06.01.E2 · riemann-surfaces / complex-analysis

Complex analysis exercise pack II (Ahlfors Ch. 5-8 supplement)

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Anchor (Master):

Formal definition of the pack Intermediate

This pack supplements the second half of Ahlfors: the residue calculus pushed into real-integral evaluation and root counting, the theory of entire and meromorphic functions through infinite products and the gamma function, harmonic-function theory and the Dirichlet problem, the Riemann mapping theorem by normal families, and analytic continuation with the monodromy theorem. Its problems exercise the units on the residue theorem 06.01.03, the argument principle and Rouché 06.01.13, harmonic functions on the plane 06.01.11, the Riemann mapping theorem 06.01.06, analytic continuation 06.01.04, the Weierstrass factorization theorem 06.01.17, the gamma function 06.01.15, and normal families and Montel's theorem 06.01.14.

The pack collects ten problems — two easy, five medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The problems are grouped by theme: residue and Rouché root-counting (easy/medium), gamma-function and infinite-product identities (medium), harmonic functions and the Poisson kernel (medium), and the Riemann mapping / monodromy circle of ideas (hard). Conventions follow Ahlfors: the argument principle counts $Z - P = \frac{1}{2 π i} \oint \frac{f ^{'}}{f} d z = n (f \circ γ, 0)$ (zeros minus poles inside $γ$ ); the gamma function is normalised by $Γ (z + 1) = z Γ (z)$ , $Γ (1) = 1$ ; the Poisson kernel on the unit disc is $P_{r} (θ) = \frac{1 - r ^{2}}{1 - 2 r c o s θ + r ^{2}}$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one problem in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead problem. Prove the reflection formula $Γ (z) Γ (1 - z) = \frac{π}{sin π z}$ for $0 < Re z < 1$ , assuming the Weierstrass product for $Γ$ and the product for $sin$ .

Solution. Use the Weierstrass product $$ \frac{1}{\Gamma(z)} = z,e^{\gamma z}\prod_{n=1}^\infty\left(1 + \frac{z}{n}\right)e^{-z/n}, $$ where $γ$ is the Euler-Mascheroni constant. Replacing $z$ by $- z$ , $$ \frac{1}{\Gamma(-z)} = -z,e^{-\gamma z}\prod_{n=1}^\infty\left(1 - \frac{z}{n}\right)e^{z/n}. $$ Multiply the two products. The $e^{\pm γ z}$ factors cancel and the $e^{\mp z / n}$ factors cancel term by term, leaving $$ \frac{1}{\Gamma(z),\Gamma(-z)} = -z^2\prod_{n=1}^\infty\left(1 - \frac{z^2}{n^2}\right). $$ Now invoke the Euler product for the sine, $sin π z = π z \prod_{n = 1}^{\infty} (1 - \frac{z ^{2}}{n ^{2}})$ , so $\prod_{n = 1}^{\infty} (1 - \frac{z ^{2}}{n ^{2}}) = \frac{s i n π z}{π z}$ . Substituting, $$ \frac{1}{\Gamma(z),\Gamma(-z)} = -z^2\cdot\frac{\sin\pi z}{\pi z} = -\frac{z\sin\pi z}{\pi}. $$ Finally use the functional equation $Γ (1 - z) = - z Γ (- z)$ , i.e. $Γ (- z) = - Γ (1 - z) / z$ . Then $$ \frac{1}{\Gamma(z),\Gamma(-z)} = \frac{1}{\Gamma(z)\cdot(-\Gamma(1-z)/z)} = -\frac{z}{\Gamma(z)\Gamma(1-z)}. $$ Equating the two expressions, $- \frac{z}{Γ ( z ) Γ ( 1 - z )} = - \frac{z s i n π z}{π}$ , hence $$ \Gamma(z),\Gamma(1-z) = \frac{\pi}{\sin\pi z}.\qquad\square $$

The reflection formula is the keystone identity of the gamma function: setting $z = \frac{1}{2}$ recovers $Γ (\frac{1}{2}) = π$ , and combined with the duplication formula it pins down $Γ$ on the whole plane. This is Ahlfors §5.2.4. The mechanism — multiply two Weierstrass products and match against the sine product — is the workhorse of classical special-function manipulation.

Exercises Intermediate

Exercise 1 (easy, numeric). Counting zeros with Rouché.

How many zeros (with multiplicity) does $p (z) = z^{4} - 5 z + 1$ have inside the unit disc $∣ z ∣ < 1$ ?

Hint

On $∣ z ∣ = 1$ , compare $p$ with the dominant term $- 5 z$ . Rouché: if $∣ f - g ∣ < ∣ g ∣$ on the boundary, then $f$ and $g$ have the same number of zeros inside.

Answer

Take $g (z) = - 5 z$ and $f (z) = p (z) = z^{4} - 5 z + 1$ , so $f - g = z^{4} + 1$ . On $∣ z ∣ = 1$ ,

∣ f (z) - g (z) ∣ = ∣ z^{4} + 1∣ \leq ∣ z ∣^{4} + 1 = 2, ∣ g (z) ∣ = 5∣ z ∣ = 5.

Since $2 < 5$ , Rouché's theorem applies: $p$ and $- 5 z$ have the same number of zeros inside $∣ z ∣ < 1$ . The function $- 5 z$ has exactly one zero there (at $z = 0$ , simple). Therefore $p$ has exactly one zero in the unit disc.

(The other three zeros of the quartic lie outside; one can check $∣ p (z) ∣$ is dominated by $z^{4}$ on a large circle, accounting for all four roots.) Ahlfors §4.4.

Exercise 3 (medium, numeric). A real integral with a simple pole on the contour, via the argument that a sine integral converges.

Evaluate $\int_{- \infty}^{\infty} \frac{sin x}{x} d x$ using a contour with a small indentation around the origin.

Hint

Consider $\oint \frac{e ^{i z}}{z} d z$ over the contour $[- R, - ε] \cup C_{ε} \cup [ε, R] \cup C_{R}$ , with $C_{ε}$ a small semicircle above the origin and $C_{R}$ the large upper semicircle. The integrand is holomorphic inside, so the total is $0$ ; take imaginary parts.

Answer

Let $f (z) = e^{i z} / z$ . Over the indented contour (real axis from $- R$ to $- ε$ , small upper semicircle $C_{ε}$ around $0$ traversed clockwise, real axis from $ε$ to $R$ , large upper semicircle $C_{R}$ ), $f$ is holomorphic inside, so by Cauchy's theorem the total integral is $0$ .

On the large semicircle $C_{R}$ , by Jordan's lemma (the $e^{i z}$ factor decays in the upper half-plane), $\int_{C_{R}} f d z \to 0$ as $R \to \infty$ .

On the small semicircle $C_{ε}$ (radius $ε$ , clockwise, half-turn around the simple pole at $0$ with residue $Res_{0} f = e^{i 0} = 1$ ), the standard small-arc lemma gives $\int_{C_{ε}} f d z \to - π i \cdot 1 = - π i$ as $ε \to 0$ (the minus sign because $C_{ε}$ is traversed clockwise, the factor $π i$ because it is a half-circle).

The two real pieces combine to the principal value $p.v. \int_{- \infty}^{\infty} \frac{e ^{i x}}{x} d x$ . Summing the four contributions to zero:

p.v. \int_{- \infty}^{\infty} \frac{e ^{i x}}{x} d x - π i = 0 ⟹ p.v. \int_{- \infty}^{\infty} \frac{e ^{i x}}{x} d x = π i .

Taking imaginary parts ( $Im e^{i x} = sin x$ , and $sin x / x$ is even with no singularity, so the principal value equals the ordinary integral):

\int_{- \infty}^{\infty} \frac{sin x}{x} d x = Im (π i) = π .

The indentation technique is the residue toolkit's handling of poles on the contour: a half-circle around a simple pole contributes $\pm π i \cdot Res$ , half of a full $2 π i \cdot Res$ . Ahlfors §4.5.

Exercise 4 (medium, proof). Hurwitz's theorem from Rouché.

Let $f_{n} \to f$ uniformly on compact subsets of a domain $Ω$ , with each $f_{n}$ holomorphic and nowhere zero. Show that $f$ is either identically zero or nowhere zero.

Hint

Suppose $f \neq \equiv 0$ but $f (z_{0}) = 0$ . Zeros of holomorphic functions are isolated, so on a small circle $∣ z - z_{0} ∣ = r$ , $f$ is bounded below by some $m > 0$ . Apply Rouché to $f$ and $f_{n}$ on that circle.

Answer

Suppose $f \neq \equiv 0$ . The limit $f$ is holomorphic (uniform limits of holomorphic functions on compacts are holomorphic). If $f$ had a zero at $z_{0}$ , then since the zeros of a nonzero holomorphic function are isolated, choose $r > 0$ so small that $f$ has no zero on the punctured disc $0 < ∣ z - z_{0} ∣ \leq r$ ; in particular $z_{0}$ is the only zero in $∣ z - z_{0} ∣ \leq r$ . On the circle $C : ∣ z - z_{0} ∣ = r$ , $f$ is continuous and nonzero, so $m := min_{C} ∣ f ∣ > 0$ .

By uniform convergence on the compact set $C$ , choose $N$ so that $∣ f_{n} - f ∣ < m \leq ∣ f ∣$ on $C$ for all $n \geq N$ . By Rouché's theorem, $f_{n}$ and $f$ have the same number of zeros inside $C$ . But $f$ has at least one zero there (at $z_{0}$ ), while $f_{n}$ has none (each $f_{n}$ is nowhere zero) — contradiction.

Therefore $f$ has no zero, i.e. $f$ is nowhere zero. So $f$ is either identically zero or nowhere zero. This is Hurwitz's theorem; it is the tool that guarantees the Riemann mapping limit is injective (a uniform limit of injective holomorphic functions is injective or constant). Ahlfors §4.4, §6.1.

Exercise 5 (medium, proof). The Poisson integral solves the Dirichlet problem on the disc.

Let $u (r e^{i θ}) = \frac{1}{2 π} \int_{0}^{2 π} P_{r} (θ - t) f (t) d t$ with $P_{r} (θ) = \frac{1 - r ^{2}}{1 - 2 r c o s θ + r ^{2}}$ and $f$ continuous on the circle. Show $u$ is harmonic in the disc.

Hint

$P_{r} (θ) = Re \frac{1 + z}{1 - z}$ with $z = r e^{i θ}$ . So $u$ is the real part of a holomorphic function of $z$ , hence harmonic.

Answer

Write $z = r e^{i θ}$ . The Poisson kernel is the real part of the Herglotz kernel:

Re \frac{1 + z e ^{- i t}}{1 - z e ^{- i t}} = \frac{1 - ∣ z ∣ ^{2}}{∣1 - z e ^{- i t} ∣ ^{2}} = \frac{1 - r ^{2}}{1 - 2 r cos ( θ - t ) + r ^{2}} = P_{r} (θ - t),

using $∣1 - z e^{- i t} ∣^{2} = 1 - 2 r cos (θ - t) + r^{2}$ .

Therefore

u (z) = \frac{1}{2 π} \int_{0}^{2 π} Re (\frac{1 + z e ^{- i t}}{1 - z e ^{- i t}}) f (t) d t = Re F (z), F (z) := \frac{1}{2 π} \int_{0}^{2 π} \frac{1 + z e ^{- i t}}{1 - z e ^{- i t}} f (t) d t .

For each fixed $t$ , the integrand $\frac{1 + z e ^{- i t}}{1 - z e ^{- i t}}$ is holomorphic in $z$ on the disc $∣ z ∣ < 1$ (its only singularity is at $z = e^{i t}$ , on the boundary). Differentiating under the integral sign (justified by uniform bounds on compact subdiscs) shows $F$ is holomorphic. The real part of a holomorphic function is harmonic, so $u = Re F$ satisfies $Δ u = 0$ in the disc.

That $u$ attains the boundary values $f$ continuously is the approximate-identity property of $P_{r}$ ; together these solve the Dirichlet problem on the disc. Ahlfors §6.3. The general-domain version comes via Perron's method 06.01.24.

Exercise 6 (medium, symbolic). Mean-value property and the maximum principle.

Show that a harmonic function $u$ on a domain satisfies the mean-value property $u (a) = \frac{1}{2 π} \int_{0}^{2 π} u (a + r e^{i θ}) d θ$ for every disc $\overline{D (a, r)} \subset Ω$ , and deduce the maximum principle.

Hint

Locally $u = Re f$ for a holomorphic $f$ ; apply the Cauchy integral formula to $f$ at the center and take real parts. For the maximum principle, an interior maximum would force $u$ constant on a disc, then on all of $Ω$ by connectedness.

Answer

Mean-value. On the simply-connected disc $D (a, R) \supset \overline{D (a, r)}$ , $u$ has a harmonic conjugate $v$ , so $f = u + i v$ is holomorphic there. By Cauchy's integral formula at the center,

f (a) = \frac{1}{2 π i} \oint_{∣ z - a ∣ = r} \frac{f ( z )}{z - a} d z = \frac{1}{2 π} \int_{0}^{2 π} f (a + r e^{i θ}) d θ,

parametrising $z = a + r e^{i θ}$ , $d z = i r e^{i θ} d θ$ . Taking real parts gives the mean-value property for $u$ .

Maximum principle. Suppose $u$ attains a maximum $M$ at an interior point $a$ . By the mean-value property over any small circle, $M = u (a) = \frac{1}{2 π} \int_{0}^{2 π} u (a + r e^{i θ}) d θ$ . Since each $u (a + r e^{i θ}) \leq M$ and the average equals $M$ , the integrand must equal $M$ for all $θ$ (a continuous function $\leq M$ with average $M$ is identically $M$ ). So $u \equiv M$ on a disc about $a$ . The set ${u = M}$ is then open; it is also closed in $Ω$ (preimage of a point under continuous $u$ ), so by connectedness $u \equiv M$ on all of $Ω$ .

Hence a non-constant harmonic function has no interior maximum (or minimum, applying the result to $- u$ ). Ahlfors §6.3.

Exercise 7 (medium, proof). Schwarz lemma and automorphisms of the disc.

Show that every holomorphic bijection $φ : D \to D$ of the unit disc has the form $φ (z) = e^{i α} \frac{z - a}{1 - a ˉ z}$ for some $α \in R$ and $a \in D$ .

Hint

Let $a = φ^{- 1} (0)$ and $ψ_{a} (z) = \frac{z - a}{1 - a ˉ z}$ , a disc automorphism with $ψ_{a} (a) = 0$ . Then $g = φ \circ ψ_{a}^{- 1}$ fixes $0$ ; apply the Schwarz lemma to $g$ and to $g^{- 1}$ .

Answer

The Blaschke factor $ψ_{a} (z) = \frac{z - a}{1 - a ˉ z}$ is a holomorphic bijection of $D$ (it maps $\partial D$ to itself since $∣ ψ_{a} (z) ∣ = 1$ for $∣ z ∣ = 1$ , and $ψ_{a} (a) = 0$ , $a \in D$ ). Let $a = φ^{- 1} (0)$ and set $g = φ \circ ψ_{a}^{- 1}$ . Then $g : D \to D$ is a holomorphic bijection with $g (0) = φ (ψ_{a}^{- 1} (0)) = φ (a) = 0$ .

Apply the Schwarz lemma to $g$ : since $g (0) = 0$ and $g : D \to D$ , $∣ g (z) ∣ \leq ∣ z ∣$ . Apply it also to $g^{- 1}$ (which fixes $0$ and maps $D \to D$ ): $∣ g^{- 1} (w) ∣ \leq ∣ w ∣$ , i.e. $∣ z ∣ \leq ∣ g (z) ∣$ . Combining, $∣ g (z) ∣ = ∣ z ∣$ for all $z$ . The equality case of the Schwarz lemma forces $g (z) = e^{i α} z$ for some real $α$ .

Therefore $φ = g \circ ψ_{a}$ , i.e.

φ (z) = e^{i α} ψ_{a} (z) = e^{i α} \frac{z - a}{1 - a ˉ z} .

This is the full automorphism group $Aut (D)$ , a three-real-parameter group ( $α$ and $a \in D$ ). It is the holomorphic isometry group of the Poincaré disc, the model behind the Schwarz-Pick theorem. Ahlfors §6.1-6.2.

Exercise 8 (hard, proof). The Riemann mapping theorem: extracting the limit.

In the proof of the Riemann mapping theorem one considers the family $F$ of injective holomorphic $f : Ω \to D$ with $f (z_{0}) = 0$ , and a maximising sequence $f_{n}$ with $∣ f_{n}^{'} (z_{0}) ∣ \to sup_{F} ∣ f^{'} (z_{0}) ∣$ . Show that a subsequence converges, locally uniformly, to an injective $f \in F$ achieving the supremum.

Hint

$F$ is uniformly bounded (into $D$ ), hence normal by Montel's theorem 06.01.14. Extract a locally-uniformly-convergent subsequence; use Hurwitz to keep injectivity and Cauchy's estimate to pass $∣ f_{n}^{'} (z_{0}) ∣$ to the limit.

Answer

Every $f \in F$ maps into $D$ , so $F$ is uniformly bounded by $1$ on $Ω$ . By Montel's theorem 06.01.14, $F$ is a normal family: the maximising sequence $f_{n}$ has a subsequence (rename it $f_{n}$ ) converging locally uniformly to a holomorphic $f : Ω \to \overline{D}$ .

The limit lands in $F$ . By Cauchy's estimate / uniform convergence of derivatives, $f_{n}^{'} (z_{0}) \to f^{'} (z_{0})$ , so $∣ f^{'} (z_{0}) ∣ = sup_{F} ∣ f^{'} (z_{0}) ∣ =: S$ . Because the supremum $S$ is positive (e.g. a Möbius map into the disc gives a competitor with nonzero derivative), $f^{'} (z_{0}) \neq = 0$ , so $f$ is nonconstant. Also $f (z_{0}) = lim f_{n} (z_{0}) = 0$ .

Injectivity. Each $f_{n}$ is injective. Fix any $w_{0} \in Ω$ and apply Hurwitz's theorem (Exercise 4) to $g_{n} (z) := f_{n} (z) - f_{n} (w_{0})$ on $Ω ∖ {w_{0}}$ : each $g_{n}$ is nowhere zero there (injectivity), so the limit $g (z) = f (z) - f (w_{0})$ is either nowhere zero on $Ω ∖ {w_{0}}$ or identically zero. The latter would make $f$ constant, excluded above. Hence $f (z) \neq = f (w_{0})$ for $z \neq = w_{0}$ ; as $w_{0}$ was arbitrary, $f$ is injective.

Range. The maximum modulus principle gives $∣ f ∣ < 1$ in the open $Ω$ (an interior boundary value $∣ f ∣ = 1$ would violate it), so $f$ maps into $D$ . Thus $f \in F$ and achieves the supremum $S$ .

A separate argument (the "square-root trick") shows this extremal $f$ is onto $D$ , completing the Riemann mapping theorem. This exercise isolates the compactness core: normal family ⟹ convergent subsequence, Hurwitz ⟹ injective limit. Ahlfors §6.1.

Exercise 9 (hard, proof). The monodromy theorem.

Let $Ω$ be simply connected and suppose a holomorphic germ at $z_{0} \in Ω$ can be analytically continued along every path in $Ω$ . Show the continuations agree, defining a single-valued holomorphic function on $Ω$ .

Hint

Two paths with the same endpoints are homotopic in a simply-connected domain. Show continuation is invariant under homotopy: the set of $t$ for which the continuations along the homotoped paths agree is open and closed.

Answer

Fix the germ $f_{z_{0}}$ at $z_{0}$ . By hypothesis it continues along every path; we must show the result of continuing to a point $w$ is independent of the path from $z_{0}$ to $w$ .

Let $γ_{0}, γ_{1}$ be two paths from $z_{0}$ to $w$ in $Ω$ . Since $Ω$ is simply connected, they are homotopic rel endpoints: there is $H : [0, 1] \times [0, 1] \to Ω$ with $H (\cdot, 0) = γ_{0}$ , $H (\cdot, 1) = γ_{1}$ , $H (0, s) = z_{0}$ , $H (1, s) = w$ . Let $f_{s}$ denote the germ at $w$ obtained by continuing $f_{z_{0}}$ along $H (\cdot, s)$ .

Claim: $s \mapsto f_{s}$ is locally constant, hence constant. Fix $s_{0}$ . Continuation along $H (\cdot, s_{0})$ is realised by a finite chain of overlapping discs on which power-series elements agree. For $s$ near $s_{0}$ , the path $H (\cdot, s)$ stays inside the same chain of discs (uniform continuity of $H$ on the compact square, Lebesgue number argument), so it produces the same terminal germ at $w$ : $f_{s} = f_{s_{0}}$ . Thus the set ${s : f_{s} = f_{s_{0}}}$ is open; the same argument shows its complement is open, so by connectedness of $[0, 1]$ it is all of $[0, 1]$ . Hence $f_{0} = f_{1}$ : continuation along $γ_{0}$ and $γ_{1}$ give the same germ at $w$ .

Therefore the value at $w$ is path-independent. Defining $F (w)$ as the common germ's value at $w$ for each $w \in Ω$ yields a single-valued function, holomorphic because it is locally given by the power-series elements. This is the monodromy theorem; its standard application is constructing a global $lo g$ or $\cdot$ on any simply-connected domain avoiding the relevant branch points. Ahlfors §8.1.

Exercise 10 (hard, numeric). Counting zeros of $e^{z} - 3 z$ in the unit disc by the argument principle.

How many zeros does $f (z) = e^{z} - 3 z$ have inside $∣ z ∣ < 1$ ?

Hint

Compare with $g (z) = - 3 z$ on $∣ z ∣ = 1$ using Rouché: $∣ f - g ∣ = ∣ e^{z} ∣ \leq e^{Re z} \leq e$ on the circle, while $∣ g ∣ = 3$ . But $e \approx 2.718 < 3$ .

Answer

Set $g (z) = - 3 z$ , so $f (z) - g (z) = e^{z}$ . On the unit circle $∣ z ∣ = 1$ , write $z = x + i y$ with $x^{2} + y^{2} = 1$ , so $x \in [- 1, 1]$ . Then

∣ f (z) - g (z) ∣ = ∣ e^{z} ∣ = e^{Re z} = e^{x} \leq e^{1} = e \approx 2.718,

while $∣ g (z) ∣ = 3∣ z ∣ = 3$ . Since $e < 3$ , we have $∣ f - g ∣ < ∣ g ∣$ on $∣ z ∣ = 1$ , so Rouché's theorem gives: $f$ and $g$ have the same number of zeros inside the unit disc.

The function $g (z) = - 3 z$ has exactly one zero inside $∣ z ∣ < 1$ (at $z = 0$ , simple). Therefore $f (z) = e^{z} - 3 z$ has exactly one zero in the unit disc.

(Numerically, the equation $e^{x} = 3 x$ has a real root near $x \approx 0.619$ , confirming a real zero in $(0, 1)$ ; Rouché shows it is the only one in the whole disc.) This is the argument principle in its Rouché packaging: count zeros of the perturbation by counting zeros of the dominant term. Ahlfors §4.4.

Exercise pack supplementing Ahlfors Chapters 5-8: residue evaluation of real integrals, the argument principle and Rouché, infinite products and the gamma function, harmonic functions and the Poisson/Dirichlet theory, the Riemann mapping theorem via normal families, and analytic continuation with monodromy.

Prerequisites

06.01.03 pending
06.01.13 pending
06.01.11 pending
06.01.06 pending
06.01.04 pending
06.01.17 pending
06.01.15 pending
06.01.14 pending

Tier anchors

beginner
intermediate: Ahlfors *Complex Analysis* (3rd ed.) exercises across Ch. 4 §4.3-4.5 (residues, argument principle, Rouché), Ch. 5 §5.1-5.4 (infinite products, Weierstrass and Mittag-Leffler, gamma function), Ch. 6 §6.1-6.5 (Riemann mapping via normal families, harmonic functions, Poisson and Dirichlet), Ch. 8 §8.1 (analytic continuation, monodromy)
master

References

Ahlfors — Complex Analysis (3rd ed., McGraw-Hill 1979) · Ch. 4 §4.4 (argument principle, Rouché), Ch. 5 §5.2 (infinite products, Weierstrass factorization), §5.2.4 (the gamma function), Ch. 6 §6.1 (Riemann mapping theorem via normal families), §6.3-6.5 (harmonic functions, Poisson integral, Dirichlet via Perron), Ch. 8 §8.1 (analytic continuation along curves, monodromy theorem)
Whittaker-Watson — A Course of Modern Analysis (Cambridge, 4th ed.) · Ch. 12 (gamma function), Ch. 7 (infinite products) — classical special-function problems
Stein-Shakarchi — Complex Analysis (Princeton 2003) · Ch. 5-8 exercises and problems (entire functions, gamma/zeta, conformal maps, harmonic functions)

Estimated time

intermediate: 5h