Cauchy-Goursat Theorem

Derivation

Cauchy-Goursat Theorem

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Statement

Let \( f \) be analytic (complex-differentiable at every point) on a simply connected open domain \( D \subseteq \mathbb{C} \), and let \( C \) be any closed rectifiable contour lying in \( D \). Then \[ \oint_C f(z)\,dz = 0 . \] Crucially, the proof requires only the existence of \( f'(z) \) at each point of \( D \) — it does not assume that \( f' \) is continuous. This is Goursat's sharpening of Cauchy's original theorem, and it is what makes analyticity such a rigid, self-reinforcing property: once \( f' \) merely exists on \( D \), the vanishing of all closed-contour integrals follows, and from it (via the Cauchy integral formula) the continuity of \( f' \), infinite differentiability, and the power-series expansion all cascade out.

Why it matters

This theorem is the load-bearing wall of complex analysis. Every downstream tool a physicist uses — the Cauchy integral formula, residue calculus, dispersion relations (Kramers–Kronig), Wick rotation of path integrals, contour evaluation of Green's functions and propagators — rests on the freedom to deform integration contours without changing the integral. That freedom is exactly the content of Cauchy–Goursat: two contours sharing endpoints (or two closed contours that can be deformed into one another within the analyticity domain) give the same integral, because their difference is a closed contour integral of an analytic function.

The "without assuming \( f' \) continuous" refinement is not pedantry. In field theory and scattering problems one often knows only that a function of a complexified energy or momentum is differentiable in some region (e.g. from a convergent integral representation), with no independent handle on the smoothness of its derivative. Goursat's argument guarantees that bare differentiability already suffices — analyticity cannot be "weakly" possessed. It is also a paradigm of a compactness-plus-local-linearity argument that reappears across analysis.

Assumptions

Assumption.\( f \) is complex-differentiable at every point of the open domain \( D \). If \( f' \) fails to exist at even one interior point enclosed by \( C \), the conclusion collapses: \( f(z) = 1/z \) is smooth as a map \( \mathbb{R}^2\setminus\{0\} \to \mathbb{R}^2 \) yet \( \oint_{|z|=1} dz/z = 2\pi i \neq 0 \) because analyticity fails at the single point \( z=0 \).

Assumption.\( D \) is simply connected (every closed loop in \( D \) can be continuously shrunk to a point within \( D \)). Drop this and the theorem fails for loops encircling a hole: on the annulus \( 1 < |z| < 3 \), \( f(z)=1/z \) is analytic everywhere, yet \( \oint_{|z|=2} dz/z = 2\pi i \). What survives on multiply connected domains is the weaker deformation theorem: the integral depends only on the homotopy class of the loop.

Assumption.\( C \) is a closed rectifiable (finite-length) contour lying entirely in \( D \). If \( C \) has infinite length, the ML-type estimates that close the proof lose all force, and the integral itself may fail to exist as a limit of Riemann–Stieltjes sums.

Assumption (what is not assumed).Continuity of \( f' \) is not required. If one does assume \( f' \) continuous, a two-line proof via Green's theorem and the Cauchy–Riemann equations suffices (shown below as the standard route) — but that proof is logically circular in the larger development of the subject, because continuity of \( f' \) is usually derived from this theorem via the Cauchy integral formula. Goursat's bisection argument breaks the circle.

Assumption (implicit).Completeness of \( \mathbb{C} \). The Goursat argument extracts a limit point from a nested sequence of compact triangles (Cantor intersection). Over a non-complete field, e.g. working only with rational-coordinate points, the nested triangles could close down on "nothing" and the proof would fail at the final localization step.

Derivation

We give two passes. Route I (standard treatment) assumes \( f' \) continuous and uses Green's theorem plus the Cauchy–Riemann equations. Route II (Goursat's bisection) removes the continuity assumption; its steps carry the rigour tag.

\[ f(z) = u(x,y) + i\,v(x,y), \qquad dz = dx + i\,dy \]

Decompose \( f \) into real and imaginary parts and expand the complex differential; pure definition, no analysis yet. A

\[ \oint_C f\,dz = \oint_C \left( u\,dx - v\,dy \right) + i \oint_C \left( v\,dx + u\,dy \right) \]

Multiply out \( (u+iv)(dx+i\,dy) \) and collect real and imaginary parts; the complex contour integral is defined as this pair of real line integrals. A

\[ \oint_C f\,dz = \iint_R \left( -\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) dA \;+\; i \iint_R \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) dA \]

Apply Green's theorem in the plane (prior result) to each real line integral, with \( R \) the region enclosed by \( C \). This step is only legal if the partial derivatives are continuous on \( R \) — this is exactly where the provisional assumption on \( f' \) enters Route I. B

\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \;\;\Longrightarrow\;\; \oint_C f\,dz = 0 \]

Insert the Cauchy–Riemann equations (prior result, valid wherever \( f' \) exists): both area integrands vanish identically. Route I complete — but only under the unwanted continuity hypothesis. A

\[ I \equiv \oint_{\partial\Delta} f(z)\,dz, \qquad I = \sum_{k=1}^{4} \oint_{\partial\Delta^{(k)}} f(z)\,dz \]

Route II begins. Reduce first to a triangle \( \Delta \subset D \). Join the midpoints of its sides, producing four congruent sub-triangles \( \Delta^{(1)},\dots,\Delta^{(4)} \). Summing the four boundary integrals, every interior segment is traversed once in each direction and cancels, leaving exactly \( I \). Pure orientation bookkeeping. A

\[ |I| \le 4\,\bigl|I^{(k^*)}\bigr| \quad \text{for the sub-triangle } \Delta_1 \equiv \Delta^{(k^*)} \text{ with largest } \bigl|I^{(k)}\bigr| \]

Triangle inequality on the four-term sum: at least one term has modulus \( \ge |I|/4 \). Select it (choice among finitely many — no axiom-of-choice subtlety). B

\[ \Delta \supset \Delta_1 \supset \Delta_2 \supset \cdots, \qquad |I| \le 4^n |I_n|, \qquad L_n = \frac{L}{2^n}, \quad d_n = \frac{d}{2^n} \]

Iterate the bisection. Each generation quarters the enclosed integral bound and halves both the perimeter \( L_n \) of \( \partial\Delta_n \) and the diameter \( d_n \) of \( \Delta_n \) (midpoint triangles are similar with ratio \( \tfrac{1}{2} \)). Induction on \( n \). B

\[ \exists!\; z_0 \in \bigcap_{n=0}^{\infty} \Delta_n \]

The \( \Delta_n \) are nested, closed, nonempty, with \( d_n \to 0 \): by the Cantor intersection theorem (completeness of \( \mathbb{C} \)) the intersection is a single point \( z_0 \in \Delta \subset D \). This is the compactness step that localizes the whole problem at one point. C

\[ f(z) = f(z_0) + f'(z_0)(z - z_0) + \eta(z)\,(z - z_0), \qquad |\eta(z)| < \varepsilon \;\text{ for } |z - z_0| < \delta \]

Differentiability of \( f \) at the single point \( z_0 \) — the only place analyticity is invoked, and only pointwise. This is the definition of \( f'(z_0) \) rewritten with an explicit remainder \( \eta(z) \equiv \frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) \), \( \eta(z_0)\equiv 0 \): for any \( \varepsilon > 0 \) there is \( \delta > 0 \) making \( |\eta| < \varepsilon \) on the disc. C

\[ \oint_{\partial\Delta_n} dz = 0, \qquad \oint_{\partial\Delta_n} (z - z_0)\,dz = 0 \;\;\Longrightarrow\;\; I_n = \oint_{\partial\Delta_n} \eta(z)(z - z_0)\,dz \]

The constant and linear pieces integrate to zero over any closed contour because they possess global primitives \( z \) and \( \tfrac{1}{2}(z-z_0)^2 \): the fundamental theorem for contour integrals needs only an antiderivative, no analyticity theory. Only the remainder term survives. B

\[ |I_n| \le \max_{z \in \partial\Delta_n} \bigl| \eta(z)(z - z_0) \bigr| \cdot L_n \le \varepsilon \, d_n L_n = \varepsilon\, \frac{d\,L}{4^n} \]

ML estimate on \( \partial\Delta_n \): for \( n \) large enough that \( d_n < \delta \) (possible since \( d_n \to 0 \) and \( z_0 \in \Delta_n \)), every \( z \in \partial\Delta_n \) satisfies \( |z - z_0| \le d_n \) and \( |\eta(z)| < \varepsilon \). C

\[ |I| \le 4^n |I_n| \le 4^n \cdot \varepsilon\,\frac{d\,L}{4^n} = \varepsilon\, d\, L \quad \forall\,\varepsilon > 0 \;\;\Longrightarrow\;\; I = 0 \]

The factor \( 4^n \) accumulated by bisection is exactly cancelled by the \( 4^{-n} \) shrinkage of \( d_n L_n \) — this precise cancellation is the engine of Goursat's proof. Since \( \varepsilon \) is arbitrary and \( d, L \) are fixed, \( |I| = 0 \). The theorem holds for every triangle in \( D \). C

\[ F(z) \equiv \int_{z_a}^{z} f(\zeta)\,d\zeta \;\;\text{(polygonal path in } D\text{)}, \qquad F'(z) = f(z) \]

Extend from triangles to arbitrary closed contours. Fix \( z_a \in D \) and define \( F \) by integrating along polygonal paths; simple connectedness of \( D \) plus the triangle result makes \( F \) path-independent (any two polygonal paths bound a region triangulable within \( D \)). The difference quotient of \( F \) then equals \( f \) plus an error controlled by continuity of \( f \) (differentiable \( \Rightarrow \) continuous), so \( F' = f \) on \( D \). B

\[ \oint_C f(z)\,dz = F\!\left(z_{\text{end}}\right) - F\!\left(z_{\text{start}}\right) = 0 \]

Fundamental theorem for contour integrals applied to the primitive \( F \): for a closed contour the endpoints coincide, so the integral vanishes. This holds for every closed rectifiable contour in \( D \), completing the proof with no continuity assumption on \( f' \). A

Result

\[ \oint_C f(z)\,dz = 0 \qquad \text{for every closed contour } C \subset D, \; f \text{ analytic on simply connected } D \]

Reading. An analytic function has no "circulation" and no "flux" around any closed loop inside its domain of analyticity: closed-loop integrals detect only obstructions — singularities or holes in the domain — and here there are none. Equivalently, contour integrals of analytic functions are path-independent, analytic functions always admit local antiderivatives, and integration contours may be freely deformed without changing the value of the integral, provided the deformation never crosses a singularity or leaves the domain.

Units check. The integral carries dimensions \( [f]\cdot[z] \). If \( z \) is a complexified length (m) and \( f \) a field amplitude (e.g. V·m\(^{-1}\)), then \( \oint f\,dz \) carries V, and the theorem asserts this quantity is exactly \( 0 \) V — a zero of the correct dimension. Both routes are dimensionally consistent throughout: in Route I each Green's-theorem integrand \( \partial u/\partial x \) carries \( [f]/[z] \) against the area element \( [z]^2 \), reproducing \( [f][z] \); in Route II the bound \( \varepsilon\,d\,L \) carries \( [f]/[z]\cdot[z]\cdot[z] = [f][z] \) (note \( \eta \), a difference quotient minus a derivative, carries \( [f]/[z] \)), matching \( |I| \).

Limiting cases

\( f' \) continuous: the theorem degenerates gracefully to Cauchy's original 1825 result, provable in two lines from Green's theorem + Cauchy–Riemann (Steps 1–4). Goursat's machinery is then optional, not wrong.
\( f \) a polynomial: an explicit primitive exists globally (\( \int z^n dz = z^{n+1}/(n+1) \)), and the result follows from Steps 13–14 alone with no bisection.
Contour shrunk to a point: \( L \to 0 \) with \( f \) bounded near the point forces \( |\oint| \le M L \to 0 \) trivially; the theorem's content lives at finite loop size.
Real-axis restriction: for \( f \) real-analytic and \( C \) a degenerate "loop" running \( a \to b \to a \) along \( \mathbb{R} \), the statement reduces to \( \int_a^b f\,dx + \int_b^a f\,dx = 0 \), the trivial orientation identity of the fundamental theorem of calculus.
Multiply connected domain, contractible loop: if \( C \) does not encircle any hole, the theorem still applies inside a simply connected sub-domain containing \( C \) — simple connectedness is needed only "as seen by the loop".

Breaks when

An enclosed singularity. \( f(z) = 1/z \) on any contour enclosing the origin: \( \oint_{|z|=R} dz/z = 2\pi i \) for every \( R > 0 \). One bad point destroys the conclusion; this failure is not a defect but the doorway to the residue theorem, which quantifies exactly how the theorem breaks.
Multiply connected domains. On the punctured plane or an annulus, \( f \) may be analytic at every point of the domain and the integral still nonzero (same \( 1/z \) example): the loop wraps a topological hole. Closed-contour integrals then classify loops by winding number — the integral becomes a homotopy invariant rather than zero.
Mere real-differentiability (smoothness without Cauchy–Riemann). \( f(z) = \bar{z} \) is infinitely differentiable as a map of \( \mathbb{R}^2 \), yet \( \oint_{|z|=1} \bar{z}\,dz = 2\pi i \). Complex differentiability, not smoothness, is the operative hypothesis.
Non-rectifiable contours. On a fractal curve of infinite length (e.g. a Koch-type loop) the ML estimates of Steps 11–12 are vacuous and the contour integral itself need not be defined; the theorem does not even parse.
Analyticity only on the contour, not inside. \( f(z) = 1/z \) is analytic on the circle \( |z| = 1 \) itself, but the theorem demands analyticity on a simply connected domain containing \( C \) — i.e. throughout the enclosed region. The hypothesis is about the region \( C \) bounds, not the curve.

Failure modes

The Green's-theorem shortcut presented as the full proof. Quoting Steps 1–4 alone silently assumes \( u, v \) have continuous partials, i.e. \( f' \) continuous — the very hypothesis Goursat eliminated. In a rigorous development this is circular, since continuity of \( f' \) is later deduced from this theorem.
"Analytic except at one point, so approximately zero." Students apply the theorem to \( 1/(z - a) \) with \( a \) inside \( C \), reasoning the exceptional point is "measure zero". The integral is \( 2\pi i \), not small: analyticity must hold at every enclosed point.
Confusing smooth with analytic. Applying the theorem to \( \bar{z} \), \( |z|^2 \), or \( \mathrm{Re}\,z \) because they "look differentiable". None satisfies Cauchy–Riemann on any open set; their closed-loop integrals are generically nonzero (e.g. \( \oint \bar z\,dz = 2i \times \text{enclosed area} \)).
Ignoring the topology of the domain. Declaring \( \oint_{|z|=2} dz/z = 0 \) "because \( 1/z \) is analytic on the annulus \( 1<|z|<3 \) containing the contour". The annulus is not simply connected; the enclosed disc contains the singularity.
Orientation and closure slips. Applying the result to an open arc, or summing sub-contour contributions with inconsistent orientation so interior segments fail to cancel in Step 5.
Misplacing where \( \varepsilon \) enters Goursat's argument. Choosing \( \varepsilon \) after \( n \), or letting \( z_0 \) depend on \( \varepsilon \). The order is: bisect and find \( z_0 \) first (Steps 5–8); then, for each \( \varepsilon \), pick \( \delta \), then \( n \) large enough that \( \Delta_n \) fits in the \( \delta \)-disc.

Discussion

Physically, Cauchy–Goursat is a conservation statement. Writing \( f = u + iv \) and reading Steps 1–2 hydrodynamically, the two real line integrals are the circulation and flux of the plane vector field \( (u, -v) \) — the Pólya field of \( f \). The Cauchy–Riemann equations say this field is simultaneously irrotational and incompressible: the flow of an ideal fluid with no sources, sinks, or vortices in the region. The theorem then asserts what any physicist expects of such a flow: zero net circulation and zero net flux through every closed loop. Every singularity of \( f \) is, in this picture, a point source, sink, or vortex, and the residue theorem is the corresponding "Gauss's law" that counts them.

The theorem is the reason contour deformation works, and contour deformation is arguably the single most-used trick in theoretical physics. Kramers–Kronig relations follow from deforming response-function integrals in the upper half of the complexified frequency plane, with causality supplying the analyticity. Wick rotation of Feynman integrals from Minkowski to Euclidean momenta is a 90-degree contour rotation legitimized by Cauchy–Goursat in the region between the contours, with the \( i\epsilon \) prescription keeping the poles out of the swept sector. Steepest-descent and saddle-point evaluations begin by deforming the contour onto the path of stationary phase — legal precisely because the integrand is analytic in the swept region.

Structurally, the theorem is the analytic seed of an equivalence that has no counterpart in real analysis: on a simply connected domain, (i) \( f \) differentiable once, (ii) \( \oint f\,dz = 0 \) for all closed contours, (iii) \( f \) has an analytic primitive, and (iv) \( f \) locally a convergent power series are all equivalent. Goursat's contribution closes the loop (i) \( \Rightarrow \) (ii); Morera's theorem is the converse (ii) \( \Rightarrow \) (i)-with-more, and the Cauchy integral formula converts (ii) into (iv). In the language of differential forms, the theorem says \( f\,dz \) is a closed 1-form (\( d(f\,dz) = \bar{\partial}f \, d\bar z \wedge dz = 0 \) by Cauchy–Riemann), and simple connectedness upgrades closed to exact — Cauchy–Goursat is thus the complex-analytic face of de Rham's theorem in degree one, with the \( 2\pi i \) of an enclosed pole as the period of a nontrivial cohomology class. Goursat's bisection argument itself is the elementary prototype of a compactness-localization proof: a global claim is reduced, by dyadic subdivision and Cantor intersection, to an infinitesimal claim at a single point, where the bare definition of the derivative is strong enough to finish.

Common misconceptions. "The theorem computes integrals" — it computes only the zero ones; nonzero contour integrals come from its failure at singularities, organized by the residue theorem. "Analyticity on the curve suffices" — the hypothesis governs the full region the curve bounds. "Goursat is just a fussier proof of the same statement" — it is a strictly stronger statement (no continuity of \( f' \) assumed), and that strength is what lets the Cauchy integral formula subsequently prove that analytic functions are automatically \( C^\infty \), rather than assuming it.

Worked examples

Example 1 — direct verification on a circle. Verify Cauchy–Goursat explicitly for \( f(z) = z^2 \) on the circle \( C: |z| = R \) with \( R = 3 \) (an entire function; the theorem predicts zero, and the parametrization must agree).

\[ z(\theta) = R e^{i\theta}, \quad dz = i R e^{i\theta} d\theta, \quad \theta \in [0, 2\pi] \]

Parametrize the contour by the polar angle; \( dz \) follows by differentiating \( z(\theta) \). A

\[ \oint_C z^2\,dz = \int_0^{2\pi} R^2 e^{2i\theta} \cdot i R e^{i\theta}\, d\theta = i R^3 \int_0^{2\pi} e^{3i\theta}\, d\theta \]

Substitute the parametrization and pull the constants \( iR^3 \) out of the integral — symbols first, the number \( R = 3 \) enters only at the end. A

\[ \int_0^{2\pi} e^{3i\theta}\, d\theta = \left[ \frac{e^{3i\theta}}{3i} \right]_0^{2\pi} = \frac{e^{6\pi i} - 1}{3i} = \frac{1 - 1}{3i} = 0 \]

Elementary antiderivative; \( e^{6\pi i} = \cos 6\pi + i \sin 6\pi = 1 \) by periodicity. A

\[ \oint_C z^2\,dz = i R^3 \cdot 0 = i\,(3)^3 \cdot 0 = 0 \]

Insert \( R = 3 \) (dimensionless complex-plane units): the \( 27i \) prefactor multiplies an exactly zero angular integral. Consistent with Cauchy–Goursat: \( z^2 \) is entire and the disc \( |z| \le 3 \) is simply connected. A

\[ \oint_{|z|=3} z^2\, dz = 0 \]

Reading. Direct computation reproduces the theorem's prediction exactly — the \( e^{3i\theta} \) phase winds three full turns and averages to zero. In dimensionless complex-plane units the result is \( 0 \); if \( z \) carried metres, the integral would carry m\(^3\) and equal \( 0 \) m\(^3\).

Units check. \( [z^2][dz] = \text{m}^2 \cdot \text{m} = \text{m}^3 \) if \( z \) is a length; the vanishing result is dimensionally unambiguous.

Example 2 — contour shifting: the displaced Gaussian. Use Cauchy–Goursat to prove that \( \int_{-\infty}^{\infty} e^{-(x + ib)^2}\, dx \) is independent of the real shift \( b \), and evaluate it for \( b = 1.5 \) (dimensionless). This identity underlies the Fourier transform of a Gaussian and the completion-of-the-square step in every Gaussian path integral.

\[ f(z) = e^{-z^2}, \qquad \oint_{\partial\mathcal{R}} f(z)\,dz = 0, \quad \mathcal{R} = \{ x + iy : |x| \le R,\; 0 \le y \le b \} \]

\( e^{-z^2} \) is entire (composition of entire functions), and the closed rectangle \( \mathcal{R} \) with corners \( \pm R \) and \( \pm R + ib \) lies in the simply connected domain \( \mathbb{C} \): Cauchy–Goursat applies to its boundary. B

\[ \int_{-R}^{R} e^{-x^2} dx \;+\; \int_0^b e^{-(R+iy)^2} i\,dy \;-\; \int_{-R}^{R} e^{-(x+ib)^2} dx \;-\; \int_0^b e^{-(-R+iy)^2} i\,dy = 0 \]

Decompose \( \partial\mathcal{R} \) into bottom, right side, top (traversed right-to-left, hence the minus sign), and left side (downward, minus sign). Orientation is counterclockwise throughout. A

\[ \left| e^{-(\pm R + iy)^2} \right| = e^{-\mathrm{Re}\left[(\pm R + iy)^2\right]} = e^{-(R^2 - y^2)} \le e^{-R^2} e^{b^2} \]

Bound the vertical-side integrands: \( (\pm R + iy)^2 = R^2 - y^2 \pm 2iRy \), and \( |e^w| = e^{\mathrm{Re}\,w} \). Each vertical side then obeys \( |\int| \le b\, e^{b^2} e^{-R^2} \to 0 \) as \( R \to \infty \) (ML estimate with length \( b \)). C

\[ \int_{-\infty}^{\infty} e^{-(x+ib)^2}\, dx = \int_{-\infty}^{\infty} e^{-x^2}\, dx = \sqrt{\pi} \]

Take \( R \to \infty \) in Step 2: the vertical sides vanish, leaving top = bottom. The remaining real Gaussian integral is the standard \( \sqrt{\pi} \). The contour has been shifted off the real axis at zero cost — pure Cauchy–Goursat. B

\[ b = 1.5: \qquad \int_{-\infty}^{\infty} e^{-(x + 1.5\,i)^2}\, dx = \sqrt{\pi} \approx 1.7725 \]

Insert the number last. The check in Step 3 is comfortably satisfied numerically: at \( R = 5 \), the vertical-side bound is \( b\,e^{b^2}e^{-R^2} = 1.5 \times e^{2.25} \times e^{-25} \approx 1.5 \times 9.488 \times 1.39\times 10^{-11} \approx 2.0 \times 10^{-10} \) — already negligible. A

\[ \int_{-\infty}^{\infty} e^{-(x + 1.5\,i)^2}\, dx = \sqrt{\pi} \approx 1.7725 \]

Reading. Shifting a Gaussian integration line by any imaginary amount leaves the integral untouched, because the integrand is entire and the connecting vertical segments die like \( e^{-R^2} \). Expanding the square, this is exactly the statement that the Fourier transform of a Gaussian is a Gaussian: \( \int e^{-x^2} e^{-2ibx} dx = \sqrt{\pi}\, e^{-b^2} \).

Units check. With \( x \) dimensionless (or measured in units of the Gaussian width \( \sigma \)), the integral and \( \sqrt{\pi} \) are both dimensionless. Restoring units, \( \int e^{-x^2/\sigma^2} dx = \sigma\sqrt{\pi} \) carries length, as an integral \( [f][dx] = 1 \times \text{m} \) must.

Problems

Verify Cauchy–Goursat directly for \( f(z) = z \) on the circle \( |z| = 2 \) by explicit parametrization.
Solution
Parametrize \( z = 2e^{i\theta} \), \( dz = 2i e^{i\theta} d\theta \), \( \theta \in [0, 2\pi] \). Then \[ \oint_{|z|=2} z\,dz = \int_0^{2\pi} 2e^{i\theta}\cdot 2i e^{i\theta}\,d\theta = 4i \int_0^{2\pi} e^{2i\theta}\, d\theta = 4i \left[ \frac{e^{2i\theta}}{2i} \right]_0^{2\pi} = 2\left( e^{4\pi i} - 1 \right) = 2(1 - 1) = 0 . \] The result is \( 0 \), as demanded: \( f(z) = z \) is entire and the disc \( |z| \le 2 \) is simply connected. Numerically, the real part \( \int_0^{2\pi} \cos 2\theta \, d\theta = 0 \) and imaginary part \( \int_0^{2\pi} \sin 2\theta\, d\theta = 0 \) each vanish by two full periods of the integrand.
Evaluate \( \displaystyle\oint_C \frac{dz}{z - 5} \) for \( C: |z| = 2 \), justifying every hypothesis of Cauchy–Goursat, and confirm by expanding the integrand in a series.
Solution
The integrand's only singularity is at \( z = 5 \), and \( |5| = 5 > 2 \), so \( f(z) = 1/(z-5) \) is analytic on the open disc \( |z| < 3.5 \) (say), which is simply connected and contains \( C \). Cauchy–Goursat gives \[ \oint_{|z|=2} \frac{dz}{z-5} = 0 . \] Series check: for \( |z| = 2 < 5 \), \[ \frac{1}{z - 5} = -\frac{1}{5} \cdot \frac{1}{1 - z/5} = -\frac{1}{5} \sum_{n=0}^{\infty} \frac{z^n}{5^n} , \] uniformly convergent on \( |z| \le 2 \) (ratio \( 2/5 = 0.4 < 1 \)), so termwise integration is legal: \[ \oint_C \frac{dz}{z-5} = -\sum_{n=0}^{\infty} \frac{1}{5^{n+1}} \oint_{|z|=2} z^n\, dz = -\sum_n \frac{1}{5^{n+1}} \cdot 0 = 0 , \] since \( \oint z^n dz = 0 \) for all \( n \ge 0 \) (each \( z^n \) has primitive \( z^{n+1}/(n+1) \)). Both routes agree: the answer is exactly \( 0 \).
Compute \( \displaystyle\oint_{|z|=1} \bar{z}\,dz \) and explain precisely which hypothesis of Cauchy–Goursat fails. Then show that for any positively oriented simple closed contour, \( \oint_C \bar{z}\,dz = 2iA \) where \( A \) is the enclosed area, and evaluate it for a circle of radius \( 1.0 \).
Solution
On \( |z| = 1 \): \( z = e^{i\theta} \), \( \bar{z} = e^{-i\theta} \), \( dz = ie^{i\theta} d\theta \), so \[ \oint_{|z|=1} \bar{z}\,dz = \int_0^{2\pi} e^{-i\theta}\, i e^{i\theta}\, d\theta = i \int_0^{2\pi} d\theta = 2\pi i \neq 0 . \] The failed hypothesis is analyticity: \( \bar z = x - iy \) has \( u = x \), \( v = -y \), so \( \partial u/\partial x = 1 \neq -1 = \partial v/\partial y \) — the Cauchy–Riemann equations fail at every point, so \( \bar z \) is nowhere complex-differentiable despite being a \( C^\infty \) map of the plane. General formula: with \( \bar{z}\,dz = (x - iy)(dx + i\,dy) \), \[ \oint_C \bar{z}\,dz = \oint_C (x\,dx + y\,dy) + i \oint_C (x\,dy - y\,dx) = 0 + i \cdot 2A = 2iA \] by Green's theorem (the first integral is \( \oint d\left(\tfrac{x^2+y^2}{2}\right) = 0 \); the second is the standard area formula). For \( R = 1.0 \): \( A = \pi R^2 = \pi \), giving \( 2iA = 2\pi i \approx 6.2832\,i \), matching the direct computation.
Use Cauchy–Goursat on the circular sector \( \{ re^{i\varphi} : 0 \le r \le R,\; 0 \le \varphi \le \pi/4 \} \) with \( f(z) = e^{-z^2} \) to evaluate the Fresnel integral \( \int_0^\infty \cos\!\left(x^2\right) dx \), and give its numerical value.
Solution
\( e^{-z^2} \) is entire, and the closed sector boundary lies in \( \mathbb{C} \) (simply connected), so \( \oint f\,dz = 0 \). Decompose the boundary: (i) real segment \( 0 \to R \); (ii) arc \( z = Re^{i\varphi} \), \( 0 \le \varphi \le \pi/4 \); (iii) ray back along \( z = t e^{i\pi/4} \), \( t: R \to 0 \). Thus \[ \int_0^R e^{-x^2} dx + I_{\text{arc}} - e^{i\pi/4} \int_0^R e^{-i t^2}\, dt = 0, \] using \( (te^{i\pi/4})^2 = t^2 e^{i\pi/2} = i t^2 \) on the ray. Arc estimate: \( |e^{-R^2 e^{2i\varphi}}| = e^{-R^2 \cos 2\varphi} \), and on \( [0, \pi/4] \) the concavity bound \( \cos 2\varphi \ge 1 - \frac{4\varphi}{\pi} \) gives \[ |I_{\text{arc}}| \le R \int_0^{\pi/4} e^{-R^2 (1 - 4\varphi/\pi)}\, d\varphi = \frac{\pi}{4R}\left( 1 - e^{-R^2} \right) \to 0 \quad (R \to \infty). \] Hence \( e^{i\pi/4} \int_0^\infty e^{-it^2} dt = \int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \), so \[ \int_0^\infty \left( \cos t^2 - i \sin t^2 \right) dt = \frac{\sqrt{\pi}}{2} e^{-i\pi/4} = \frac{\sqrt{\pi}}{2}\left( \frac{\sqrt 2}{2} - i \frac{\sqrt 2}{2} \right). \] Real part: \[ \int_0^\infty \cos\!\left(x^2\right) dx = \frac{\sqrt{\pi}}{2}\cdot\frac{\sqrt{2}}{2} = \sqrt{\frac{\pi}{8}} \approx 0.6267 \] (dimensionless; with \( x \) in \( \sqrt{\text{m}} \)-type units the integral would carry those units). The imaginary part gives \( \int_0^\infty \sin(x^2)\,dx = \sqrt{\pi/8} \approx 0.6267 \) as well.
Goursat bookkeeping. A triangle \( \Delta \) has perimeter \( L = 6.0 \) and diameter \( d = 2.5 \) (dimensionless units), and \( f \) is analytic on a domain containing \( \Delta \). (a) After \( n = 8 \) bisections, what are \( L_8 \), \( d_8 \), and the accumulated prefactor \( 4^8 \)? (b) If differentiability at the limit point \( z_0 \) gives \( |\eta| < \varepsilon = 10^{-3} \) on the relevant disc, bound \( \left|\oint_{\partial\Delta} f\,dz\right| \) and state the final conclusion.
Solution
(a) Each bisection halves lengths: \( L_8 = L/2^8 = 6.0/256 = 0.0234 \), \( d_8 = d/2^8 = 2.5/256 = 0.00977 \). The prefactor is \( 4^8 = 65\,536 \). (b) The chain of estimates (Steps 7, 11, 12 of the derivation): \[ \left| \oint_{\partial\Delta} f\,dz \right| \le 4^8 \left| I_8 \right| \le 4^8 \cdot \varepsilon\, d_8 L_8 = 4^8 \cdot \varepsilon \cdot \frac{d}{2^8} \cdot \frac{L}{2^8} = \varepsilon\, d\, L . \] The \( 4^8 = 65\,536 \) growth is exactly cancelled by \( d_8 L_8 = dL/4^8 \): numerically \( d_8 L_8 = 0.00977 \times 0.0234 = 2.29 \times 10^{-4} = (2.5)(6.0)/65\,536 \). With \( \varepsilon = 10^{-3} \): \[ \left| \oint_{\partial\Delta} f\,dz \right| \le 10^{-3} \times 2.5 \times 6.0 = 1.5 \times 10^{-2} . \] But \( \varepsilon \) was arbitrary: differentiability at \( z_0 \) supplies, for every \( \varepsilon > 0 \), an \( n \) making the same bound \( \varepsilon dL \) hold. A nonnegative number below \( 15\,\varepsilon \times 10^{-3} \)… more precisely below \( \varepsilon \cdot 15 \) for all \( \varepsilon > 0 \) must be \( 0 \): \[ \oint_{\partial\Delta} f\,dz = 0 . \] This is the entire engine of Goursat's proof in one number: the bisection cost \( 4^n \) and the geometric shrinkage \( 4^{-n} \) cancel identically, leaving only the arbitrarily small \( \varepsilon \).