Complex Analysis — Kreyszig

Chapter 0: Why Complex Analysis?

Here is a real integral that is notoriously difficult with real methods alone:

∫₀^∞ cos(x)/(x² + 1) dx = π/(2e)

No amount of integration-by-parts or trigonometric substitution makes this easy. But with the residue theorem from complex analysis, it falls out in three lines. Complex analysis is not just an abstract exercise — it is a power tool for real-world computation.

What makes complex analysis special: In real calculus, a function can be differentiable without being smooth (e.g., |x| at x = 0, or differentiable but not twice-differentiable functions). In complex analysis, if a function is differentiable once, it is differentiable infinitely many times, has a convergent power series, and satisfies a web of extraordinary identities. The complex derivative is far more restrictive than the real one, and this rigidity is the source of all the power.

Complex analysis also gives us:

• Conformal mappings that solve 2D boundary value problems (electrostatics, fluid flow, heat)

• Contour integration for evaluating "impossible" real integrals

• Power series and Laurent series for understanding singularities

• The mathematical foundation of quantum mechanics and signal processing

What makes complex differentiability so much stronger than real differentiability?

Complex numbers are larger than real numbers If a complex function is differentiable once, it is automatically infinitely differentiable and has a power series Complex derivatives are easier to compute

Chapter 1: Complex Numbers

A complex number z = x + iy has a real part x and an imaginary part y, where i² = −1. We can represent z as a point (x, y) in the complex plane (also called the Argand plane).

The polar form is often more useful:

z = r e^iθ = r(cos θ + i sin θ)

where r = |z| = √(x² + y²) is the modulus (distance from origin) and θ = arg(z) is the argument (angle from the positive real axis).

Euler's formula: e^iθ = cos θ + i sin θ. This is perhaps the most important equation in mathematics. It unifies exponentials, trigonometry, and complex numbers. Setting θ = π gives Euler's identity: e^iπ + 1 = 0, linking five fundamental constants.

Multiplication in polar form is beautiful: z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)}. Multiply the moduli, add the arguments. Multiplication is rotation + scaling.

Roots: zⁿ = w has exactly n roots, equally spaced around a circle. For example, the cube roots of 1 are 1, e^i2π/3, e^i4π/3 — three points forming an equilateral triangle.

Complex Plane Explorer

Click to place a complex number z (orange dot). Teal dot shows z². The unit circle is shown. Observe how squaring doubles the angle and squares the distance.

Click on the plane to set z.

In polar form, what does multiplying two complex numbers do geometrically?

Multiplies the moduli and adds the arguments (scale + rotate) Adds the moduli and multiplies the arguments Reflects across the real axis

Chapter 2: Analytic Functions

A function f(z) of a complex variable z = x + iy is analytic (or holomorphic) at a point z₀ if it has a complex derivative there:

f'(z₀) = lim_h→0 [f(z₀ + h) − f(z₀)] / h

This looks identical to the real definition, but h is now a complex number approaching 0 from every direction. This is an enormously stronger requirement.

The key difference: In real calculus, the limit (f(x+h)−f(x))/h only approaches from left and right. In complex calculus, h can approach 0 from any direction in the plane. For the limit to exist and be the same regardless of approach direction, the function must satisfy the Cauchy-Riemann equations (next chapter).

Examples of analytic functions: Polynomials, e^z, sin z, cos z, rational functions (away from poles). These are the workhorses of complex analysis.

Non-analytic: f(z) = z (the conjugate) is not analytic anywhere, despite being perfectly smooth as a real function of (x, y). Also |z| is not analytic. Complex differentiability is restrictive.

Why is z not analytic? Write f(z) = z = x − iy. Then u = x, v = −y. Check Cauchy-Riemann: u_x = 1 but v_y = −1. They are not equal. The function fails the Cauchy-Riemann test at every point.

If f is analytic in a region, then it automatically has derivatives of all orders, and it has a convergent Taylor series. It also satisfies the maximum modulus principle: |f(z)| cannot attain a maximum in the interior of the region. Like the maximum principle for harmonic functions, extremes are always on the boundary.

Important analytic functions:

Function	Where analytic	Key property
zⁿ (polynomial)	Everywhere (entire)	n zeros counting multiplicity
e^z	Everywhere (entire)	Never zero; period 2πi
sin z, cos z	Everywhere (entire)	Zeros on real axis; unbounded on imaginary axis
1/z	z ≠ 0	Simple pole at origin
Log z	z ≠ 0, branch cut	Multi-valued; needs branch choice

Why is f(z) = z (the complex conjugate) not analytic?

Because the limit [f(z+h)−f(z)]/h gives different values depending on the direction h approaches 0 Because conjugation is not continuous Because it is a real-valued function

Chapter 3: Cauchy-Riemann Equations

Write f(z) = u(x, y) + iv(x, y), splitting into real and imaginary parts. The condition for f to be analytic is:

∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x

These are the Cauchy-Riemann equations. They are the bridge between complex differentiability and real partial derivatives.

Where they come from: Approach z₀ along the real axis (h = Δx) and the imaginary axis (h = iΔy). The complex derivative must be the same either way. Along the real axis: f' = u_x + iv_x. Along the imaginary axis: f' = v_y − iu_y. Setting these equal gives the Cauchy-Riemann equations.

Consequence: If f = u + iv is analytic, then both u and v are harmonic (satisfy Laplace's equation ∇²u = 0 and ∇²v = 0). This is because:

u_xx + u_yy = v_yx − v_xy = 0

So the real and imaginary parts of any analytic function automatically solve the Laplace equation. This is why complex analysis and potential theory are inseparable.

Harmonic conjugates: If u is harmonic, we can (locally) find a v such that f = u + iv is analytic. The function v is the harmonic conjugate of u. The level curves of u and v form orthogonal families — one family gives equipotential lines, the other gives streamlines in fluid flow.

Example: f(z) = z² = (x + iy)² = (x² − y²) + i(2xy). So u = x² − y², v = 2xy. Check: u_x = 2x = v_y. u_y = −2y = −v_x. Cauchy-Riemann satisfied.

Polar form: For f(z) = u(r, θ) + iv(r, θ) in polar coordinates, the Cauchy-Riemann equations become:

u_r = (1/r) v_θ, v_r = −(1/r) u_θ

This form is useful for functions naturally expressed in polar coordinates, like zⁿ = rⁿ e^inθ.

If f(z) = u + iv is analytic, what PDE do u and v each satisfy?

Laplace's equation ∇²u = 0 (they are harmonic functions) The heat equation The wave equation

Chapter 4: Contour Integrals

A contour integral is an integral of a complex function along a curve (contour) C in the complex plane:

∫_C f(z) dz

If C is parameterized by z(t) = x(t) + iy(t) for a ≤ t ≤ b, then:

∫_C f(z) dz = ∫_a^b f(z(t)) z'(t) dt

The fundamental result is Cauchy's integral theorem (next chapter): if f is analytic inside and on a simple closed contour C, then ∮_C f(z) dz = 0. The integral around any closed loop is zero for analytic functions.

But not for all functions: Consider f(z) = 1/z and C = the unit circle. Then ∮_C (1/z) dz = 2πi. This is nonzero because 1/z is not analytic at z = 0 (there is a singularity inside the contour). The value 2πi is not accidental — it comes from the residue of 1/z at z = 0.

Key example: ∮_C zⁿ dz around the unit circle. For n ≠ −1, the integral is 0. For n = −1, the integral is 2πi. This single fact is the foundation of residue calculus.

Why 2πi? Parameterize the unit circle: z = e^iθ, dz = ie^iθ dθ. Then ∮ (1/z) dz = ∫₀^2π (1/e^iθ) · ie^iθ dθ = ∫₀^2π i dθ = 2πi. The factor of i comes from dz, and 2π comes from going around the full circle.

ML inequality: |∫_C f(z) dz| ≤ M · L, where M = max|f(z)| on C and L = length of C. This bound is essential for proving that certain contour integrals vanish — particularly the arcs at infinity used in residue calculations.

Path independence: If f is analytic in a simply connected domain, ∫_C f(z) dz depends only on endpoints. This is the complex analog of conservative fields: analytic functions have an antiderivative F(z) with F' = f, and the integral is F(z₂) − F(z₁).

What is ∮_C (1/z) dz when C is the unit circle traversed counterclockwise?

2πi 0 π

Chapter 5: Cauchy's Theorem & Formula

Cauchy's Integral Theorem

If f(z) is analytic in a simply connected domain D, then for any closed contour C in D:

∮_C f(z) dz = 0

Analytic functions have zero circulation. This is the complex-analysis version of "conservative fields are path-independent."

Cauchy's Integral Formula

If f is analytic inside and on C, and z₀ is inside C:

f(z₀) = (1/2πi) ∮_C f(z)/(z − z₀) dz

This is astonishing: The value of f at any interior point is completely determined by its values on the boundary. You can reconstruct the entire function from its boundary values. Nothing like this exists in real analysis. It is as if knowing the temperature on the walls of a room tells you the temperature at every point inside — which is exactly what happens for harmonic functions (Laplace's equation).

Derivatives too: Differentiating under the integral sign:

f⁽ⁿ⁾(z₀) = (n!/2πi) ∮_C f(z)/(z − z₀)ⁿ⁺¹ dz

This proves that analytic functions have derivatives of all orders. One derivative implies infinitely many. This is the rigidity that makes complex analysis so powerful.

Liouville's theorem: A bounded entire function (analytic everywhere) must be constant. This seemingly innocent result implies the fundamental theorem of algebra: every nonconstant polynomial has a root. The proof: if p(z) has no root, then 1/p(z) is entire and bounded, hence constant — contradiction.

Morera's theorem (converse of Cauchy): If f is continuous and ∮_C f(z) dz = 0 for every closed contour C, then f is analytic. So zero integrals around all loops characterizes analyticity.

Contour Integral Demo

The unit circle contour (orange) around the origin. For f(z) = 1/z, the contour integral is 2πi. Drag the contour center to see how the integral changes when the singularity is inside vs. outside.

Click to move the contour center.

What does Cauchy's integral formula say, in plain language?

The value of an analytic function at any interior point is determined by its values on any enclosing contour All contour integrals equal zero Analytic functions are always real-valued

Chapter 6: Power & Laurent Series

Taylor (Power) Series

If f is analytic at z₀, it has a convergent Taylor series:

f(z) = Σ_n=0^∞ a_n(z − z₀)ⁿ, a_n = f⁽ⁿ⁾(z₀)/n!

The series converges in the largest disk centered at z₀ in which f is analytic. The radius of convergence is the distance from z₀ to the nearest singularity.

Laurent Series

Near a singularity, the Taylor series breaks down. The Laurent series allows negative powers:

f(z) = Σ_n=−∞^∞ a_n(z − z₀)ⁿ

The part with negative powers, Σ_n=−∞⁻¹ a_n(z − z₀)ⁿ, is the principal part. It encodes the behavior near the singularity.

Types of singularities:
• Removable singularity: No negative powers (the function can be made analytic by defining f(z₀) properly). Example: sin(z)/z at z = 0.
• Pole of order m: Finitely many negative powers, down to (z − z₀)^−m. Example: 1/z² has a pole of order 2 at z = 0.
• Essential singularity: Infinitely many negative powers. Example: e^1/z at z = 0. Near an essential singularity, f takes every complex value (with at most one exception) — Picard's theorem.

The coefficient a₋₁ of (z − z₀)⁻¹ in the Laurent series is the residue of f at z₀. This single number is the key to the residue theorem.

Example: f(z) = e^1/z = 1 + 1/z + 1/(2!z²) + 1/(3!z³) + ... The Laurent series at z = 0 has infinitely many negative powers — essential singularity. The residue is a₋₁ = 1.

Example: f(z) = 1/(z² + 1) = 1/((z − i)(z + i)). At z = i, the Laurent expansion starts with 1/(2i(z − i)). The residue is 1/(2i).

Singularity type	Principal part	Example
Removable	None (no negative powers)	sin(z)/z at z = 0
Pole (order m)	Finitely many terms to (z−z₀)^−m	1/z³ at z = 0
Essential	Infinitely many negative powers	e^1/z at z = 0

What is the residue of f at a singularity z₀?

The coefficient a₋₁ of (z − z₀)⁻¹ in the Laurent series The value of f at z₀ The derivative f'(z₀)

Chapter 7: The Residue Theorem

This is the most powerful computational tool in complex analysis:

∮_C f(z) dz = 2πi Σ Res(f, z_k)

The integral of f around a closed contour C equals 2πi times the sum of residues at all singularities z_k inside C. The residue captures exactly how much a singularity contributes to the contour integral.

Computing residues:
• Simple pole (order 1): Res(f, z₀) = lim_z→z₀ (z − z₀) f(z)
• Pole of order m: Res(f, z₀) = (1/(m−1)!) lim_z→z₀ d^m−1/dz^m−1 [(z − z₀)^m f(z)]
• If f = p/q with simple zero of q at z₀: Res(f, z₀) = p(z₀)/q'(z₀)

Evaluating Real Integrals

The residue theorem turns real integrals into algebra. The strategy:

1. Embed the real integral into a contour integral in the complex plane.

2. Choose a contour that includes the real axis segment and closes in the upper or lower half-plane.

3. Show the integral over the closing arc vanishes (via the ML inequality).

4. Compute the residues inside the contour.

Example: ∫_−∞^∞ 1/(x² + 1) dx. The integrand has poles at z = ±i. Close in the upper half-plane, capturing z = i. Res(1/(z²+1), i) = 1/(2i). Result: 2πi · 1/(2i) = π.

Common contour types: (1) Semicircular contour in the upper half-plane for ∫_−∞^∞ f(x) dx. (2) Wedge or sector contours for integrals involving x^α. (3) Rectangular contours for integrals involving exponentials. (4) Keyhole contours for branch cuts.

Trigonometric integrals: For ∫₀^2π R(cos θ, sin θ) dθ, substitute z = e^iθ, so cos θ = (z + 1/z)/2, sin θ = (z − 1/z)/(2i), dθ = dz/(iz). The real integral becomes a contour integral around the unit circle.

Example: ∫₀^2π dθ/(2 + cos θ). With z = e^iθ: the integral becomes ∮ 2dz/(iz(2 + (z + 1/z)/2)) = ∮ 4dz/(i(z² + 4z + 1)). The poles inside the unit circle are at z = −2 + √3. Computing the residue gives the result: 2π/√3.

How does the residue theorem help evaluate real integrals?

Embed the real integral in a contour integral, compute residues of poles inside the contour, and the real integral equals 2πi times the sum of residues By converting all real functions to complex polynomials It only works for polynomial integrands

Chapter 8: Complex Function Lab

Visualize how complex functions transform the plane. These "domain coloring" plots assign a color to each point based on the value of f(z): hue encodes the argument (angle), brightness encodes the modulus (magnitude).

Complex Function Visualizer

Domain coloring of f(z). Hue = arg(f(z)), brightness = |f(z)|. Zeros appear as dark spots where all colors meet. Poles appear as bright spots. Choose a function.

f(z) = z²

Conformal Mapping

Conformal maps are analytic functions that preserve angles. They transform one region into another while keeping the angle between any two curves unchanged. This is invaluable for solving Laplace's equation: transform a hard domain into a simple one (like a disk or half-plane), solve there, and map back.

Conformal Mapping Visualizer

A grid in the z-plane (left) mapped through f(z) to the w-plane (right). Notice how the grid lines remain orthogonal — angles are preserved.

Why conformality matters: Laplace's equation is invariant under conformal mapping. If u is harmonic in the w-plane, then u(f(z)) is harmonic in the z-plane. This means you can solve the Laplace equation on a complicated domain by mapping it to a simple one — solving there — and mapping the solution back. Electrical engineers use this to design waveguides and analyze electromagnetic fields.

Chapter 9: Connections

This lesson	Where it leads
Complex numbers	Quantum mechanics (wave functions are complex), AC circuits (impedance)
Analytic functions	2D potential theory, fluid dynamics (potential flow)
Cauchy-Riemann	Harmonic functions, Laplace equation (Ch 6), electrostatics
Cauchy's theorem	Green's theorem in disguise (Ch 5), topological invariants
Laurent series	Classification of singularities, algebraic geometry
Residues	Evaluating Fourier/Laplace integrals, transfer functions (controls)
Conformal mapping	Airfoil design (Joukowski), waveguide analysis, general relativity

The grand unified picture: Complex analysis is the bridge between algebra, geometry, and analysis. The Cauchy-Riemann equations link complex derivatives to harmonic functions (PDEs, Ch 6). Cauchy's theorem is Green's theorem in complex form (vector calculus, Ch 5). Laurent series connect algebra to topology (winding numbers). And conformal maps turn geometry into computation. No other branch of mathematics connects so many fields so elegantly.

"The shortest path between two truths in the real domain passes through the complex domain." — Jacques Hadamard

What is the connection between Cauchy's integral theorem and Green's theorem?

Cauchy's theorem is Green's theorem applied to the Cauchy-Riemann equations — the 2D curl and divergence vanish for analytic functions They are unrelated Green's theorem is a special case of Cauchy's theorem for real functions