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49  Complex integration

Cauchy’s theorem, the integral formula, and their consequences

49.1 Contours and contour integrals

A contour (or path) in the complex plane is a piecewise smooth curve \(C\): \(z = z(t)\), \(a \leq t \leq b\), where \(z(t) = x(t) + iy(t)\) and \(z'(t)\) is piecewise continuous. Piecewise smooth means the curve can have finitely many corners, but no cusps or fractal wiggles — at each smooth piece the tangent direction is well-defined.

The contour integral of \(f\) along \(C\) is:

\[\int_C f(z)\,dz = \int_a^b f(z(t))\,z'(t)\,dt\]

This is just a real line integral in disguise. Writing \(f = u + iv\) and \(dz = dx + idy\), multiply out the product: \[(u + iv)(dx + i\,dy) = u\,dx + iu\,dy + iv\,dx + i^2 v\,dy = (u\,dx - v\,dy) + i(v\,dx + u\,dy)\]

Separating real and imaginary parts:

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

Properties. - Linearity: \(\int_C (\alpha f + \beta g)\,dz = \alpha\int_C f\,dz + \beta\int_C g\,dz\). - Reversal: \(\int_{-C}f\,dz = -\int_C f\,dz\) (reversing the direction negates the integral). - Additivity: if \(C = C_1 + C_2\), then \(\int_C = \int_{C_1} + \int_{C_2}\).

49.1.1 The ML inequality

The most useful estimate for contour integrals:

\[\left|\int_C f(z)\,dz\right| \leq M \cdot L\]

where \(M = \max_{z \in C}|f(z)|\) and \(L\) is the arc length of \(C\). This is the complex analogue of \(|\int_a^b f\,dx| \leq \max|f| \cdot (b-a)\).

49.1.2 A fundamental example

\[\oint_{|z|=r} \frac{dz}{z^{n+1}} = \begin{cases} 2\pi i & n = 0 \\ 0 & n \neq 0 \end{cases}\]

Proof for \(n=0\): parametrise \(z = re^{i\theta}\), \(dz = ire^{i\theta}d\theta\): \[\int_0^{2\pi}\frac{ire^{i\theta}}{re^{i\theta}}\,d\theta = \int_0^{2\pi}i\,d\theta = 2\pi i\]

This single result — that the integral of \(1/z\) around the origin is \(2\pi i\) — is the seed from which the residue theorem grows.

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49.2 Cauchy’s theorem

Cauchy-Goursat theorem. If \(f\) is analytic throughout a simply connected domain \(D\) (no holes — defined in Chapter 1), then for every closed contour \(C\) in \(D\):

\[\oint_C f(z)\,dz = 0\]

Intuition. Because the domain has no holes, any closed curve can be continuously shrunk to a point. The integral shrinks with it — to zero.

Consequence: path independence. If \(f\) is analytic in a simply connected domain and \(C_1\), \(C_2\) are two paths from \(z_1\) to \(z_2\) in that domain, then \(\int_{C_1}f\,dz = \int_{C_2}f\,dz\). The integral depends only on the endpoints, not the path.

Deformation principle. If \(f\) is analytic everywhere between two contours \(C_1\) and \(C_2\) (with the same orientation), then \(\int_{C_1}f\,dz = \int_{C_2}f\,dz\). Contours can be freely deformed through regions where \(f\) is analytic, without changing the integral.

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49.3 Cauchy’s integral formula

Cauchy’s integral formula. Let \(f\) be analytic in and on a simple closed contour \(C\) (traversed counterclockwise). Then for any \(z_0\) interior to \(C\):

\[f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}\,dz\]

Equivalently:

\[\oint_C \frac{f(z)}{z - z_0}\,dz = 2\pi i\,f(z_0)\]

Proof sketch. Because \(f\) is analytic everywhere inside \(C\) except at \(z_0\), the deformation principle applies to the annular region between \(C\) and any circle around \(z_0\): the integrand \(f(z)/(z-z_0)\) is analytic there. Shrink \(C\) to a small circle \(C_\varepsilon\) of radius \(\varepsilon\) around \(z_0\) without changing the integral. As \(\varepsilon \to 0\), \(f(z) \to f(z_0)\) uniformly on \(C_\varepsilon\), and \(\oint_{C_\varepsilon} dz/(z-z_0) = 2\pi i\).

Significance. The value of an analytic function at any interior point is completely determined by its values on the boundary. This has no analogue in real analysis.

49.3.1 Derivatives of all orders

Differentiating the integral formula under the integral sign (justified by uniform convergence):

\[f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz\]

Consequence. An analytic function has derivatives of all orders, all of which are themselves analytic. In real analysis, once-differentiable functions need not be twice differentiable. In complex analysis, once differentiable means infinitely differentiable — a qualitative difference.

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49.4 Major consequences

49.4.1 Liouville’s theorem

A bounded entire function is constant.

Proof. If \(|f(z)| \leq M\) everywhere, the ML inequality applied to the derivative formula gives \(|f'(z_0)| \leq M/R\) for any circle of radius \(R\). Letting \(R \to \infty\) gives \(f'(z_0) = 0\).

49.4.2 Fundamental theorem of algebra

Every non-constant polynomial \(p(z)\) of degree \(n \geq 1\) has exactly \(n\) complex roots (counting multiplicity).

Proof sketch. If \(p\) had no roots, \(1/p(z)\) would be entire and bounded (since \(|p(z)| \to \infty\)), contradicting Liouville’s theorem.

49.4.3 Morera’s theorem

If \(f\) is continuous on a domain \(D\) and \(\oint_C f\,dz = 0\) for every closed triangle \(C\) in \(D\), then \(f\) is analytic in \(D\). (The converse of Cauchy’s theorem.)

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49.5 Application: real integrals via contour integration

The tools developed so far can already evaluate some real integrals. The general strategy — closing a real integral with a contour in the complex plane and applying Cauchy’s theorem — will be developed fully in the next chapter. The key example here:

Example. \(\displaystyle\int_{-\infty}^{\infty} \frac{dx}{1+x^2}\).

The integrand \(f(z) = 1/(1+z^2)\) has poles at \(z = \pm i\). Close the real-line integral with a large semicircle \(\Gamma_R\) of radius \(R\) in the upper half-plane, forming a closed contour \(C = [-R,R] + \Gamma_R\).

The semicircle contribution vanishes. On \(\Gamma_R\), \(|z| = R\), so \(|1+z^2| \geq R^2 - 1\) for large \(R\). The ML inequality gives: \[\left|\int_{\Gamma_R}\frac{dz}{1+z^2}\right| \leq \frac{1}{R^2-1}\cdot \pi R \to 0 \quad\text{as } R\to\infty\]

Applying Cauchy’s integral formula. Factor the denominator: \(1/(1+z^2) = [1/(z+i)]/(z-i)\). This has the form \(f(z)/(z-z_0)\) with \(f(z) = 1/(z+i)\) and \(z_0 = i\). Since \(z_0 = i\) lies inside the upper half-plane contour and \(f(z) = 1/(z+i)\) is analytic there:

\[\oint_C \frac{dz}{1+z^2} = \oint_C \frac{dz}{(z-i)(z+i)} = 2\pi i \cdot f(i) = 2\pi i \cdot \frac{1}{2i} = \pi\]

Since the semicircle contributes nothing, the real-line integral equals \(\pi\) — confirming the known result \(\arctan(x)\big|_{-\infty}^{\infty} = \pi\).

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    intDiv.style.cssText = "margin-top:0.5rem; font-family:sans-serif; font-size:0.88em; padding:0.5rem; background:#f0f9ff; border-radius:4px;";
    const intStr = tVal > 0.98
      ? `∮ dz/z = ${finalIntIm > 0 ? "+" : ""}${finalIntIm.toFixed(2)}i = 2\u03c0i × ${def.nWinding}`
      : `Running: ${intRe.toFixed(3)} + ${intIm.toFixed(3)}i`;
    intDiv.innerHTML = `<strong>Contour integral:</strong> <span style="color:#1e40af">${intStr}</span>`;

    const wrap = document.createElement("div");
    wrap.appendChild(svg);
    wrap.appendChild(intDiv);
    return wrap;
  }

  const container = document.createElement("div");
  container.style.cssText = "border:1px solid #e5e7eb; border-radius:8px; padding:1rem; margin:1rem 0;";
  const title = document.createElement("div");
  title.style.cssText = "font-weight:600; margin-bottom:0.5rem; font-family:sans-serif;";
  title.textContent = "Winding number = contour integral: \u222e dz/z = 2\u03c0i \xd7 n";
  container.appendChild(title);
  container.appendChild(pathSelector2);
  container.appendChild(tSlider2);

  const vizDiv2 = document.createElement("div");
  container.appendChild(vizDiv2);

  const note = document.createElement("div");
  note.style.cssText = "margin-top:0.5rem; font-size:0.82em; color:#6b7280; font-style:italic;";
  note.textContent = "Move the slider to trace the path. When the loop closes: \u222e dz/z = 2\u03c0i times the winding number. The red \u00d7 marks the singularity at the origin. This is the seed from which the residue theorem grows.";
  container.appendChild(note);

  function update2() {
    vizDiv2.innerHTML = "";
    vizDiv2.appendChild(render2(pathSelector2.value, tSlider2.value));
  }
  pathSelector2.addEventListener("input", update2);
  tSlider2.addEventListener("input", update2);
  update2();

  return container;
}

---

49.6 Exercises

katex = require("katex@0.16")

function makeStepperHTML(steps) {
  let current = 0;
  const totalSteps = steps.length;
  const container = document.createElement("div");
  container.style.cssText = "border:1px solid #e5e7eb; border-radius:8px; padding:1rem 1.25rem; margin:0.75rem 0; font-family:inherit;";
  const stepsDiv = document.createElement("div");
  stepsDiv.style.marginBottom = "0.75rem";
  function renderSteps() {
    stepsDiv.innerHTML = "";
    for (let i = 0; i < current; i++) {
      const s = steps[i];
      const row = document.createElement("div");
      row.style.cssText = "display:grid; grid-template-columns:220px 1fr; gap:0.35rem 1rem; align-items:baseline; padding:0.35rem 0; border-top:1px solid #f3f4f6;";
      const opCell = document.createElement("span");
      opCell.style.cssText = "font-size:0.85em; color:#6b7280; font-style:italic;";
      opCell.textContent = s.op;
      const eqCell = document.createElement("span");
      eqCell.innerHTML = katex.renderToString(s.eq, { throwOnError: false, displayMode: false });
      row.appendChild(opCell); row.appendChild(eqCell);
      if (s.note) {
        const noteRow = document.createElement("div");
        noteRow.style.cssText = "grid-column:2; font-size:0.82em; color:#6b7280; padding-bottom:0.2rem;";
        noteRow.textContent = s.note;
        row.appendChild(noteRow);
      }
      stepsDiv.appendChild(row);
    }
    if (current === totalSteps) {
      const done = document.createElement("div");
      done.style.cssText = "margin-top:0.5rem; font-size:0.9em; color:#059669; font-weight:500;";
      done.textContent = "Solution complete.";
      stepsDiv.appendChild(done);
    }
  }
  const controls = document.createElement("div");
  controls.style.cssText = "display:flex; gap:0.5rem; align-items:center;";
  const nextBtn = document.createElement("button");
  nextBtn.textContent = "Next step →";
  nextBtn.style.cssText = "padding:0.35rem 0.85rem; border:1px solid #d1d5db; border-radius:4px; background:#fff; cursor:pointer; font-size:0.9em;";
  const resetBtn = document.createElement("button");
  resetBtn.textContent = "Reset";
  resetBtn.style.cssText = "padding:0.35rem 0.75rem; border:1px solid #e5e7eb; border-radius:4px; background:#f9fafb; cursor:pointer; font-size:0.9em; color:#6b7280;";
  const counter = document.createElement("span");
  counter.style.cssText = "font-size:0.82em; color:#9ca3af; margin-left:0.25rem;";
  function updateButtons() {
    nextBtn.disabled = current === totalSteps;
    nextBtn.style.opacity = current === totalSteps ? "0.4" : "1";
    counter.textContent = current === 0 ? "Click to reveal steps" : `Step ${current} of ${totalSteps}`;
  }
  nextBtn.onclick = () => { if (current < totalSteps) { current++; renderSteps(); updateButtons(); } };
  resetBtn.onclick = () => { current = 0; renderSteps(); updateButtons(); };
  controls.appendChild(nextBtn); controls.appendChild(resetBtn); controls.appendChild(counter);
  container.appendChild(stepsDiv); container.appendChild(controls);
  renderSteps(); updateButtons();
  return container;
}

---

49.6.1 Exercise 1: Direct computation of a contour integral

Compute \(\displaystyle\int_C \bar{z}\,dz\) where \(C\) is the straight line from \(0\) to \(1+i\).

makeStepperHTML([
  {
    op: "Parametrise C: z(t) = t(1+i), 0 ≤ t ≤ 1",
    eq: "z(t) = t + it, \\quad dz = (1+i)\\,dt, \\quad \\bar{z}(t) = t - it = t(1-i)"
  },
  {
    op: "Substitute into the integral",
    eq: "\\int_C \\bar{z}\\,dz = \\int_0^1 t(1-i)(1+i)\\,dt = \\int_0^1 t(1-i^2)\\,dt = \\int_0^1 2t\\,dt"
  },
  {
    op: "Evaluate",
    eq: "\\int_C \\bar{z}\\,dz = 2\\cdot\\frac{1}{2} = 1"
  },
  {
    op: "Compare with ∫_C z dz (which IS path-independent)",
    eq: "\\int_C z\\,dz = \\left[\\frac{z^2}{2}\\right]_0^{1+i} = \\frac{(1+i)^2}{2} = \\frac{2i}{2} = i",
    note: "z dz = d(z²/2) since z² is analytic. But z̄ is not analytic, so z̄ dz is path-dependent and gives a real answer 1 here."
  }
])

---

49.6.2 Exercise 2: Cauchy’s theorem — zero integral

Show that \(\displaystyle\oint_C \frac{e^z}{z^2+4}\,dz = 0\) where \(C\) is the circle \(|z| = 1\).

makeStepperHTML([
  {
    op: "Locate the singularities of the integrand",
    eq: "z^2 + 4 = 0 \\implies z = \\pm 2i"
  },
  {
    op: "Check whether singularities lie inside |z| = 1",
    eq: "|2i| = 2 > 1 \\quad \\text{and} \\quad |-2i| = 2 > 1",
    note: "Both poles are outside the contour C."
  },
  {
    op: "Apply Cauchy-Goursat theorem",
    eq: "f(z) = \\frac{e^z}{z^2+4} \\text{ is analytic everywhere inside } |z|=1",
    note: "No singularities inside C, so the domain enclosed is simply connected and f is analytic there."
  },
  {
    op: "Conclude",
    eq: "\\oint_{|z|=1}\\frac{e^z}{z^2+4}\\,dz = 0"
  }
])

---

49.6.3 Exercise 3: Cauchy’s integral formula

Evaluate \(\displaystyle\oint_C \frac{e^z}{z - 1}\,dz\) where \(C\) is the circle \(|z| = 2\) (counterclockwise).

makeStepperHTML([
  {
    op: "Identify z₀ and check it is inside C",
    eq: "\\text{Integrand} = \\frac{f(z)}{z-z_0} \\text{ with } f(z) = e^z,\\; z_0 = 1",
    note: "|z₀| = 1 < 2, so z₀ = 1 is inside |z| = 2. f(z) = e^z is analytic everywhere."
  },
  {
    op: "Apply Cauchy's integral formula",
    eq: "\\oint_C \\frac{e^z}{z-1}\\,dz = 2\\pi i\\,f(1) = 2\\pi i\\,e"
  }
])

---

49.6.4 Exercise 4: Derivative formula

Evaluate \(\displaystyle\oint_C \frac{\cos z}{(z - \pi)^2}\,dz\) where \(C: |z| = 4\) (counterclockwise).

makeStepperHTML([
  {
    op: "Recognise the derivative formula with n=1",
    eq: "\\oint_C \\frac{f(z)}{(z-z_0)^{n+1}}\\,dz = \\frac{2\\pi i}{n!}\\,f^{(n)}(z_0)"
  },
  {
    op: "Identify f(z) = cos z, z₀ = π, n = 1",
    eq: "\\text{Need } f'(\\pi) = -\\sin(\\pi) = 0"
  },
  {
    op: "Apply the formula",
    eq: "\\oint_C \\frac{\\cos z}{(z-\\pi)^2}\\,dz = \\frac{2\\pi i}{1!}\\,f'(\\pi) = 2\\pi i \\cdot 0 = 0",
    note: "The integral is zero not because the integrand is analytic inside C (it has a pole at π), but because the numerator's derivative vanishes at the pole."
  }
])

---

49.6.5 Exercise 5: ML inequality estimate

Without evaluating it exactly, show that \(\displaystyle\left|\oint_C \frac{dz}{z^3}\right| \leq \frac{\pi}{4}\), where \(C\) is the upper semicircle \(|z| = 2\) from \(z = 2\) to \(z = -2\).

makeStepperHTML([
  {
    op: "Identify M = max|f(z)| on C",
    eq: "f(z) = z^{-3}, \\quad |f(z)| = |z|^{-3} = 2^{-3} = \\tfrac{1}{8} \\text{ on } |z|=2"
  },
  {
    op: "Compute the arc length L",
    eq: "C \\text{ is a semicircle of radius } 2 \\implies L = \\pi\\cdot 2 = 2\\pi"
  },
  {
    op: "Apply the ML inequality",
    eq: "\\left|\\oint_C \\frac{dz}{z^3}\\right| \\leq M\\cdot L = \\frac{1}{8}\\cdot 2\\pi = \\frac{\\pi}{4}",
    note: "The actual value (by a direct computation or the residue theorem) is 0, since z^{−3} = d(−z^{−2}/2)/dz is analytic away from the origin and the origin is not enclosed."
  }
])