi/2}\log\sin + 2\int_0^{\pi/2}\log\cos = \pi\log 2 + 2J$ (each half equals $J/2$ by the symmetries of $\sin$), so $J = -\pi\log 2$ (the improper integrals converge, $\log\sin$ being [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) at the endpoints): the mean is $\log2 - \log2 = 0$. The formula survives roots on the circle. **20.** $\frac{\eu^z - 1}z = \sum_{j\geq0}\frac{z^j}{(j+1)!}$ is entire and equals $1$ at $0$: its reciprocal is [holomorphic](#def-b3-holomorphic-holo) near $0$ (on $\abs z < 2\pi$ in fact, the nearest other zeros of $\eu^z - 1$ being $\pm2\iu\pi$), so $\frac z{\eu^z-1}$ is analytic at $0$. Multiplying the two series and reading the coefficient of $z^n$, $n \geq 1$, in $(\frac{\eu^z-1}z)\cdot(\frac z{\eu^z-1}) = 1$: $$ \sum_{k=0}^{n}\frac{B_k}{k!\,(n+1-k)!} = 0 \quad\Longleftrightarrow\quad \sum_{k=0}^{n}\binom{n+1}{k}B_k = 0 . $$ Successively: $B_0 = 1$, $B_1 = -\frac12$, $B_2 = \frac16$, $B_3 = 0$, $B_4 = -\frac1{30}$, $B_5 = 0$, $B_6 = \frac1{42}$. Evenness: with $F(z) = \frac z{\eu^z-1} + \frac z2$, $$ F(-z) = \frac{-z}{\eu^{-z} - 1} - \frac z2 = \frac{z\,\eu^z}{\eu^z - 1} - \frac z2 = z + \frac{z}{\eu^z - 1} - \frac z2 = F(z) : $$ $F$ is even, so $B_{2k+1} = 0$ for $k \geq 1$ (the lone odd coefficient $B_1$ was absorbed by $+\frac z2$). Forward pointer: $\cot w = \iu + \frac{2\iu}{\eu^{2\iu w}-1}$ gives $w\cot w = 1 + \sum_{k\geq1}\frac{B_{2k}}{(2k)!}(2\iu w)^{2k}$, so the Laurent coefficients of the cotangent — hence, by [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues), every $\zeta(2k)$ — are priced by Bernoulli numbers: $$ \zeta(2k) = (-1)^{k+1}\,\frac{(2\pi)^{2k}B_{2k}}{2\,(2k)!} . $$ **21.** Write $f = \sum_nc_nz^n$ (radius $\infty$); then $g(z) = \overline{f(\bar z)} = \sum_n\bar c_nz^n$ is entire. On $\R$: $g(x) = \overline{f(x)} = f(x)$, so $g$ and $f$ agree on a set with accumulation points in the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\C$: [Theorem 16.13](#thm-b3-holomorphic-identity) gives $g \equiv f$, that is $f(\bar z) = \overline{f(z)}$ (equivalently: all $c_n$ are real). For a real polynomial $P$: $P(\bar\alpha) = \overline{P(\alpha)} = 0$, and the same identity applied to the real derivatives $P', P'', \dots$ preserves multiplicities: nonreal roots pair up — question 8’s pairing, reproved analytically. **22.** The ladder, assembled: bounded $\Rightarrow$ constant (11); dominated by $A + B\abs z^m$ $\Rightarrow$ polynomial (12); real part bounded above $\Rightarrow$ constant (13); doubly periodic $\Rightarrow$ constant (14); range missing a disc $\Rightarrow$ constant (15); proper $\Rightarrow$ polynomial (16). Every rung is the Cauchy formula: the value at a point is a circle average, so all Taylor coefficients are priced by the size of $f$ on large circles, and a growth cap annihilates coefficients wholesale. Nothing of the sort constrains real $\mathcal C^\infty$ functions: a [bump function](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#def-b3-lp-mollifier) ([Theorem 12.9](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-regularization)) is bounded, compactly supported and wildly nonconstant, and its derivatives at any point outside the support all vanish without the function vanishing anywhere near. Smoothness couples the derivatives at distinct points not at all; [holomorphy](#def-b3-holomorphic-holo) chains every derivative to a single integral over a distant circle. A local condition with a global informer — that is why entire functions obey law and order. **23.** Expanding $\abs{P(\eu^{\iu\theta})}^2 = \sum_{k,l}a_k\bar a_l \eu^{\iu(k-l)\theta}$ and averaging kills every term $k \neq l$: the mean is $\sum_k\abs{a_k}^2 =: N$. Suppose first that $P$ has no zero on the unit circle, so $\theta \mapsto \log\abs{P(\eu^{\iu\theta})}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). The bound $\log t \leq t - 1$ applied to $t = \abs P^2/N$ gives, after averaging, $$ \frac1{2\pi}\int_0^{2\pi}\log\abs{P(\eu^{\iu\theta})}^2 \dd\theta - \log N \;\leq\; \frac1N\cdot N - 1 = 0, $$ so the geometric mean of $\abs P$ is at most $\sqrt N$; question 19 identifies that geometric mean as $\abs c\prod_i\max(1, \abs{\alpha_i})$: Landau’s inequality. Roots on the circle: pick $r > 1$ distinct from every $\abs{\alpha_i}$; the polynomial $P(rX)$, with roots $\alpha_i/r$ off the unit circle and coefficients $a_kr^k$, satisfies the inequality; both sides are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) in $r$, and letting $r \to 1^+$ gives the general case. On $X^2 - X$: roots $0$ and $1$, so the left side is $1$, and the right side is $\sqrt{1 + 1} = \sqrt2$: true, with room. **24.** Write $\eu^z - 1 = z\,g(z)$ with $g(z) = \sum_{k\geq0}\frac{z^k}{(k+1)!}$ entire, $g(0) = 1$. Since $\eu^z = 1$ exactly on $2\pi\iu\Z$, $g$ has no zero in $D(0, 2\pi)$ (for $0 < \abs z < 2\pi$ because $\eu^z - 1 \neq 0$, at $0$ by $g(0) = 1$), so $h = 1/g$ is [holomorphic](#def-b3-holomorphic-holo) on $D(0,2\pi)$ and its Taylor series at $0$ — by definition $\sum\frac{B_n}{n!}z^n$ — converges on the whole disc: $\rho \geq 2\pi$. If $\rho > 2\pi$, the sum $S$ would be [holomorphic](#def-b3-holomorphic-holo) on $D(0,\rho)$, and it agrees with $z \mapsto z/(\eu^z - 1)$ on $0 < \abs z < 2\pi$; both are [holomorphic](#def-b3-holomorphic-holo) 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) $D(0,\rho) \setminus 2\pi\iu\Z$, so by the identity theorem they agree there. But as $z \to 2\pi\iu$, $\abs{z/(\eu^z - 1)} \to \infty$ (numerator $\to 2\pi$, denominator $\to 0$) while $S$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $2\pi\iu$: contradiction. Hence $\rho = 2\pi$ exactly, and Hadamard’s formula gives $\limsup_n\abs{B_n/n!}^{1/n} = \frac1{2\pi}$; the odd coefficients being zero from $n = 3$ on, the $\limsup$ is carried by the even indices, which is the stated formula with $n = 2k$. Numerically at $k = 6$: $(2\pi)^{12} \approx 3.7858\cdot10^9$ and $2\cdot12! = 958\,003\,200$, so $2\,(2k)!/(2\pi)^{2k} \approx 0.25305$, against $\abs{B_{12}} = \frac{691}{2730} \approx 0.25311$. The ratio, $1.00025$, is exactly $\zeta(12)$ to the digits shown: the residue-calculus formula $\zeta(2k) = (-1)^{k+1}\frac{(2\pi)^{2k}B_{2k}}{2\,(2k)!}$ of [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues) explains both the factor $2$ and the tiny excess. **25.** Cauchy bound: if $\abs z > 1 + M$ with $M = \max_k\abs{q_k}$, then $$ \Bigl|\sum_{k 0$: all roots lie in $\overline D(0, 1+M)$. The coefficients of the $P_j$ converge, hence are bounded by some $M$: all roots of all $P_j$ (and of $P$) lie in the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K = \overline D(0, 1 + M)$. Let $v_j \in K^n$ be a vector listing the roots of $P_j$ with multiplicity (question 7). Every subsequence of $(v_j)$ has a further subsequence converging to some $(\beta_1, \dots, \beta_n)$; the coefficients of $\prod_i(X - \alpha_i^{(j)})$ are, up to sign, the elementary symmetric functions of $v_j$ — [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) — so along that subsequence they converge to the coefficients of $\prod_i(X - \beta_i)$; but they converge to those of $P$ by hypothesis, so $\prod_i(X - \beta_i) = P$: every subsequential limit of $(v_j)$ is a permutation of the root vector of $P$. If the matching distance $\delta_j = \min_\sigma\max_i\, \abs{\alpha_i^{(j)} - \alpha_{\sigma(i)}}$ did not tend to $0$, a subsequence would keep $\delta_j \geq \varepsilon$ while its root vectors converge to a permutation of the roots of $P$ — forcing $\delta_j \to 0$ along it: contradiction. So the root multisets converge. Sharpness: $P_\varepsilon = (X - 1)^2 + \varepsilon$ has roots $1 \pm \iu\sqrt\varepsilon$: the double root moves by $\sqrt\varepsilon$, e.g. by $10^{-2}$ for $\varepsilon = 10^{-4}$. In general, if $\alpha$ is an $m$-fold root, then near $\alpha$ one has $\abs{P(z)} \asymp \abs{z - \alpha}^m$, so a size-$\varepsilon$ perturbation displaces the cluster of roots by about $\varepsilon^{1/m}$: Hölder [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of exponent $\frac1m$ and no better — which is why a numerical solver near a double root retains only half the working digits.