06.01.E1 · riemann-surfaces / complex-analysis

Complex analysis exercise pack I (Ahlfors Ch. 1-4 supplement)

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Formal definition of the pack Intermediate

This pack supplements the first four chapters of Ahlfors: the geometry of the extended plane, holomorphy and the Cauchy-Riemann equations, conformal and Möbius maps, power and Laurent series, and the Cauchy theory in its integral-formula and residue forms. Its problems exercise the units on holomorphic functions 06.01.01, the Cauchy-Riemann equations and harmonic conjugates 06.01.10, Möbius transformations 06.01.08, the Riemann sphere 06.01.07, power and Laurent series 06.01.27, the Cauchy integral formula 06.01.02, the index of a closed curve 06.01.28, and the residue theorem 06.01.03.

The pack collects ten problems — three easy, four 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 Ahlfors chapter: the algebra and geometry of holomorphy and the sphere (easy), conformal/Möbius maps and power-series manipulation (medium), and Cauchy's theorem with its contour-integration consequences (hard). Conventions follow Ahlfors: $C_{\infty} = C \cup {\infty}$ is the Riemann sphere, the cross-ratio is $(z_{1}, z_{2}, z_{3}, z_{4}) = \frac{z _{1} - z _{3}}{z _{1} - z _{4}} \cdot \frac{z _{2} - z _{4}}{z _{2} - z _{3}}$ , and the winding number (index) of a closed curve $γ$ about $a \in / γ$ is $n (γ, a) = \frac{1}{2 π i} \int_{γ} \frac{d z}{z - a}$ .

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. Evaluate $\int_{0}^{2 π} \frac{d θ}{a + cos θ}$ for $a > 1$ by contour integration.

Solution. Substitute $z = e^{i θ}$ , so $d z = i z d θ$ and $cos θ = \frac{1}{2} (z + z^{- 1})$ . As $θ$ runs over $[0, 2 π)$ , $z$ traverses the unit circle $∣ z ∣ = 1$ once positively. Then $d θ = d z / (i z)$ and

\int_{0}^{2 π} \frac{d θ}{a + cos θ} = \oint_{∣ z ∣ = 1} \frac{1}{a + \frac{1}{2} ( z + z ^{- 1} )} \cdot \frac{d z}{i z} = \oint_{∣ z ∣ = 1} \frac{2 d z}{i ( z ^{2} + 2 a z + 1 )} .

The denominator $z^{2} + 2 a z + 1$ has roots $z_{\pm} = - a \pm a^{2} - 1$ . Since $a > 1$ , both roots are real and negative with $z_{+} z_{-} = 1$ , so exactly one root lies inside the unit disc, namely $z_{+} = - a + a^{2} - 1$ (it satisfies $∣ z_{+} ∣ < 1$ because $∣ z_{-} ∣ = 1/∣ z_{+} ∣ > 1$ ). The residue of the integrand at $z_{+}$ is

z = z_{+} Res \frac{2}{i ( z - z _{+} ) ( z - z _{-} )} = \frac{2}{i ( z _{+} - z _{-} )} = \frac{2}{i \cdot 2 a ^{2} - 1} = \frac{1}{i a ^{2} - 1} .

By the residue theorem the integral is $2 π i$ times this residue:

\int_{0}^{2 π} \frac{d θ}{a + cos θ} = 2 π i \cdot \frac{1}{i a ^{2} - 1} = \frac{2 π}{a ^{2} - 1} . □

The substitution $z = e^{i θ}$ converting a trigonometric integral over $[0, 2 π)$ into a contour integral over $∣ z ∣ = 1$ is the prototype of the residue-evaluation toolkit. Every rational-in- $cos, sin$ integral reduces this way; the only work is locating which poles fall inside the unit disc. This is Ahlfors §4.3, the residue calculus applied to real integrals.

Exercises Intermediate

Exercise 1 (easy, proof). Cauchy-Riemann forces holomorphy of $z \mapsto z^{2}$ but not of $z \mapsto \overset{z}{ˉ}$ .

Verify directly from the Cauchy-Riemann equations that $f (z) = z^{2}$ is holomorphic on all of $C$ , while $g (z) = \overset{z}{ˉ}$ is holomorphic nowhere.

Hint

Write $z = x + i y$ and separate real and imaginary parts. The Cauchy-Riemann equations are $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ .

Answer

For $f (z) = z^{2} = (x^{2} - y^{2}) + i (2 x y)$ we have $u = x^{2} - y^{2}$ , $v = 2 x y$ . Then $u_{x} = 2 x = v_{y}$ and $u_{y} = - 2 y = - v_{x}$ , so the Cauchy-Riemann equations hold at every point, and the partials are continuous, so $f$ is holomorphic on $C$ with $f^{'} (z) = u_{x} + i v_{x} = 2 x + i 2 y = 2 z$ .

For $g (z) = \overset{z}{ˉ} = x - i y$ we have $u = x$ , $v = - y$ . Then $u_{x} = 1$ but $v_{y} = - 1$ , so $u_{x} = v_{y}$ fails everywhere. Hence $g$ is holomorphic at no point, even though it is smooth as a map $R^{2} \to R^{2}$ . This is the basic Ahlfors §2.1 contrast: real-differentiability is necessary but the Cauchy-Riemann constraint is the extra condition that defines complex differentiability.

Exercise 2 (easy, numeric). Chordal distance on the Riemann sphere.

Under stereographic projection from the north pole, the chordal (spherical) distance between the images of $z, w \in C$ on the unit sphere is $d (z, w) = \frac{2∣ z - w ∣}{1 + ∣ z ∣ ^{2} 1 + ∣ w ∣ ^{2}}$ . Compute the chordal distance between $z = 0$ and $w = \infty$ , and between $z = 1$ and $w = - 1$ .

Hint

For $w = \infty$ , take the limit of the formula as $∣ w ∣ \to \infty$ : $d (z, \infty) = \frac{2}{1 + ∣ z ∣ ^{2}}$ .

Answer

For $z = 0$ , $w = \infty$ : $d (0, \infty) = \frac{2}{1 + 0} = 2$ . This is the diameter of the unit sphere — $0$ maps to the south pole, $\infty$ to the north pole, and they are antipodal.

For $z = 1$ , $w = - 1$ : $d (1, - 1) = \frac{2 \cdot ∣1 - ( - 1 ) ∣}{1 + 1 1 + 1} = \frac{2 \cdot 2}{2} = 2$ . The points $1$ and $- 1$ map to antipodal points on the equator, again at chordal distance $2$ .

These confirm the chordal metric is bounded by $2$ (the sphere's diameter), unlike the Euclidean metric on $C$ . Ahlfors §1.4.

Exercise 4 (medium, proof). Möbius maps preserve the cross-ratio.

Show that any Möbius transformation $T (z) = \frac{a z + b}{cz + d}$ (with $a d - b c \neq = 0$ ) preserves the cross-ratio: $(T z_{1}, T z_{2}, T z_{3}, T z_{4}) = (z_{1}, z_{2}, z_{3}, z_{4})$ .

Hint

Compute $T z_{i} - T z_{j} = \frac{( a d - b c ) ( z _{i} - z _{j} )}{( c z _{i} + d ) ( c z _{j} + d )}$ , then form the cross-ratio; every denominator factor and the $(a d - b c)$ cancel.

Answer

For any two points,

T z_{i} - T z_{j} = \frac{a z _{i} + b}{c z _{i} + d} - \frac{a z _{j} + b}{c z _{j} + d} = \frac{( a z _{i} + b ) ( c z _{j} + d ) - ( a z _{j} + b ) ( c z _{i} + d )}{( c z _{i} + d ) ( c z _{j} + d )} .

Expanding the numerator, the $a c z_{i} z_{j}$ and $b d$ terms cancel and the rest collect to $(a d - b c) (z_{i} - z_{j})$ . Hence

T z_{i} - T z_{j} = \frac{( a d - b c ) ( z _{i} - z _{j} )}{( c z _{i} + d ) ( c z _{j} + d )} .

Now form the cross-ratio $(T z_{1}, T z_{2}, T z_{3}, T z_{4}) = \frac{T z _{1} - T z _{3}}{T z _{1} - T z _{4}} \cdot \frac{T z _{2} - T z _{4}}{T z _{2} - T z _{3}}$ . Substituting the formula, each of the four differences contributes a factor $(a d - b c)$ — two in the numerator and two in the denominator, cancelling — and the linear denominators $(c z_{i} + d)$ each appear once in the numerator and once in the denominator across the product, cancelling as well. What remains is

\frac{z _{1} - z _{3}}{z _{1} - z _{4}} \cdot \frac{z _{2} - z _{4}}{z _{2} - z _{3}} = (z_{1}, z_{2}, z_{3}, z_{4}) .

The cross-ratio is the fundamental Möbius invariant: it is the unique (up to convention) projective invariant of four points on $C_{\infty}$ , and invariance characterises Möbius maps among the homographies. Ahlfors §2.4.

Exercise 5 (medium, proof). Cauchy's estimate and Liouville's theorem.

Let $f$ be entire and suppose $∣ f (z) ∣ \leq M$ for all $z$ . Use the Cauchy integral formula for $f^{'}$ to derive Cauchy's estimate on a circle of radius $R$ , then let $R \to \infty$ to conclude $f$ is constant (Liouville).

Hint

$f^{'} (a) = \frac{1}{2 π i} \oint_{∣ z - a ∣ = R} \frac{f ( z )}{( z - a ) ^{2}} d z$ . Bound the integral by (length) $\times$ (max integrand).

Answer

By the Cauchy integral formula for the derivative,

f^{'} (a) = \frac{1}{2 π i} \oint_{∣ z - a ∣ = R} \frac{f ( z )}{( z - a ) ^{2}} d z .

On the circle $∣ z - a ∣ = R$ we have $∣ z - a ∣^{2} = R^{2}$ and $∣ f (z) ∣ \leq M$ , and the circle has length $2 π R$ . The standard ML-estimate gives

∣ f^{'} (a) ∣ \leq \frac{1}{2 π} \cdot \frac{M}{R ^{2}} \cdot 2 π R = \frac{M}{R} .

This is Cauchy's estimate. Since $f$ is entire the formula holds for every $R > 0$ , and the bound $M$ is independent of $R$ . Letting $R \to \infty$ forces $∣ f^{'} (a) ∣ = 0$ . As $a$ was arbitrary, $f^{'} \equiv 0$ , so $f$ is constant. This is Liouville's theorem, and it immediately yields the fundamental theorem of algebra: a nonconstant polynomial $p$ with no zero would make $1/ p$ a bounded entire function, hence constant — contradiction. Ahlfors §4.2-4.3.

Exercise 6 (medium, symbolic). Laurent expansion in an annulus.

Find the Laurent expansion of $f (z) = \frac{1}{( z - 1 ) ( z - 2 )}$ valid in the annulus $1 < ∣ z ∣ < 2$ .

Hint

Partial fractions: $f (z) = \frac{1}{z - 2} - \frac{1}{z - 1}$ . In the annulus, expand $\frac{1}{z - 1}$ in powers of $1/ z$ (since $∣ z ∣ > 1$ ) and $\frac{1}{z - 2}$ in powers of $z$ (since $∣ z ∣ < 2$ ).

Answer

Partial fractions give $f (z) = \frac{1}{z - 2} - \frac{1}{z - 1}$ .

For the first term, since $∣ z ∣ < 2$ , write $\frac{1}{z - 2} = - \frac{1}{2} \cdot \frac{1}{1 - z /2} = - \frac{1}{2} \sum_{n = 0}^{\infty} (\frac{z}{2})^{n} = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{2 ^{n + 1}}$ (geometric, $∣ z /2∣ < 1$ ).

For the second term, since $∣ z ∣ > 1$ , write $\frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/ z} = \frac{1}{z} \sum_{m = 0}^{\infty} \frac{1}{z ^{m}} = \sum_{m = 1}^{\infty} \frac{1}{z ^{m}}$ (geometric, $∣1/ z ∣ < 1$ ).

Therefore in $1 < ∣ z ∣ < 2$ ,

f (z) = - n = 0 \sum \infty \frac{z ^{n}}{2 ^{n + 1}} - m = 1 \sum \infty \frac{1}{z ^{m}} .

The expansion has both a regular part (nonnegative powers, from the pole at $2$ outside the annulus) and a principal part (negative powers, from the pole at $1$ inside the annulus). A different annulus ( $0 < ∣ z ∣ < 1$ or $∣ z ∣ > 2$ ) would give a different Laurent series for the same function — the series is annulus-dependent. Ahlfors §3.1, §4.3.

Exercise 7 (medium, proof). Möbius map sending the upper half-plane to the disc.

Find a Möbius transformation carrying the upper half-plane $H = {Im z > 0}$ conformally onto the unit disc $D = {∣ w ∣ < 1}$ , and verify it sends the real axis to the unit circle.

Hint

Try $w = T (z) = \frac{z - i}{z + i}$ . Check that $T (i) = 0$ (center to center) and that real $z$ gives $∣ w ∣ = 1$ .

Answer

Take $T (z) = \frac{z - i}{z + i}$ , with $a d - b c = i - (- i) = 2 i \neq = 0$ , so it is a genuine Möbius map.

It sends $i \mapsto 0$ , placing the interior point $i \in H$ at the center of $D$ . For real $z = x$ ,

∣ T (x) ∣ = \frac{∣ x - i ∣}{∣ x + i ∣} = \frac{x ^{2} + 1}{x ^{2} + 1} = 1,

so the real axis (the boundary $\partial H$ ) maps onto the unit circle $\partial D$ . A Möbius map is a bijection of $C_{\infty}$ and is continuous, so it carries the connected region $H$ onto one of the two components of $C_{\infty} ∖ \partial D$ ; since the interior point $i \mapsto 0 \in D$ , it maps $H$ onto $D$ . Möbius maps are conformal where defined (the derivative is nonzero), so this is a conformal equivalence $H \sim D$ .

This is the Cayley transform, the bridge between the upper-half-plane and disc models used throughout hyperbolic geometry and the Schwarz-Pick theory. Ahlfors §2.4, §3.1.

Exercise 8 (hard, proof). Homotopy form of Cauchy's theorem.

Let $f$ be holomorphic on an open set $Ω$ and let $γ_{0}, γ_{1}$ be two closed curves in $Ω$ that are freely homotopic in $Ω$ through closed curves. Show $\int_{γ_{0}} f d z = \int_{γ_{1}} f d z$ .

Hint

Let $H (s, t)$ be the homotopy, $H (\cdot, 0) = γ_{0}$ , $H (\cdot, 1) = γ_{1}$ . The function $I (t) = \int_{H (\cdot, t)} f d z$ is differentiable in $t$ ; differentiate under the integral sign and use that $f$ has a local primitive.

Answer

Parametrise the homotopy as a continuous (we may smooth it) map $H : [0, 1] \times [0, 1] \to Ω$ with $H (s, 0) = γ_{0} (s)$ , $H (s, 1) = γ_{1} (s)$ , and $H (0, t) = H (1, t)$ for all $t$ (closed curves). Define

I (t) = \int_{0}^{1} f (H (s, t)) \partial_{s} H (s, t) d s .

Differentiating in $t$ and using $\partial_{t} \partial_{s} H = \partial_{s} \partial_{t} H$ ,

I^{'} (t) = \int_{0}^{1} [f^{'} (H) \partial_{t} H \partial_{s} H + f (H) \partial_{s} \partial_{t} H] d s = \int_{0}^{1} \partial_{s} [f (H) \partial_{t} H] d s,

since $\partial_{s} [f (H) \partial_{t} H] = f^{'} (H) \partial_{s} H \partial_{t} H + f (H) \partial_{s} \partial_{t} H$ . By the fundamental theorem of calculus this is $[f (H) \partial_{t} H]_{s = 0}^{s = 1}$ . Because the homotopy keeps the curves closed, $H (0, t) = H (1, t)$ and $\partial_{t} H (0, t) = \partial_{t} H (1, t)$ , so the boundary term vanishes: $I^{'} (t) = 0$ .

Hence $I$ is constant, so $I (0) = I (1)$ , i.e. $\int_{γ_{0}} f d z = \int_{γ_{1}} f d z$ . This is the homotopy form of Cauchy's theorem; specialising $γ_{1}$ to a constant curve recovers the statement that the integral of a holomorphic function over a null-homotopic loop is zero. The homological form (Ahlfors §4.1, Dixon's proof) is the refinement that replaces "homotopic" with "homologous" via winding numbers. Ahlfors §3.4-4.1.

Exercise 9 (hard, numeric). A rational improper integral by a semicircular contour.

Evaluate $\int_{- \infty}^{\infty} \frac{d x}{x ^{4} + 1}$ by closing the contour in the upper half-plane.

Hint

The poles in the upper half-plane are $e^{iπ /4}$ and $e^{3 iπ /4}$ . The large semicircle contributes nothing as $R \to \infty$ because the integrand decays like $R^{- 4}$ .

Answer

The poles of $1/ (z^{4} + 1)$ are the fourth roots of $- 1$ : $z_{k} = e^{iπ (2 k + 1) /4}$ for $k = 0, 1, 2, 3$ . The two in the upper half-plane are $z_{0} = e^{iπ /4}$ and $z_{1} = e^{3 iπ /4}$ . At a simple pole $z_{j}$ of $1/ (z^{4} + 1)$ , the residue is $1/ (4 z_{j}^{3}) = z_{j} / (4 z_{j}^{4}) = z_{j} / (4 \cdot (- 1)) = - z_{j} /4$ (using $z_{j}^{4} = - 1$ ).

So the residues are $- z_{0} /4 = - \frac{1}{4} e^{iπ /4}$ and $- z_{1} /4 = - \frac{1}{4} e^{3 iπ /4}$ . Their sum:

- \frac{1}{4} (e^{iπ /4} + e^{3 iπ /4}) = - \frac{1}{4} (\frac{2}{2} (1 + i) + \frac{2}{2} (- 1 + i)) = - \frac{1}{4} \cdot 2 i = - \frac{2}{4} i .

Closing with the upper semicircle $Γ_{R}$ of radius $R$ : on $Γ_{R}$ the integrand is $O (R^{- 4})$ and the arc has length $π R$ , so its contribution is $O (R^{- 3}) \to 0$ . Thus

\int_{- \infty}^{\infty} \frac{d x}{x ^{4} + 1} = 2 π i \sum (residues) = 2 π i \cdot (- \frac{2}{4} i) = \frac{π 2}{2} = \frac{π}{2} .

The semicircular-contour method works for any rational $P / Q$ with $de g Q \geq de g P + 2$ and no real poles: the arc decays, and the integral is $2 π i$ times the sum of upper-half-plane residues. Ahlfors §4.3.

Exercise 10 (hard, proof). The winding number is a constant integer on each component of the complement.

Let $γ$ be a piecewise- $C^{1}$ closed curve in $C$ and let $a \in / γ$ . Show that $n (γ, a) = \frac{1}{2 π i} \int_{γ} \frac{d z}{z - a}$ is an integer, and that $a \mapsto n (γ, a)$ is constant on each connected component of $C ∖ γ$ and vanishes on the unbounded component.

Hint

For integrality, parametrise $γ : [0, 1] \to C$ and consider $h (t) = \int_{0}^{t} \frac{γ ^{'} ( s )}{γ ( s ) - a} d s$ ; show $e^{- h (t)} (γ (t) - a)$ is constant, forcing $e^{h (1)} = 1$ . For local constancy, the integral depends continuously on $a$ , and a continuous integer-valued function is locally constant.

Answer

Integrality. Let $γ : [0, 1] \to C$ parametrise the curve with $γ (0) = γ (1)$ , and set $h (t) = \int_{0}^{t} \frac{γ ^{'} ( s )}{γ ( s ) - a} d s$ , so $h (1) = 2 π i n (γ, a)$ . Consider $φ (t) = e^{- h (t)} (γ (t) - a)$ . Then

φ^{'} (t) = e^{- h (t)} (- h^{'} (t) (γ (t) - a) + γ^{'} (t)) = e^{- h (t)} (- γ^{'} (t) + γ^{'} (t)) = 0,

using $h^{'} (t) = γ^{'} (t) / (γ (t) - a)$ . So $φ$ is constant: $φ (1) = φ (0)$ , i.e. $e^{- h (1)} (γ (1) - a) = γ (0) - a$ . Since $γ (1) = γ (0) \neq = a$ , this gives $e^{- h (1)} = 1$ , hence $h (1) \in 2 π i Z$ and $n (γ, a) \in Z$ .

Local constancy. As a function of $a$ , the integrand $1/ (z - a)$ is continuous in $a$ uniformly for $z \in γ$ (a compact set) as long as $a$ stays off $γ$ , so $a \mapsto n (γ, a)$ is continuous on $C ∖ γ$ . A continuous integer-valued function is locally constant, hence constant on each connected component.

Vanishing at infinity. On the unbounded component pick $∣ a ∣$ very large; then $∣1/ (z - a) ∣ \leq 1/ (∣ a ∣ - max_{γ} ∣ z ∣) \to 0$ , so $∣ n (γ, a) ∣ \leq \frac{1}{2 π} \cdot length (γ) \cdot \frac{1}{∣ a ∣ - m a x ∣ z ∣} < 1$ for $∣ a ∣$ large. Being an integer of modulus less than $1$ , it is $0$ ; by constancy it is $0$ on the whole unbounded component.

This is the basic topology of the index that underlies the homological Cauchy theorem and the argument principle. Ahlfors §3.4, §4.1.

Exercise pack supplementing Ahlfors Chapters 1-4: the Riemann sphere and chordal metric, Cauchy-Riemann and conformality, Möbius transformations and the cross-ratio, power and Laurent series, and the Cauchy theory in integral-formula and residue form.

Prerequisites

06.01.01 pending
06.01.10 pending
06.01.08 pending
06.01.07 pending
06.01.27 pending
06.01.02 pending
06.01.28 pending
06.01.03 pending

Tier anchors

beginner
intermediate: Ahlfors *Complex Analysis* (3rd ed.) exercises across Ch. 1 §1.4 (Riemann sphere, chordal metric), Ch. 2 §2.1-2.4 (Cauchy-Riemann, conformality, Möbius and cross-ratio), Ch. 3 §3.1-3.4 (power series, Cauchy's theorem, index), Ch. 4 §4.1-4.3 (homological Cauchy, residues, real-integral warm-ups)
master

References

Ahlfors — Complex Analysis (3rd ed., McGraw-Hill 1979) · Ch. 1 §1.4 (stereographic projection, chordal metric); Ch. 2 §2.1 (analytic functions, Cauchy-Riemann), §2.4 (Möbius transformations, cross-ratio, symmetry); Ch. 3 §3.1 (power series), §3.4 (Cauchy's theorem, index of a closed curve); Ch. 4 §4.1 (general Cauchy theorem), §4.3 (residue calculus)
Needham — Visual Complex Analysis (Oxford 1997) · Ch. 3-5 (Möbius transformations, winding numbers) — geometric companion problems
Stein-Shakarchi — Complex Analysis (Princeton 2003) · Ch. 1-3 exercises and problems (Cauchy theory, contour integration)

Estimated time

intermediate: 5h