Book 5A

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Course 5Book 5A: Complex Analysis and Conformal GeometryChapter 1

Holomorphic Functions Are Conformal

Complex derivatives as rotations and scalings, from airfoils to the Mercator map.

21 min read · Updated Oct 2, 2026

Read with Stein and Shakarchi, Complex Analysis, chapter 1, "Preliminaries to Complex Analysis" (complex numbers, holomorphic functions and the Cauchy–Riemann equations, power series, integration along curves). The conformal point of view is developed further in their chapter 8.

In this chapter · 6 sections
  1. 1.1A wing from a circle
  2. 1.2Complex differentiability
  3. 1.3Holomorphic maps preserve angles
  4. 1.4The basic examples
  5. 1.5History
  6. 1.6Exercises

This book is an optional side track. Nothing in the main line, from Book 6A to Perelman, depends on it: harmonic functions, the maximum principle and the two-dimensional Ricci flow all have their own canonical homes (6A.2 Harmonic Functions, 6A.4 Maximum Principles, 11A.7 Ricci Flow on Surfaces). It is here because complex analysis offers a second, very beautiful way of seeing those things, and because it ends at the uniformization theorem, the two-dimensional ancestor of the geometrization conjecture that Ricci flow proved in three dimensions.

The starting point is a single observation. A function ff of a complex variable is complex differentiable when its derivative, viewed as a 2×22 \times 2 real matrix, is a rotation followed by a scaling. So wherever f′(z)≠0f'(z) \neq 0, a holomorphic map preserves angles: it is conformal. Holomorphic functions are, geometrically, the angle-preserving maps of the plane, and that is why they solve problems in fluid flow, electrostatics and cartography, and why they lead to curvature.

By the end of this chapter you will be able to:

  • define complex differentiability and derive the Cauchy–Riemann equations;
  • explain why the derivative of a holomorphic function is a rotation plus a scaling, and why holomorphic maps preserve angles where f′≠0f' \neq 0;
  • work with the basic examples: powers, eze^z, the logarithm, Möbius transformations;
  • see the Mercator projection as a conformal map, and compute its area distortion;
  • use the Joukowsky map to turn flow past a circle into flow past a wing.

A wing from a circle

In the world In use The Joukowsky airfoil

Flow of an ideal fluid past a circular cylinder is easy to describe exactly. Flow past a wing is not. In the first decade of the twentieth century, Martin Kutta (1902) and Nikolai Joukowsky (Zhukovsky, 1906) used conformal maps to turn the first problem into the second. The Joukowsky map

J(z)=z+1zJ(z) = z + \frac1z

sends the unit circle onto the segment [−2,2][-2, 2] (since J(eiθ)=2cos⁡θJ(e^{i\theta}) = 2\cos\theta), and sends a slightly larger circle passing through z=1z = 1, with its centre shifted a little to the left and up, onto a smooth curve with a rounded nose and a sharp tail: a Joukowsky airfoil (Figure 1.1). Because JJ is holomorphic and conformal away from z=±1z = \pm1, it carries the streamlines of flow around the circle to streamlines of flow around the airfoil. The sharp tail comes from the point z=1z = 1, where J′(1)=0J'(1) = 0 and angles are doubled.

Two facts from this theory became the foundation of aerodynamics. The lift per unit span of a wing in a stream of density ρ\rho and speed VV is ρVΓ\rho V\Gamma, where Γ\Gamma is the circulation of the flow around it (the Kutta–Joukowski theorem), and the circulation is fixed by requiring the flow to leave the sharp trailing edge smoothly (the Kutta condition). Joukowsky airfoils are not the shapes of modern wings, but they were the first wing sections whose lift could be computed exactly, and they are still the standard first example in aerodynamics courses.

Figure 1.1. Left: a circle through z=1z = 1 with centre −0.08+0.08i-0.08 + 0.08i. Right: its image under J(z)=z+1/zJ(z) = z + 1/z (computed), a Joukowsky airfoil with its sharp trailing edge at J(1)=2J(1) = 2. The unit circle (dashed) is flattened onto the segment [−2,2][-2, 2].

Complex differentiability

Write z=x+iyz = x + iy and a function of zz as f=u+ivf = u + iv with u,vu, v real. Complex numbers multiply by rotating and scaling: multiplying by reiθre^{i\theta} rotates by θ\theta and scales by rr (2B.6 Power Series, Exponentials and Bump Functions).

Definition 1.1 Holomorphic function

Let Ω⊆C\Omega \subseteq \mathbb{C} be open. A function f:Ω→Cf : \Omega \to \mathbb{C} is complex differentiable at z0∈Ωz_0 \in \Omega if

f′(z0)=lim⁡h→0f(z0+h)−f(z0)hf'(z_0) = \lim_{h\to0}\frac{f(z_0 + h) - f(z_0)}{h}

exists, the limit being taken over complex h→0h \to 0. It is holomorphic on Ω\Omega if it is complex differentiable at every point.

The definition looks like the one of 2A.10 Derivatives, but the limit is over all directions in the plane at once, and that is a very strong condition. Polynomials in zz are holomorphic, with the usual derivatives; so are power series inside their disc of convergence (2B.6 Power Series, Exponentials and Bump Functions), in particular eze^z; so are quotients where the denominator doesn't vanish. But f(z)=zˉf(z) = \bar z is not: the difference quotient hˉh\frac{\bar h}{h} is 11 for real hh and −1-1 for imaginary hh.

Theorem 1.2 The Cauchy–Riemann equations

f=u+ivf = u + iv is complex differentiable at z0z_0 if and only if it is differentiable there as a map R2→R2\mathbb{R}^2 \to \mathbb{R}^2 (2B.8 Calculus in Several Variables) and

∂u∂x=∂v∂y,∂u∂y=−∂v∂x.\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.

Then f′(z0)=ux+ivxf'(z_0) = u_x + iv_x, and the real derivative is the matrix

Df(z0)=(uxuyvxvy)=(a−bba),f′(z0)=a+ib.Df(z_0) = \begin{pmatrix} u_x & u_y \\ v_x & v_y\end{pmatrix} = \begin{pmatrix} a & -b\\ b & a\end{pmatrix}, \qquad f'(z_0) = a + ib.

Proof. If f′(z0)=a+ibf'(z_0) = a + ib exists, then f(z0+h)=f(z0)+(a+ib)h+o(∣h∣)f(z_0 + h) = f(z_0) + (a + ib)h + o(|h|). Multiplication by a+iba + ib, written in real coordinates, sends (h1,h2)(h_1, h_2) to (ah1−bh2, bh1+ah2)(ah_1 - bh_2,\ bh_1 + ah_2): that is the matrix above, so ff is real differentiable with ux=vy=au_x = v_y = a, vx=−uy=bv_x = -u_y = b. Conversely, if Df(z0)Df(z_0) has this form, it is multiplication by a+iba + ib, and the real differentiability statement is the complex one.

The matrix (a−bba)\begin{pmatrix} a & -b\\ b & a\end{pmatrix} equals r(cos⁡θ−sin⁡θsin⁡θcos⁡θ)r\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix} with a+ib=reiθa + ib = re^{i\theta}: the derivative of a holomorphic function is a rotation by arg⁡f′(z0)\arg f'(z_0) followed by a scaling by ∣f′(z0)∣|f'(z_0)|. Its determinant is a2+b2=∣f′(z0)∣2≥0a^2 + b^2 = |f'(z_0)|^2 \geq 0, so holomorphic maps preserve orientation, and the area of a small region is multiplied by ∣f′∣2|f'|^2 (3A.5 Product Measures and Change of Variables).

A second consequence: if uu and vv are C2C^2, then Δu=uxx+uyy=vyx−vxy=0\Delta u = u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0, and likewise Δv=0\Delta v = 0. The real and imaginary parts of a holomorphic function are harmonic. (We will see in 5A.2 Cauchy’s Theorem and Its Consequences that holomorphic functions are automatically smooth.) This is the bridge from complex analysis to the Laplace equation, developed in 5A.4 Harmonic Functions and Conformal Mapping.

Holomorphic maps preserve angles

Proposition 1.3 Conformality

Let ff be holomorphic near z0z_0 with f′(z0)≠0f'(z_0) \neq 0. If two smooth curves cross at z0z_0 at angle ϕ\phi, their images cross at f(z0)f(z_0) at the same angle ϕ\phi, with the same orientation. Infinitesimal circles are mapped to infinitesimal circles.

Proof. A curve γ\gamma with γ(0)=z0\gamma(0) = z_0 has image f∘γf\circ\gamma, with tangent (f∘γ)′(0)=f′(z0)γ′(0)(f\circ\gamma)'(0) = f'(z_0)\gamma'(0) (chain rule, 2B.8 Calculus in Several Variables). Multiplying every tangent vector by the same non-zero complex number rotates them all by the same angle and scales them by the same factor, so the angle between two tangent vectors is unchanged. A small circle is mapped, to first order, by a rotation-scaling, which maps circles to circles.

A map with this property is conformal. Conversely, an orientation-preserving C1C^1 map of a plane region whose derivative is everywhere a non-zero multiple of a rotation is holomorphic, by Theorem 1.2. So in the plane, conformal = holomorphic with non-vanishing derivative. (Orientation-reversing conformal maps are the conjugates of holomorphic ones.)

Where f′(z0)=0f'(z_0) = 0 conformality fails, in a specific way: if f(z)−f(z0)f(z) - f(z_0) vanishes to order kk at z0z_0, angles at z0z_0 are multiplied by kk. For f(z)=z2f(z) = z^2 at 00, the two coordinate axes, at right angles, go to the positive and negative real axis, at a straight angle (Figure 1.2). This is the mechanism behind the Joukowsky airfoil's cusp.

Figure 1.2. A grid in the first quadrant and its image under z↦z2z \mapsto z^2 (computed). Horizontal and vertical lines become families of parabolas that still cross at right angles, because z2z^2 is conformal away from 00. At 00, where the derivative vanishes, the right angle of the quadrant is doubled to a straight angle.

The basic examples

Powers and exponentials. znz^n maps the sector of angle 2πn\frac{2\pi}n onto the plane minus a ray, multiplying angles at 00 by nn. The exponential ez=ex(cos⁡y+isin⁡y)e^z = e^x(\cos y + i\sin y) (2B.6 Power Series, Exponentials and Bump Functions) maps horizontal lines to rays from 00 and vertical lines to circles around 00, conformally, since (ez)′=ez≠0(e^z)' = e^z \neq 0. It maps every horizontal strip of height 2π2\pi onto the plane minus 00, which is the planar version of the observation in 2B.9 The Inverse and Implicit Function Theorems that a locally invertible map need not be globally injective.

Logarithms. On a region that doesn't wind around 00, such as the plane minus the negative real axis, eze^z has a holomorphic inverse, a branch of the logarithm: log⁡z=log⁡∣z∣+iarg⁡z\log z = \log|z| + i\arg z with arg⁡z∈(−π,π)\arg z \in (-\pi, \pi). It maps rays to horizontal lines and circles to vertical segments. There is no continuous logarithm on the whole punctured plane, because going once around 00 increases the argument by 2π2\pi; this is the first appearance of a topological obstruction, the winding number (7A.4 The Fundamental Group), in analysis.

Möbius transformations. For ad−bc≠0ad - bc \neq 0, T(z)=az+bcz+dT(z) = \frac{az + b}{cz + d} is holomorphic except at z=−d/cz = -d/c, with T′(z)=ad−bc(cz+d)2≠0T'(z) = \frac{ad - bc}{(cz + d)^2} \neq 0. These maps send circles and lines to circles and lines (Exercise 1.6), and compose like 2×22 \times 2 matrices. Two of them are used constantly: the Cayley map z↦z−iz+iz \mapsto \frac{z - i}{z + i}, a conformal bijection from the upper half-plane onto the unit disc, and the disc automorphisms z↦eiθz−a1−aˉzz \mapsto e^{i\theta}\frac{z - a}{1 - \bar az} with ∣a∣<1|a| < 1, which map the unit disc conformally onto itself. In 5A.4 Harmonic Functions and Conformal Mapping the Schwarz lemma shows these are all the conformal bijections of the disc.

In the world Data The Mercator projection is conformal

Gerardus Mercator's world map of 1569 places a point at longitude λ\lambda and latitude ϕ\phi at

X=λ,Y=log⁡tan⁡(π4+ϕ2).X = \lambda, \qquad Y = \log\tan\Big(\frac\pi4 + \frac\phi2\Big).

It was designed for navigation: a course of constant compass bearing (a rhumb line) crosses every meridian at the same angle, so on a map that preserves angles it must be a straight line (2B.1 Metric Spaces's figure). The Mercator projection is exactly such a map. In complex terms it is the composition of stereographic projection from the north pole, which maps the sphere conformally to the plane, with the logarithm, which maps the punctured plane conformally to a strip, followed by a reflection that swaps the two axes (Exercise 1.8). All three preserve angles, hence so does the composition.

The price is area. At latitude ϕ\phi the map stretches lengths in every direction by the same factor sec⁡ϕ\sec\phi, so areas by sec⁡2ϕ\sec^2\phi: a factor of 44 at 60°60° and about 10.510.5 at Greenland's typical latitude of 72°72°. Greenland (about 2.22.2 million km²) appears on a Mercator map roughly as large as Africa (about 3030 million km²), which is fourteen times larger. Gauss's Theorema Egregium (8A.9 The Curvature of Surfaces) explains why no map can avoid this: a map of a curved surface onto the plane cannot preserve both angles and areas, because the sphere's curvature is not zero. In 5A.5 Uniformization and the Two-Dimensional Ricci Flow the same stretching factor appears as the conformal factor e2ue^{2u} of a metric, and the curvature is computed from it.

Where this goes Conformal maps and conformal metrics

A holomorphic map ff pulls back the Euclidean metric ∣dw∣2|dw|^2 to ∣f′(z)∣2∣dz∣2|f'(z)|^2|dz|^2: a metric that measures lengths by the factor ∣f′∣|f'|, the same in all directions. Metrics of this form, e2u∣dz∣2e^{2u}|dz|^2, are called conformal metrics, and every metric on a surface can be written this way in suitable local coordinates (isothermal coordinates). In 5A.5 Uniformization and the Two-Dimensional Ricci Flow the Gaussian curvature of e2u∣dz∣2e^{2u}|dz|^2 is computed to be −e−2uΔu-e^{-2u}\Delta u, which turns the geometric problem of finding a metric of constant curvature into a nonlinear PDE for uu. And under the two-dimensional Ricci flow, a conformal metric stays conformal, and the flow becomes a single heat-type equation for uu (5A.5 Uniformization and the Two-Dimensional Ricci Flow, 11A.7 Ricci Flow on Surfaces).

History

Jean d'Alembert (1752) and Leonhard Euler met the Cauchy–Riemann equations in hydrodynamics, as the conditions for a two-dimensional flow to be incompressible and irrotational. Augustin-Louis Cauchy developed complex integration from 1814, and Bernhard Riemann's 1851 thesis made conformal mapping and the equations named after both men the foundation of the subject. Gauss's 1822 prize essay for the Copenhagen Academy treated conformal maps between surfaces in general, motivated by geodesy. Mercator's map dates from 1569; stereographic projection goes back to antiquity (Hipparchus and Ptolemy). Kutta's and Joukowsky's work on lift dates from 1902 and 1906.

Recall Where we stand

A function of a complex variable is holomorphic when its complex derivative exists; equivalently, its real derivative is a rotation followed by a scaling, which is the content of the Cauchy–Riemann equations. Holomorphic maps preserve angles and orientation wherever f′≠0f' \neq 0, and multiply angles where f′f' vanishes. Their real and imaginary parts are harmonic. Powers, exponentials, logarithms and Möbius transformations are the basic conformal maps. The Mercator projection is conformal and distorts area by sec⁡2ϕ\sec^2\phi; conformal metrics e2u∣dz∣2e^{2u}|dz|^2 are where curvature enters. 5A.2 Cauchy’s Theorem and Its Consequences integrates holomorphic functions along curves and finds that they are far more rigid than real differentiable ones.

Exercises

Exercise 1.4 Checking the Cauchy–Riemann equations

Decide which are holomorphic, and where: (a) z3z^3; (b) ∣z∣2|z|^2; (c) ezˉe^{\bar z}; (d) 1z\frac{1}{z}; (e) x2−y2+2ixyx^2 - y^2 + 2ixy; (f) x−iyx - iy. For the holomorphic ones, compute f′f' from ux+ivxu_x + iv_x.

Solution

(a) Everywhere, 3z23z^2. (b) Only at 00 (the equations 2x=02x = 0, 2y=02y = 0). (c) Nowhere. (d) On C∖{0}\mathbb{C}\setminus\{0\}, −1/z2-1/z^2. (e) This is z2z^2: everywhere, 2z2z. (f) This is zˉ\bar z: nowhere.

Exercise 1.5 Angles at a zero of the derivative

Show that if f(z)−f(z0)=(z−z0)kg(z)f(z) - f(z_0) = (z - z_0)^kg(z) with gg holomorphic and g(z0)≠0g(z_0) \neq 0, then curves meeting at z0z_0 at angle ϕ\phi have images meeting at angle kϕk\phi.

Exercise 1.6 Möbius transformations map circles to circles

(a) Show that every Möbius transformation is a composition of translations z↦z+bz \mapsto z + b, scalings z↦azz \mapsto az and the inversion z↦1/zz \mapsto 1/z. (b) Show that 1/z1/z maps the family of circles and lines {A∣z∣2+Bˉz+Bzˉ+C=0}\{A|z|^2 + \bar Bz + B\bar z + C = 0\} (A,CA, C real) to itself. (c) Show that the Cayley map z↦z−iz+iz \mapsto \frac{z - i}{z + i} sends the real axis onto the unit circle and the upper half-plane into the disc.

Exercise 1.7 Disc automorphisms

For ∣a∣<1|a| < 1 and ϕa(z)=z−a1−aˉz\phi_a(z) = \frac{z - a}{1 - \bar az}, show that ∣ϕa(z)∣=1|\phi_a(z)| = 1 when ∣z∣=1|z| = 1, that ϕa\phi_a maps the disc into itself, and that ϕa−1=ϕ−a\phi_a^{-1} = \phi_{-a}. Compute ϕa′(0)\phi_a'(0) and ϕa′(a)\phi_a'(a).

Solution

For ∣z∣=1|z| = 1, ∣1−aˉz∣=∣zˉ−aˉ∣=∣z−a∣|1 - \bar az| = |\bar z - \bar a| = |z - a|. By the maximum modulus principle (5A.2 Cauchy’s Theorem and Its Consequences) or a direct computation, ∣ϕa∣<1|\phi_a| < 1 inside. ϕa′(z)=1−∣a∣2(1−aˉz)2\phi_a'(z) = \frac{1 - |a|^2}{(1 - \bar az)^2}, so ϕa′(0)=1−∣a∣2\phi_a'(0) = 1 - |a|^2 and ϕa′(a)=11−∣a∣2\phi_a'(a) = \frac{1}{1 - |a|^2}.

Exercise 1.8 Mercator as a logarithm

Stereographic projection from the north pole onto the equatorial plane sends a point of the unit sphere at longitude λ\lambda and latitude ϕ\phi to the complex number w=tan⁡(π4+ϕ2)eiλw = \tan\big(\frac\pi4 + \frac\phi2\big)e^{i\lambda} (check this with a little trigonometry, or take it as given). Show that log⁡w=Y+iX\log w = Y + iX in the notation of the Mercator formula, where X=λX = \lambda and Y=log⁡tan⁡(π4+ϕ2)Y = \log\tan(\frac\pi4 + \frac\phi2), so Mercator's map is the logarithm followed by the reflection (a,b)↦(b,a)(a, b) \mapsto (b, a). Then show that dYdϕ=sec⁡ϕ\frac{dY}{d\phi} = \sec\phi, and deduce that the map stretches lengths by sec⁡ϕ\sec\phi in both directions (a parallel at latitude ϕ\phi has length 2πcos⁡ϕ2\pi\cos\phi on the unit sphere but 2π2\pi on the map).

Solution

log⁡w=log⁡tan⁡(π4+ϕ2)+iλ=Y+iX\log w = \log\tan(\frac\pi4 + \frac\phi2) + i\lambda = Y + iX. Differentiating, dYdϕ=12sec⁡2(π4+ϕ2)tan⁡(π4+ϕ2)=12sin⁡(π4+ϕ2)cos⁡(π4+ϕ2)=1sin⁡(π2+ϕ)=sec⁡ϕ\frac{dY}{d\phi} = \frac{\frac12\sec^2(\frac\pi4 + \frac\phi2)}{\tan(\frac\pi4 + \frac\phi2)} = \frac{1}{2\sin(\frac\pi4 + \frac\phi2)\cos(\frac\pi4 + \frac\phi2)} = \frac{1}{\sin(\frac\pi2 + \phi)} = \sec\phi. North–south, a length dϕd\phi on the sphere becomes sec⁡ϕ dϕ\sec\phi\,d\phi; east–west, cos⁡ϕ dλ\cos\phi\,d\lambda becomes dλd\lambda: the same factor sec⁡ϕ\sec\phi.

Exercise 1.9 The Joukowsky map

(a) Show that J(z)=z+1/zJ(z) = z + 1/z maps the circle ∣z∣=R>1|z| = R > 1 onto an ellipse with semi-axes R+1RR + \frac1R and R−1RR - \frac1R, and the unit circle onto [−2,2][-2, 2]. (b) Show JJ is conformal except at z=±1z = \pm1, and that it maps the exterior of the unit disc bijectively onto C∖[−2,2]\mathbb{C}\setminus[-2, 2].

Exercise 1.10 Rehearsal: the conformal factor of a holomorphic map is harmonic

Let ff be holomorphic on Ω\Omega with f′≠0f' \neq 0, and let u=log⁡∣f′∣u = \log|f'|, so that the pullback of the flat metric is e2u∣dz∣2e^{2u}|dz|^2. Show that uu is harmonic. (Locally f′=egf' = e^g for a holomorphic gg, by taking a branch of the logarithm, and u=Re⁡gu = \operatorname{Re}g.) In 5A.5 Uniformization and the Two-Dimensional Ricci Flow the curvature of e2u∣dz∣2e^{2u}|dz|^2 is −e−2uΔu-e^{-2u}\Delta u; this exercise says the pullback of a flat metric is flat, as it must be, since a conformal map is a local isometry between e2u∣dz∣2e^{2u}|dz|^2 and the flat plane.

Solution

Near any point, f′f' is non-vanishing and holomorphic, so (on a disc) f′=egf' = e^g with gg holomorphic (5A.2 Cauchy’s Theorem and Its Consequences gives the existence of such a gg on simply connected regions). Then ∣f′∣=eRe⁡g|f'| = e^{\operatorname{Re}g} and u=Re⁡gu = \operatorname{Re}g, the real part of a holomorphic function, which is harmonic.

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