Cauchy's integral theorem

Theorem

Cauchy's integral theorem

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Statement

Let \(\Omega \subseteq \mathbb{C}\) be a non-empty, open, simply connected set, and let \(f:\Omega \to \mathbb{C}\) be holomorphic on \(\Omega\) (complex-differentiable at every point of \(\Omega\)). Let \(\gamma:[a,b]\to\Omega\) be a closed, piecewise-\(C^1\) (rectifiable) curve in \(\Omega\), i.e. \(\gamma(a)=\gamma(b)\). Then \[ \oint_\gamma f(z)\,dz \;=\; 0. \]

Why it matters

This is the load-bearing result of complex analysis: essentially every other structural theorem about holomorphic functions — the Cauchy integral formula, the power-series (Taylor) expansion of holomorphic functions, Cauchy's estimates, Liouville's theorem, the maximum modulus principle, the residue theorem — is derived from it, directly or via one further step. The theorem says something that has no real-variable analogue of comparable strength: complex differentiability, a purely local and seemingly weak hypothesis, forces a rigid global constraint (path-independence of integrals) on the function.

It also explains, retrospectively, why complex analysis is so much more rigid than real analysis: a real \(C^1\) function need satisfy no such vanishing condition on any closed loop of its graph, yet a merely-once-complex-differentiable function is already forced into this global straightjacket, and (as a consequence proved later, via Cauchy's formula) is automatically infinitely differentiable and analytic.

Hypotheses

f is holomorphic on the open set \(\Omega\). Counterexample: \(f(z)=\bar z\) is continuous but nowhere holomorphic (it fails the Cauchy–Riemann equations everywhere). Parametrising the unit circle by \(z=e^{i\theta}\), \(0\le\theta\le 2\pi\): \(\oint_{|z|=1}\bar z\,dz=\int_0^{2\pi} e^{-i\theta}\cdot ie^{i\theta}\,d\theta=\int_0^{2\pi} i\,d\theta=2\pi i\neq 0\).

\(\Omega\) is simply connected (every closed curve in \(\Omega\) is null-homotopic in \(\Omega\)). Counterexample: \(f(z)=1/z\) is holomorphic on the punctured plane \(\Omega=\mathbb{C}\setminus\{0\}\), which is open but not simply connected (a loop around the puncture cannot be contracted without leaving \(\Omega\)). Yet \(\oint_{|z|=1} dz/z = 2\pi i \neq 0\). Holomorphy alone, without a topological restriction on \(\Omega\), is not enough.

\(\gamma\) is closed and lies entirely inside \(\Omega\) (including wherever \(f\) fails to be holomorphic). If any point "inside" the loop, or on it, is excluded from \(\Omega\) — even a single singularity — the conclusion can fail: take \(\Omega=\mathbb{C}\setminus\{0\}\) again (not simply connected because of exactly this missing point) and \(\gamma\) the unit circle; the integral of \(1/z\) is \(2\pi i\), not \(0\).

Proof

The proof proceeds in three stages: (I) Goursat's lemma — the theorem is proved first for the boundary of a triangle, using no topology beyond compactness of the plane; (II) a primitive of \(f\) is manufactured locally on any disc using Goursat's lemma; (III) the local primitives are patched into one single-valued global primitive on \(\Omega\) using simple connectivity, after which the theorem is a one-line consequence of the Fundamental Theorem of Calculus for contour integrals.

\textbf{Goursat's Lemma.} Let \(U\) be open, \(f\) holomorphic on \(U\), and let \(\Delta\subseteq U\) be a closed (solid) triangle with boundary \(\partial\Delta\). Then \(\displaystyle\int_{\partial\Delta} f(z)\,dz = 0\).

Stated first as it is the computational engine of the whole theorem; it is proved by bisection, using only continuity/differentiability at a single limit point, not any global hypothesis on \(f\) or \(U\). B

Let \(\eta=\int_{\partial\Delta} f\,dz\). Bisect the sides of \(\Delta\) at their midpoints to split \(\Delta\) into four congruent sub-triangles \(\Delta^{(1)},\dots,\Delta^{(4)}\), each similar to \(\Delta\) with half its diameter and half its perimeter. Orient each \(\partial\Delta^{(i)}\) positively; then \[\int_{\partial\Delta} f\,dz=\sum_{i=1}^4 \int_{\partial\Delta^{(i)}} f\,dz,\] because every interior edge introduced by the bisection is traversed exactly twice, once in each direction, by two adjacent sub-triangles, so those contributions cancel and only the outer edges of \(\Delta\) survive.

Direct computation from the additivity of the contour integral over a common edge traversed with opposite orientation; no property of \(f\) beyond integrability (continuity) is used. A

By the triangle inequality, \(|\eta|\le \sum_i \bigl|\int_{\partial\Delta^{(i)}} f\bigr|\le 4\max_i \bigl|\int_{\partial\Delta^{(i)}} f\bigr|\), so some sub-triangle \(\Delta_1\in\{\Delta^{(1)},\dots,\Delta^{(4)}\}\) satisfies \(\bigl|\int_{\partial\Delta_1} f\bigr|\ge |\eta|/4\). Iterating this bisection produces a nested sequence \(\Delta=\Delta_0\supseteq \Delta_1\supseteq \Delta_2\supseteq\cdots\) with \(\operatorname{diam}(\Delta_n)=2^{-n}\operatorname{diam}(\Delta)\), \(\operatorname{perim}(\Delta_n)=2^{-n}\operatorname{perim}(\Delta)\), and \(\bigl|\int_{\partial\Delta_n} f\bigr|\ge 4^{-n}|\eta|\).

Pigeonhole among four terms bounding a sum; induction sets up the geometric decay used in Step 5. B

The sets \(\Delta_n\) are non-empty, compact and nested with \(\operatorname{diam}(\Delta_n)\to 0\), so by the Cantor intersection property (nested compact sets in a complete metric space) there is a unique point \(z_0\in\bigcap_n \Delta_n\subseteq \Delta\subseteq U\).

Cantor's nested compact set theorem, valid because \((\mathbb{C},|\cdot|)\) is a complete metric space. A

Fix \(\varepsilon\gt 0\). Since \(f\) is complex-differentiable at \(z_0\), there is \(\delta\gt 0\) such that \(D(z_0,\delta)\subseteq U\) and \[ |f(z)-f(z_0)-f'(z_0)(z-z_0)|\le \varepsilon|z-z_0| \qquad \text{for } |z-z_0|\lt \delta. \] Choose \(n\) large enough that \(\Delta_n\subseteq D(z_0,\delta)\) (possible since \(\operatorname{diam}(\Delta_n)\to0\) and \(z_0\in\Delta_n\)). Write \(f(z)=\bigl[f(z_0)+f'(z_0)(z-z_0)\bigr]+R(z)\) with \(|R(z)|\le\varepsilon|z-z_0|\) on \(\Delta_n\).

Definition of complex differentiability at a point (the \(o(|z-z_0|)\) remainder estimate), applied at the single point \(z_0\) produced by Step 4. B

The affine function \(a+bz\) (with \(a=f(z_0)-f'(z_0)z_0,\ b=f'(z_0)\)) has the explicit primitive \(az+bz^2/2\), so by direct antiderivative evaluation \(\int_{\partial\Delta_n}(a+bz)\,dz=0\). Hence \[ \int_{\partial\Delta_n} f\,dz=\int_{\partial\Delta_n} R(z)\,dz. \]

A closed contour integral of a function with an explicit global antiderivative on a neighbourhood of the contour vanishes: parametrise each edge and telescope, exactly as in ordinary calculus. A

By the standard \(ML\)-estimate, \(\bigl|\int_{\partial\Delta_n} R\,dz\bigr|\le \Bigl(\sup_{z\in\partial\Delta_n}|R(z)|\Bigr)\cdot\operatorname{perim}(\Delta_n) \le \varepsilon\,\operatorname{diam}(\Delta_n)\cdot\operatorname{perim}(\Delta_n) = \varepsilon\cdot 4^{-n}\operatorname{diam}(\Delta)\operatorname{perim}(\Delta).\) Combined with Step 3, \(4^{-n}|\eta|\le \bigl|\int_{\partial\Delta_n} f\bigr| \le \varepsilon\cdot4^{-n}\operatorname{diam}(\Delta)\operatorname{perim}(\Delta)\), so \(|\eta|\le \varepsilon\,\operatorname{diam}(\Delta)\operatorname{perim}(\Delta)\). As \(\varepsilon\gt0\) was arbitrary, \(\eta=0\), proving Goursat's Lemma.

\(ML\)-inequality for contour integrals (\(|\int_\gamma g\,dz|\le \sup_\gamma|g|\cdot\operatorname{length}(\gamma)\)); then let \(\varepsilon\to0\). C

\textbf{Local primitive.} Fix \(a\in\Omega\) and a disc \(D(a,r)\subseteq\Omega\) (exists since \(\Omega\) open). Define \(F(z)=\int_{[a,z]} f(w)\,dw\) for \(z\in D(a,r)\), the integral along the straight segment \([a,z]\subseteq D(a,r)\) (a disc is convex, so this segment stays in \(\Omega\)). For \(z,z+h\in D(a,r)\), the solid triangle with vertices \(a,z,z+h\) lies in the convex set \(D(a,r)\subseteq\Omega\), so Goursat's Lemma (Step 7) gives \[ \int_{[a,z]}f+\int_{[z,z+h]}f+\int_{[z+h,a]}f=0,\quad\text{i.e.}\quad F(z+h)-F(z)=\int_{[z,z+h]} f(w)\,dw. \]

Applies Goursat's Lemma (Step 7) to the specific triangle \(a,z,z+h\), which is admissible since \(D(a,r)\) is convex and contained in \(\Omega\). B

\[ \left|\frac{F(z+h)-F(z)}{h}-f(z)\right| = \left|\frac{1}{h}\int_{[z,z+h]}\bigl(f(w)-f(z)\bigr)\,dw\right| \le \sup_{w\in[z,z+h]}|f(w)-f(z)| \xrightarrow[h\to0]{} 0, \] using the \(ML\)-estimate on a segment of length \(|h|\) and continuity of \(f\) at \(z\). Hence \(F\) is holomorphic on \(D(a,r)\) with \(F'=f\) there.

\(ML\)-inequality again, plus continuity of \(f\) (itself a consequence of complex differentiability). B

\textbf{Global primitive via simple connectivity.} Fix \(z_0\in\Omega\). Cover \(\Omega\) by discs on each of which a local primitive exists as in Steps 8–9. For any \(z\in\Omega\), pick a path \(\sigma\) from \(z_0\) to \(z\) in \(\Omega\), cover \(\sigma\) by finitely many such discs \(D_1,\dots,D_k\) in order along \(\sigma\), and chain the local primitives \(F_1,\dots,F_k\) by adjusting each by an additive constant so that \(F_i=F_{i+1}\) on the (connected) overlap \(D_i\cap D_{i+1}\) — legitimate because \((F_i-F_{i+1})'=f-f=0\) on a connected open set forces \(F_i-F_{i+1}\) constant there. This produces a value \(F(z)\).

Uses that a holomorphic function with zero derivative on a connected open set is constant (integrate the real/imaginary parts, or apply the Mean Value Inequality along a path of segments). C

\textbf{Monodromy Theorem} (named lemma, standard): if \(\Omega\) is simply connected, analytic continuation of a local primitive along any two paths in \(\Omega\) between the same endpoints \(z_0,z\) yields the same value, because the two paths are homotopic in \(\Omega\) and the chained local primitives vary continuously (in fact stay constant, since derivatives all agree with \(f\)) as the path is deformed through the homotopy. Consequently \(F(z)\) from Step 10 is well defined independently of the path \(\sigma\) and the disc cover chosen, and \(F:\Omega\to\mathbb{C}\) is a single holomorphic function with \(F'=f\) on all of \(\Omega\).

Monodromy Theorem for analytic continuation — the point at which simple connectivity of \(\Omega\) is used essentially; without it (e.g. \(\Omega=\mathbb{C}\setminus\{0\}\)) continuation of \(\log\) around the puncture does not return to its starting value. C

\textbf{Conclusion.} Let \(\gamma:[a,b]\to\Omega\) be closed and piecewise \(C^1\), say \(C^1\) on each of \([t_{j-1},t_j]\), \(a=t_0\lt t_1\lt\cdots\lt t_m=b\). On each piece, the chain rule gives \(\frac{d}{dt}\bigl(F(\gamma(t))\bigr)=F'(\gamma(t))\gamma'(t)=f(\gamma(t))\gamma'(t)\), so by the ordinary (real-variable) Fundamental Theorem of Calculus applied to the real and imaginary parts, \[ \int_{t_{j-1}}^{t_j} f(\gamma(t))\gamma'(t)\,dt = F(\gamma(t_j))-F(\gamma(t_{j-1})). \] Summing over \(j=1,\dots,m\) telescopes to \(\oint_\gamma f\,dz = F(\gamma(b))-F(\gamma(a))\). Since \(\gamma\) is closed, \(\gamma(a)=\gamma(b)\), so \(\oint_\gamma f\,dz = 0\). \(\blacksquare\)

Fundamental Theorem of Calculus for contour integrals (definition of \(\int_\gamma f\,dz\) as \(\int_a^b f(\gamma(t))\gamma'(t)\,dt\), piecewise chain rule, telescoping sum). A

Result

\(f\) holomorphic on simply connected \(\Omega\), \(\gamma\) closed in \(\Omega\) \(\;\Longrightarrow\;\) \(\displaystyle\oint_\gamma f(z)\,dz=0\)

Reading. If you never leave a region where a function is complex-differentiable, and that region has no "holes" the loop could wind around, then integrating the function all the way round any closed loop always gives exactly zero — regardless of the loop's shape or how many times it doubles back on itself.

Scope. Applies to any \(f\) holomorphic on any open simply connected \(\Omega\subseteq\mathbb{C}\) (in particular: entire functions on all of \(\mathbb{C}\); any holomorphic function on a disc, half-plane, or star-shaped domain; any holomorphic function on a simply connected \(\Omega\) even if \(f\) has singularities strictly outside \(\Omega\)) and any closed rectifiable curve \(\gamma\) lying in \(\Omega\). It does not apply, without modification, to multiply connected domains (annuli, punctured planes) — there the residue theorem, of which this is the residue-free special case, is needed.

Corollaries & converses

Path independence. If \(f\) is holomorphic on simply connected \(\Omega\) and \(\gamma_1,\gamma_2\) are two piecewise-\(C^1\) paths in \(\Omega\) from \(z_0\) to \(z_1\), then \(\int_{\gamma_1} f = \int_{\gamma_2} f\) (apply the theorem to the closed loop \(\gamma_1\) followed by \(\gamma_2\) reversed).
Existence of primitives. Every \(f\) holomorphic on a simply connected \(\Omega\) has a holomorphic antiderivative \(F\) on \(\Omega\) with \(F'=f\) — this was in fact constructed inside the proof (Steps 8–11) and is logically equivalent to the theorem itself.
Downstream results. The Cauchy Integral Formula, Cauchy's estimates on derivatives, Liouville's theorem, the Fundamental Theorem of Algebra, and the identity theorem all follow from this theorem (typically via one further application to \(g(z)=f(z)/(z-a)\) or similar).
Partial converse — Morera's Theorem. If \(f\) is continuous on an open set \(\Omega\) and \(\oint_{\partial\Delta} f\,dz = 0\) for every closed triangle \(\Delta\subseteq\Omega\), then \(f\) is holomorphic on \(\Omega\). (Proof: the same Steps 8–9 construct a local primitive \(F\) from continuity plus the triangle-vanishing hypothesis alone; \(F\) is then holomorphic, hence — by a theorem proved from Cauchy's Integral Formula — infinitely differentiable, so \(F'=f\) is itself holomorphic.) The full converse (integral vanishing on every closed curve implies holomorphic) is also true and reduces to this triangle version.

Fails without

Holomorphy dropped: \(f(z)=\bar z\) on \(\Omega=\mathbb{C}\) (simply connected, but \(f\) not holomorphic): \(\oint_{|z|=1}\bar z\,dz=2\pi i\neq0\).
Simple connectivity dropped: \(f(z)=1/z\) holomorphic on \(\Omega=\mathbb{C}\setminus\{0\}\) (not simply connected): \(\oint_{|z|=1} dz/z=2\pi i\neq0\).
Curve leaves the domain of holomorphy: \(f(z)=1/(z^2+1)\) is holomorphic on \(\mathbb{C}\setminus\{\pm i\}\); if \(\gamma\) is a circle of radius \(2\) centred at \(0\) there is no simply connected \(\Omega\) containing \(\gamma\) and its interior on which \(f\) is holomorphic (both \(\pm i\) lie inside), and indeed \(\oint_{|z|=2}\frac{dz}{z^2+1}=2\pi i\left(\operatorname{Res}_{z=i}+\operatorname{Res}_{z=-i}\right)=0\) only by an unrelated symmetry cancellation — replacing the integrand by \(1/(z-i)\) instead gives \(\oint_{|z|=2}\frac{dz}{z-i}=2\pi i\neq 0\), showing the theorem's hypotheses (not merely its numerical conclusion) genuinely fail.

Common errors

Applying the theorem to \(f(z)=1/z\) (or any function with a pole) on a loop that encircles the pole, forgetting that "holomorphic on \(\Omega\)" must hold at every point of \(\Omega\), including the interior enclosed by \(\gamma\), not just on the trace of \(\gamma\) itself.
Treating "\(\Omega\) simply connected" as automatic for any open connected set; annuli and punctured discs are connected but not simply connected, and the theorem can fail there.
Quoting the theorem for a curve that is not closed and expecting zero; for open \(\gamma\) the correct statement is \(\int_\gamma f\,dz = F(\text{endpoint})-F(\text{start})\), which is generally non-zero.
Confusing this theorem with the Cauchy Integral Formula \(f(a)=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-a}\,dz\) — the latter is a different (later, stronger) statement about a function divided by \(z-a\), not the vanishing of \(\oint_\gamma f\).
Believing the theorem requires \(\gamma\) to be a simple (non-self-intersecting) curve; it holds for any closed piecewise-\(C^1\) curve, self-intersecting or winding round multiple times, as long as it stays in \(\Omega\).

Discussion

Augustin-Louis Cauchy first stated a version of this result in 1825, under the extra (and at the time unavoidable) hypothesis that \(f'\) itself be continuous, which allowed him to invoke Green's theorem: writing \(f=u+iv\), \(dz=dx+i\,dy\), the real and imaginary parts of \(\oint_\gamma f\,dz\) become exactly the line integrals appearing in Green's theorem, and the Cauchy–Riemann equations make both resulting double integrals vanish identically. That argument is genuinely illuminating but logically weaker than the one given above, since it presupposes \(f\in C^1\) as a real map, a fact not yet known to follow from mere complex differentiability. Édouard Goursat closed this gap in 1900 by proving the triangle case using only the definition of the derivative and a bisection argument — no continuity of \(f'\) assumed — which is why the lemma bears his name and why the modern theorem is often called the Cauchy–Goursat theorem.

The deepest content of the theorem, in retrospect, is the equivalence between three things that look completely different: (i) a local, pointwise differential condition (holomorphy), (ii) a global integral condition (vanishing of loop integrals), and (iii) the existence of a primitive. Steps 8–11 of the proof show (i) \(\Rightarrow\) (iii), and Step 12 shows (iii) \(\Rightarrow\) (ii); Morera's theorem supplies the missing arrow (ii) \(\Rightarrow\) (i) (via triangles), closing the loop of equivalences. This triad is the structural skeleton on which the rest of elementary complex analysis (Cauchy's formula, Taylor series, residues) is built.

The role of simple connectivity is best understood homotopically rather than purely topologically: the theorem generalises to the statement that \(\oint_{\gamma_0} f=\oint_{\gamma_1}f\) whenever \(\gamma_0\) and \(\gamma_1\) are freely homotopic closed curves within the (not necessarily simply connected) domain of holomorphy of \(f\) — a fact sometimes called the homotopy invariance of the contour integral, and provable by exactly the chaining-and-monodromy argument of Steps 10–11 applied along the homotopy. Simple connectivity of \(\Omega\) is precisely the condition guaranteeing every closed curve is homotopic to a constant curve (which trivially has integral \(0\)), so the theorem as stated is the special case of homotopy invariance where the target loop is a point.

Common misconception: that the theorem says holomorphic functions integrate to zero because "the antiderivative undoes the derivative", as though this were a triviality analogous to the real 1-D Fundamental Theorem of Calculus. It is not trivial: on the real line every continuous function trivially has a primitive (via \(\int_a^x\)), but a complex-differentiable function need not have a well-defined single-valued primitive unless the domain is simply connected — \(1/z\) is the standing counterexample, complex-differentiable everywhere it is defined, yet with no single-valued antiderivative on \(\mathbb{C}\setminus\{0\}\) (any candidate primitive is a branch of \(\log z\), which cannot be made continuous, let alone holomorphic, all the way around the origin).

Worked examples

Compute \(\displaystyle\oint_{|z|=2} \frac{z^2+1}{z^2+9}\,dz\).

Identify singularities of the integrand. A

\(z^2+9=0 \iff z=\pm 3i\), so \(f(z)=\dfrac{z^2+1}{z^2+9}\) is holomorphic on \(\mathbb{C}\setminus\{3i,-3i\}\). Since \(|3i|=|-3i|=3\gt 2\), both singular points lie outside the closed disc \(\{|z|\le 2\}\).

Direct check of modulus against the radius of the contour. A

Take \(\Omega=D(0,2.5)\), an open disc (hence simply connected) that avoids both \(\pm3i\); then \(f\) is holomorphic on all of \(\Omega\), and \(\gamma=\{|z|=2\}\) is a closed curve in \(\Omega\). All hypotheses of Cauchy's Integral Theorem hold.

Explicit exhibition of the simply connected domain required by the theorem. A

\(\displaystyle\oint_{|z|=2} \frac{z^2+1}{z^2+9}\,dz = 0\)

Evaluate the real Gaussian-type integral \(\displaystyle I(b)=\int_{-\infty}^{\infty} e^{-x^2}\cos(2bx)\,dx\) for fixed \(b\in\mathbb{R}\), given the standard fact \(\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}\).

Set up a rectangular contour to exploit that \(e^{-z^2}\) is entire. B

Let \(f(z)=e^{-z^2}\), entire (holomorphic on \(\Omega=\mathbb{C}\), simply connected). For \(R\gt0\) let \(\gamma_R\) be the boundary of the rectangle with vertices \(-R,\,R,\,R+ib,\,-R+ib\), traversed counterclockwise; \(\gamma_R\) is a closed piecewise-\(C^1\) curve in \(\Omega\). By Cauchy's Integral Theorem, \(\oint_{\gamma_R} f(z)\,dz=0\).

Direct application of the theorem proved above: entire function, any closed contour. A

Split \(\gamma_R\) into bottom, right, top (reversed), left (reversed) edges: \[ \int_{-R}^{R} e^{-x^2}dx \;+\; \int_0^b e^{-(R+iy)^2}i\,dy \;-\; \int_{-R}^{R} e^{-(x+ib)^2}dx \;-\; \int_0^b e^{-(-R+iy)^2}i\,dy \;=\;0. \]

Definition of the contour integral as a sum over the four oriented edges of the rectangle. A

On the right edge, \(|e^{-(R+iy)^2}|=e^{-(R^2-y^2)}\le e^{-R^2+b^2}\to0\) as \(R\to\infty\) (uniformly for \(y\in[0,b]\)), so that edge's integral \(\to0\) by the \(ML\)-estimate (length \(b\), bound \(\to0\)); identically for the left edge. Hence, letting \(R\to\infty\): \[ \int_{-\infty}^\infty e^{-x^2}\,dx = \int_{-\infty}^\infty e^{-(x+ib)^2}\,dx. \]

\(ML\)-estimate applied to the two vertical edges, then passing to the limit in the identity from Step 3. B

Expand \(e^{-(x+ib)^2}=e^{-x^2+b^2}e^{-2ibx}=e^{b^2}e^{-x^2}\bigl(\cos(2bx)-i\sin(2bx)\bigr)\). The left side of Step 4 is \(\sqrt{\pi}\) (given). The imaginary part of the right side integrates to \(0\) by oddness of \(e^{-x^2}\sin(2bx)\), so equating real parts: \[ \sqrt{\pi} = e^{b^2}\int_{-\infty}^\infty e^{-x^2}\cos(2bx)\,dx = e^{b^2}\,I(b). \]

Algebraic expansion, symmetry (odd integrand over \(\mathbb{R}\) has zero integral), and the given value of the real Gaussian integral. B

\(\displaystyle I(b)=\int_{-\infty}^\infty e^{-x^2}\cos(2bx)\,dx = \sqrt{\pi}\,e^{-b^2}\)

Problems

Compute \(\displaystyle\oint_{|z|=3} \sin(z)\,dz\).
Solution
\(\sin z\) is entire, hence holomorphic on \(\Omega=\mathbb{C}\) (simply connected), and \(\{|z|=3\}\) is a closed curve in \(\Omega\). By Cauchy's Integral Theorem the integral is \(0\).
Compute \(\displaystyle\oint_{|z-1|=1/2} \frac{dz}{z-3}\).
Solution
The integrand \(f(z)=1/(z-3)\) is holomorphic on \(\mathbb{C}\setminus\{3\}\). The disc \(\Omega=D(1,2)\) is open, simply connected, avoids \(z=3\) (since \(|3-1|=2\) is on the boundary — choose instead, say, \(\Omega=D(1,1.9)\) to be safely inside), and contains the curve \(\{|z-1|=1/2\}\) together with its interior. All hypotheses hold on \(\Omega\), so the integral is \(0\).
Let \(f\) be holomorphic on an open set containing the closed annulus \(\{1\le|z|\le2\}\). Using Cauchy's Integral Theorem (not the residue theorem), show \(\displaystyle\oint_{|z|=2} f\,dz=\oint_{|z|=1} f\,dz\) (both counterclockwise).
Solution
Cut the annulus along the segment \([1,2]\) on the positive real axis (approached from just above and just below) to form a simply connected "keyhole" region \(K\): the boundary of \(K\) consists of the outer circle \(|z|=2\) traversed counterclockwise, the segment from \(2\) to \(1\) just below the cut, the inner circle \(|z|=1\) traversed clockwise, and the segment from \(1\) to \(2\) just above the cut. \(K\) is simply connected and \(f\) is holomorphic on (a neighbourhood of) \(\overline{K}\), so by Cauchy's Integral Theorem the integral of \(f\) around \(\partial K\) is \(0\). As the width of the keyhole cut shrinks to zero, the two straight segments are traversed in opposite directions with the same integrand values on them (by continuity of \(f\)) and their contributions cancel in the limit, leaving \[ \oint_{|z|=2} f\,dz \;-\; \oint_{|z|=1} f\,dz = 0, \] i.e. the two circle integrals (both counterclockwise) are equal.
Let \(\Omega\) be simply connected and \(f:\Omega\to\mathbb{C}\) holomorphic with \(f(z)\neq0\) for all \(z\in\Omega\). Show there exists a holomorphic \(g:\Omega\to\mathbb{C}\) with \(e^{g(z)}=f(z)\) for all \(z\in\Omega\) (a holomorphic branch of \(\log f\)).
Solution
Since \(f\) is holomorphic and non-vanishing on \(\Omega\), the function \(h(z)=f'(z)/f(z)\) is holomorphic on \(\Omega\) (quotient of holomorphic functions with non-vanishing denominator). Because \(\Omega\) is simply connected, Cauchy's Integral Theorem applies to \(h\) (Steps 8–11 of the proof), giving a holomorphic primitive \(G\) on \(\Omega\) with \(G'=h=f'/f\). Fix \(z_0\in\Omega\) and adjust \(G\) by an additive constant so that \(e^{G(z_0)}=f(z_0)\) (possible since \(f(z_0)\neq0\)); set \(g=G\). Consider \(\phi(z)=f(z)e^{-G(z)}\): \(\phi'(z)=f'(z)e^{-G(z)}-f(z)G'(z)e^{-G(z)}=e^{-G(z)}\bigl(f'(z)-f(z)\cdot f'(z)/f(z)\bigr)=0\) on the connected set \(\Omega\), so \(\phi\) is constant, equal to \(\phi(z_0)=f(z_0)e^{-G(z_0)}=1\). Hence \(f(z)=e^{G(z)}=e^{g(z)}\) for all \(z\in\Omega\), as required.
Assuming Cauchy's Integral Theorem, derive the Cauchy Integral Formula: if \(f\) is holomorphic on simply connected \(\Omega\), \(\gamma\) is a closed curve in \(\Omega\) that winds once counterclockwise around \(a\in\Omega\setminus\gamma\) and \(0\) times around every point outside a simply connected region containing \(a\) and bounded appropriately by \(\gamma\) (take \(\gamma\) a circle \(|z-a|=r\) for concreteness), show \(\displaystyle\oint_\gamma \frac{f(z)}{z-a}\,dz = 2\pi i f(a)\).
Solution
Define \(g(z)=\dfrac{f(z)-f(a)}{z-a}\) for \(z\in\Omega\setminus\{a\}\) and \(g(a)=f'(a)\). Since \(f\) is holomorphic at \(a\), \(\lim_{z\to a} g(z)=f'(a)=g(a)\), so \(g\) is continuous on \(\Omega\); and \(g\) is holomorphic on \(\Omega\setminus\{a\}\) as a quotient of holomorphic functions with non-vanishing denominator there. One checks (e.g. via Morera's theorem applied to small triangles, splitting off the case where \(a\) is a vertex and using continuity of \(g\) at \(a\) to bound that piece) that \(g\) is in fact holomorphic at \(a\) too, hence holomorphic on all of \(\Omega\). Now for \(0\lt\rho\lt r\), the annulus \(\rho\le|z-a|\le r\) lies in \(\Omega\setminus\{a\}\) where... instead apply Problem 3's result directly to \(g\) (holomorphic on all of \(\Omega\), including \(a\)): \(\oint_{|z-a|=r} g\,dz = \oint_{|z-a|=\rho} g\,dz\) for every \(\rho\in(0,r]\); but by Cauchy's Integral Theorem applied directly to \(g\) on the disc \(\Omega\) (simply connected, \(g\) holomorphic everywhere on it), \(\oint_{|z-a|=r} g\,dz=0\) outright. Hence \[ 0=\oint_{|z-a|=r} \frac{f(z)-f(a)}{z-a}\,dz = \oint_{|z-a|=r}\frac{f(z)}{z-a}\,dz \;-\; f(a)\oint_{|z-a|=r}\frac{dz}{z-a}. \] Parametrising \(z=a+re^{i\theta}\), \(\oint_{|z-a|=r} dz/(z-a) = \int_0^{2\pi} i\,d\theta = 2\pi i\). Substituting: \(0=\oint_{|z-a|=r}\frac{f(z)}{z-a}dz - 2\pi i f(a)\), i.e. \(\oint_{|z-a|=r}\frac{f(z)}{z-a}\,dz = 2\pi i f(a)\), as required.