50 Series and residues

Laurent series, isolated singularities, and the residue theorem

50.1 Taylor series for analytic functions

If \(f\) is analytic in a disk \(|z - z_0| < R\), it has a Taylor series that converges in that disk:

\[f(z) = \sum_{n=0}^{\infty} c_n(z-z_0)^n, \qquad c_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz\]

The series converges absolutely for \(|z - z_0| < R\) and uniformly on any closed subdisk. The radius of convergence \(R\) equals the distance from \(z_0\) to the nearest singularity of \(f\).

Key Taylor series (centred at 0, valid for all \(z\)):

\[e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}, \qquad \sin z = \sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}, \qquad \cos z = \sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n)!}\]

\[\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n \quad (|z| < 1)\]

These are exactly the real Taylor series with \(x\) replaced by \(z\).

50.2 Laurent series

At a point \(z_0\) where \(f\) is not analytic, the Taylor series does not converge. The Laurent series extends the power series to include negative powers:

\[f(z) = \sum_{n=-\infty}^{\infty} c_n(z-z_0)^n\]

where the coefficients are:

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

This converges in an annulus \(r < |z - z_0| < R\) (with \(r\) possibly 0 and \(R\) possibly \(\infty\)).

The principal part consists of the negative-power terms \(\sum_{n=1}^{\infty} c_{-n}(z-z_0)^{-n}\). The analytic part is the non-negative power series \(\sum_{n=0}^{\infty}c_n(z-z_0)^n\). How many terms the principal part has — zero, finitely many, or infinitely many — classifies the type of singularity, as the next section shows.

Code

{
  const W = 380, H = 340, CX = W/2, CY = H/2, SCALE = 62;
  const svgNS = "http://www.w3.org/2000/svg";

  const funcSel9 = Inputs.select(["1/z", "1/(z(z-1))", "e^{1/z} (essential)"], {
    value: "1/z", label: "f(z)"
  });
  const rInnerSl = Inputs.range([0, 1.4], {value: 0.3, step: 0.05, label: "Inner radius r"});
  const rOuterSl = Inputs.range([0.5, 3.0], {value: 1.5, step: 0.05, label: "Outer radius R"});
  const nTermsSl = Inputs.range([1, 15], {value: 5, step: 1, label: "Terms N"});

  function laurentPartialSum(fname, re, im, N) {
    // Returns approximate f(z) via N-term Laurent expansion at z0=0
    switch(fname) {
      case "1/z": {
        // Exact: 1/z
        const d=re*re+im*im; if(d<1e-10) return [NaN,NaN];
        return [re/d,-im/d];
      }
      case "1/(z(z-1))": {
        // = -1/z - 1 - z - z^2 - ... for |z|<1
        // = 1/z^2 * 1/(1 - 1/z) = 1/z^2 * sum (1/z)^n for |z|>1
        const d2=re*re+im*im; if(d2<1e-10) return [NaN,NaN];
        // Use exact formula
        const dr=re-1,di=im,d3=dr*dr+di*di; if(d3<1e-10) return [NaN,NaN];
        // 1/(z(z-1)) = 1/(z-1) - 1/z
        const p1r=dr/d3, p1i=-di/d3;
        const p2r=re/d2, p2i=-im/d2;
        return [p1r-p2r, p1i-p2i];
      }
      case "e^{1/z} (essential)": {
        // Partial sum of e^{1/z} = sum (1/z)^n / n!
        const d=re*re+im*im; if(d<1e-10) return [NaN,NaN];
        const invRe=re/d, invIm=-im/d;
        let sumRe=1, sumIm=0, powRe=1, powIm=0, fact=1;
        for(let k=1;k<=Math.min(N,12);k++) {
          const newPowRe=powRe*invRe-powIm*invIm;
          const newPowIm=powRe*invIm+powIm*invRe;
          powRe=newPowRe; powIm=newPowIm; fact*=k;
          sumRe+=powRe/fact; sumIm+=powIm/fact;
        }
        return [sumRe,sumIm];
      }
      default: return [1,0];
    }
  }

  function hslToRgb(h,s,l) {
    h=((h%1)+1)%1;
    const c=(1-Math.abs(2*l-1))*s,x=c*(1-Math.abs((h*6)%2-1)),m=l-c/2;
    let r=0,g=0,b=0;
    if(h<1/6){r=c;g=x;}else if(h<2/6){r=x;g=c;}else if(h<3/6){g=c;b=x;}
    else if(h<4/6){g=x;b=c;}else if(h<5/6){r=x;b=c;}else{r=c;b=x;}
    return [Math.round((r+m)*255),Math.round((g+m)*255),Math.round((b+m)*255)];
  }

  function toScreen(re,im) { return [CX+re*SCALE, CY-im*SCALE]; }

  function renderLaurent(fname, rInner, rOuter, N) {
    const canvas=document.createElement("canvas");
    canvas.width=W; canvas.height=H;
    const ctx=canvas.getContext("2d");
    const img=ctx.createImageData(W,H); const data=img.data;

    for (let px=0;px<W;px++) {
      for (let py=0;py<H;py++) {
        const re=(px-CX)/SCALE, im=-(py-CY)/SCALE;
        const dist=Math.sqrt(re*re+im*im);
        const inAnnulus = dist > rInner && dist < rOuter;

        if(inAnnulus) {
          const [wr,wi]=laurentPartialSum(fname,re,im,N);
          if(!isFinite(wr)||!isFinite(wi)||isNaN(wr)) {
            const idx=(py*W+px)*4;
            data[idx]=255;data[idx+1]=255;data[idx+2]=255;data[idx+3]=255;
            continue;
          }
          const arg=Math.atan2(wi,wr),mag=Math.sqrt(wr*wr+wi*wi);
          const hue=(arg/(2*Math.PI)+1)%1;
          const logmag=Math.log(mag+0.001);
          const brightness=0.5+0.35*Math.tanh(logmag/2);
          const [r,g,b]=hslToRgb(hue,0.8,brightness);
          const idx=(py*W+px)*4; data[idx]=r;data[idx+1]=g;data[idx+2]=b;data[idx+3]=255;
        } else {
          // Outside annulus: dim grey
          const idx=(py*W+px)*4;
          data[idx]=220;data[idx+1]=220;data[idx+2]=220;data[idx+3]=255;
        }
      }
    }
    ctx.putImageData(img,0,0);

    // Axes
    const [ox,oy]=toScreen(0,0);
    ctx.strokeStyle="#9ca3af"; ctx.lineWidth=0.8;
    ctx.beginPath(); ctx.moveTo(0,oy); ctx.lineTo(W,oy); ctx.stroke();
    ctx.beginPath(); ctx.moveTo(ox,0); ctx.lineTo(ox,H); ctx.stroke();

    // Annulus circles
    ctx.strokeStyle="rgba(30,64,175,0.8)"; ctx.lineWidth=2; ctx.setLineDash([5,3]);
    ctx.beginPath(); ctx.arc(ox,oy,rInner*SCALE,0,2*Math.PI); ctx.stroke();
    ctx.strokeStyle="rgba(5,150,105,0.8)";
    ctx.beginPath(); ctx.arc(ox,oy,rOuter*SCALE,0,2*Math.PI); ctx.stroke();
    ctx.setLineDash([]);

    // Singularity marker
    ctx.strokeStyle="#ef4444"; ctx.lineWidth=2;
    ctx.beginPath(); ctx.moveTo(ox-6,oy-6); ctx.lineTo(ox+6,oy+6); ctx.stroke();
    ctx.beginPath(); ctx.moveTo(ox+6,oy-6); ctx.lineTo(ox-6,oy+6); ctx.stroke();

    // Labels
    ctx.fillStyle="rgba(30,64,175,0.9)"; ctx.font="10px sans-serif";
    ctx.fillText(`r = ${rInner.toFixed(2)}`,ox+rInner*SCALE+3,oy-4);
    ctx.fillStyle="rgba(5,150,105,0.9)";
    ctx.fillText(`R = ${rOuter.toFixed(2)}`,ox+rOuter*SCALE+3,oy-4);
    ctx.fillStyle="#374151"; ctx.font="11px sans-serif";
    ctx.fillText(`${fname}, N=${N} terms`,5,14);
    ctx.fillStyle="#6b7280"; ctx.font="10px sans-serif";
    ctx.fillText("Convergence annulus shown in colour",5,H-6);

    // Singularity type label
    let singType="";
    if(fname==="1/z") singType="Simple pole (order 1)";
    else if(fname==="1/(z(z-1))") singType="Poles at 0 and 1";
    else singType="Essential singularity";

    const infoDiv=document.createElement("div");
    infoDiv.style.cssText="margin-top:0.4rem; font-family:sans-serif; font-size:0.85em; color:#374151;";
    infoDiv.innerHTML=`<strong>Singularity type:</strong> <span style="color:#1e40af">${singType}</span> — series converges in the coloured annulus ${rInner.toFixed(2)} &lt; |z| &lt; ${rOuter.toFixed(2)}`;

    const wrap=document.createElement("div");
    wrap.appendChild(canvas);
    wrap.appendChild(infoDiv);
    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="Laurent series: convergence in an annulus";
  container.appendChild(title);
  container.appendChild(funcSel9);
  container.appendChild(rInnerSl);
  container.appendChild(rOuterSl);
  container.appendChild(nTermsSl);

  const vizDiv9=document.createElement("div");
  container.appendChild(vizDiv9);

  const note9=document.createElement("div");
  note9.style.cssText="margin-top:0.5rem; font-size:0.82em; color:#6b7280; font-style:italic;";
  note9.textContent="The domain colouring (hue = arg f, brightness = log|f|) is shown only inside the annulus r < |z| < R — where the Laurent series converges. Grey regions lie outside. Adjust the radii to see how the convergence region changes. The singularity at the origin is marked ×.";
  container.appendChild(note9);

  function update9() {
    // Ensure outer > inner
    if(rOuterSl.value <= rInnerSl.value) rOuterSl.value = rInnerSl.value + 0.2;
    vizDiv9.innerHTML="";
    vizDiv9.appendChild(renderLaurent(funcSel9.value, rInnerSl.value, rOuterSl.value, nTermsSl.value));
  }
  funcSel9.addEventListener("input", update9);
  rInnerSl.addEventListener("input", update9);
  rOuterSl.addEventListener("input", update9);
  nTermsSl.addEventListener("input", update9);
  update9();

  return container;
}

Computing Laurent series in practice. For functions that are products, quotients, or compositions of known series, it is usually more efficient to manipulate the known Taylor series directly rather than computing coefficients by integration.

Example. \(\sin z / z^3\) near \(z = 0\):

\[\frac{\sin z}{z^3} = \frac{1}{z^3}\left(z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots\right) = \frac{1}{z^2} - \frac{1}{6} + \frac{z^2}{120} - \cdots\]

The principal part has only one negative-power term, \(1/z^2\).

50.3 Classification of isolated singularities

An isolated singularity of \(f\) at \(z_0\) is a point where \(f\) is not analytic but is analytic in some punctured disk \(0 < |z - z_0| < \delta\). The Laurent series reveals the type:

Removable singularity. The principal part is absent (no negative-power terms). \(f\) can be assigned a value at \(z_0\) to make it analytic. Example: \(\sin z / z\) at \(z = 0\) — the Laurent series is \(1 - z^2/6 + \cdots\), which extends continuously to \(z = 0\) with value 1.

Pole of order \(m\). The principal part has exactly \(m\) terms: \[f(z) = \frac{c_{-m}}{(z-z_0)^m} + \cdots + \frac{c_{-1}}{z-z_0} + c_0 + c_1(z-z_0) + \cdots, \quad c_{-m} \neq 0\] \(|f(z)| \to \infty\) as \(z \to z_0\). A simple pole is a pole of order 1.

Essential singularity. The principal part has infinitely many terms. The behaviour near \(z_0\) is wild. Example: \(e^{1/z}\) at \(z = 0\) — as \(z \to 0\) along different directions, \(e^{1/z}\) oscillates with arbitrarily large magnitude and argument, taking values dense in the entire complex plane. There is no finite or infinite limit.

50.4 Residues

The residue of \(f\) at an isolated singularity \(z_0\) is the Laurent coefficient \(c_{-1}\):

\[\text{Res}[f, z_0] = c_{-1}\]

Why it matters. Chapter 2 (§ A fundamental example) showed that \(\oint_{|z|=r} dz/z = 2\pi i\), and that \(\oint_{|z|=r} z^n\,dz = 0\) for all other integers \(n\). The same parametrisation argument extends to \((z-z_0)^n\) around any circle centred at \(z_0\). So when integrating the Laurent series term by term around a small circle enclosing \(z_0\), every term vanishes except the \(c_{-1}\) term: \[\oint_C \sum_{n=-\infty}^{\infty} c_n(z-z_0)^n\,dz = c_{-1}\cdot 2\pi i\]

50.4.1 Computing residues

At a simple pole: \(\text{Res}[f, z_0] = \lim_{z \to z_0}(z-z_0)f(z)\).

At a simple pole of \(p/q\) where \(q(z_0)=0\), \(q'(z_0)\neq 0\): \[\text{Res}\!\left[\frac{p}{q}, z_0\right] = \frac{p(z_0)}{q'(z_0)}\]

This is just L’Hôpital applied to the simple-pole limit. Quick check: \(\text{Res}[1/(z^2-1),\,1]\) — here \(p = 1\), \(q = z^2-1\), \(q'(z) = 2z\), so the residue is \(1/(2\cdot 1) = 1/2\).

At a pole of order \(m\): \[\text{Res}[f, z_0] = \frac{1}{(m-1)!}\lim_{z \to z_0}\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^m f(z)\right]\]

50.5 The residue theorem

Residue theorem. Let \(f\) be analytic on and inside a simple closed contour \(C\) (counterclockwise) except for isolated singularities \(z_1, \ldots, z_k\) inside \(C\). Then:

\[\oint_C f(z)\,dz = 2\pi i\sum_{j=1}^{k}\text{Res}[f, z_j]\]

This is the master formula of complex analysis. Cauchy’s integral formula is the special case where \(f\) has only simple poles.

50.6 Evaluating real integrals

50.6.1 Type 1: trigonometric integrals \(\int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta\)

Substitute \(z = e^{i\theta}\): \(\cos\theta = (z+z^{-1})/2\), \(\sin\theta = (z-z^{-1})/(2i)\), \(d\theta = dz/(iz)\). The integral becomes a contour integral on \(|z| = 1\).

50.6.2 Type 2: improper integrals \(\int_{-\infty}^{\infty} R(x)\,dx\)

Close the contour with a large semicircle in the upper (or lower) half-plane. Jordan’s lemma ensures the semicircle’s contribution vanishes as the radius grows, provided the degree of the denominator of \(R\) exceeds the degree of the numerator by at least 1. The residue theorem gives the real integral as \(2\pi i\) times the sum of residues in the upper half-plane.

Example. \(\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+x^2}\). Poles of \(1/(1+z^2)\): \(z = \pm i\). Only \(z = i\) is in the upper half-plane. Residue: \(1/(2i)\). Integral: \(2\pi i \cdot 1/(2i) = \pi\).

Code

{
  const W = 400, H = 360, CX = W/2, CY = H/2, SCALE = 70;
  const svgNS = "http://www.w3.org/2000/svg";

  const funcData = {
    "1/(1+z²)": {
      poles: [{re:0,im:1,res:{re:0,im:-0.5},label:"i"},{re:0,im:-1,res:{re:0,im:0.5},label:"−i"}],
      realIntegral: "π",
      realValue: Math.PI,
      eval: (re,im) => {
        const dr=re, di=im+1, d=dr*dr+di*di;
        if(d<1e-8) return [NaN,NaN];
        const dr2=re, di2=im-1, d2=dr2*dr2+di2*di2;
        // 1/(z²+1) = 1/((z+i)(z-i))
        // just colour: use |f|
        // f = 1/(1+z^2): real = (1-y^2+x^2-2x^2)/... let's compute directly
        const r2 = 1 + re*re - im*im; const i2 = 2*re*im;
        const den = r2*r2 + i2*i2;
        if(den<1e-10) return [NaN,NaN];
        return [r2/den, -i2/den];
      }
    },
    "1/((z²+1)(z²+4))": {
      poles: [
        {re:0,im:1,res:{re:0,im:-1/6},label:"i"},
        {re:0,im:-1,res:{re:0,im:1/6},label:"−i"},
        {re:0,im:2,res:{re:0,im:1/12},label:"2i"},
        {re:0,im:-2,res:{re:0,im:-1/12},label:"−2i"}
      ],
      realIntegral: "π/2",
      realValue: Math.PI/2,
      eval: (re,im) => {
        // f = 1/((z^2+1)(z^2+4))
        const a1=re*re-im*im+1, b1=2*re*im, a2=re*re-im*im+4, b2=2*re*im;
        const ra=a1*a2-b1*b2, ia=a1*b2+a2*b1;
        const d=ra*ra+ia*ia; if(d<1e-10) return [NaN,NaN];
        return [ra/d,-ia/d];
      }
    },
    "z²/((z²−1)(z+2))": {
      poles: [
        {re:1,im:0,res:{re:1/3,im:0},label:"1"},
        {re:-1,im:0,res:{re:1/3,im:0},label:"−1"},
        {re:-2,im:0,res:{re:4/3,im:0},label:"−2"}
      ],
      realIntegral: "(sum of residues)",
      realValue: null,
      eval: (re,im) => {
        const z2r=re*re-im*im, z2i=2*re*im;
        const d1r=z2r-1,d1i=z2i;
        const d2r=re+2,d2i=im;
        const d12r=d1r*d2r-d1i*d2i, d12i=d1r*d2i+d1i*d2r;
        const nd=d12r*d12r+d12i*d12i; if(nd<1e-10) return [NaN,NaN];
        return [(z2r*d12r+z2i*d12i)/nd,(z2i*d12r-z2r*d12i)/nd];
      }
    }
  };

  const funcSelector = Inputs.select(Object.keys(funcData), {
    value: "1/(1+z²)", label: "f(z)"
  });
  const radiusSlider = Inputs.range([0.1, 4.5], {value: 1.5, step: 0.05, label: "Contour radius R"});

  function hslToRgb(h,s,l) {
    h=((h%1)+1)%1;
    const c=(1-Math.abs(2*l-1))*s,x=c*(1-Math.abs((h*6)%2-1)),m=l-c/2;
    let r=0,g=0,b=0;
    if(h<1/6){r=c;g=x;}else if(h<2/6){r=x;g=c;}else if(h<3/6){g=c;b=x;}
    else if(h<4/6){g=x;b=c;}else if(h<5/6){r=x;b=c;}else{r=c;b=x;}
    return [Math.round((r+m)*255),Math.round((g+m)*255),Math.round((b+m)*255)];
  }

  function toScreen(re,im) { return [CX+re*SCALE, CY-im*SCALE]; }

  function render3(fname, R) {
    const fd = funcData[fname];

    const canvas = document.createElement("canvas");
    canvas.width=W; canvas.height=H;
    const ctx=canvas.getContext("2d");
    const img=ctx.createImageData(W,H); const data=img.data;

    for (let px=0;px<W;px++) {
      for (let py=0;py<H;py++) {
        const re=(px-CX)/SCALE, im=-(py-CY)/SCALE;
        const [wr,wi]=fd.eval(re,im);
        if(!isFinite(wr)||!isFinite(wi)) {
          const idx=(py*W+px)*4;
          data[idx]=255;data[idx+1]=255;data[idx+2]=255;data[idx+3]=255;
          continue;
        }
        const arg=Math.atan2(wi,wr), mag=Math.sqrt(wr*wr+wi*wi);
        const hue=(arg/(2*Math.PI)+1)%1;
        const logmag=Math.log(mag+0.001);
        const brightness=0.5+0.35*Math.tanh(logmag/2);
        const [r,g,b]=hslToRgb(hue,0.75,brightness);
        const idx=(py*W+px)*4;
        data[idx]=r;data[idx+1]=g;data[idx+2]=b;data[idx+3]=255;
      }
    }
    ctx.putImageData(img,0,0);

    // Contour
    const [cx0,cy0]=toScreen(0,0);
    ctx.strokeStyle="rgba(255,255,255,0.9)"; ctx.lineWidth=2;
    ctx.beginPath(); ctx.arc(cx0,cy0,R*SCALE,0,2*Math.PI); ctx.stroke();
    ctx.fillStyle="rgba(255,255,255,0.8)"; ctx.font="11px sans-serif";
    ctx.fillText(`C: |z| = ${R.toFixed(2)}`,5,14);

    // Pole markers and labels
    const enclosed = [];
    fd.poles.forEach(p => {
      const [psx,psy]=toScreen(p.re,p.im);
      const inC = Math.sqrt(p.re*p.re+p.im*p.im) < R;
      if(inC) enclosed.push(p);
      ctx.strokeStyle=inC?"#fbbf24":"rgba(255,255,255,0.7)";
      ctx.lineWidth=2;
      ctx.beginPath(); ctx.moveTo(psx-7,psy-7); ctx.lineTo(psx+7,psy+7); ctx.stroke();
      ctx.beginPath(); ctx.moveTo(psx+7,psy-7); ctx.lineTo(psx-7,psy+7); ctx.stroke();
      ctx.fillStyle=inC?"#fbbf24":"rgba(255,255,255,0.7)";
      ctx.font="10px sans-serif";
      ctx.fillText(p.label,psx+9,psy+4);
    });

    // Residue sum
    let sumRe=0, sumIm=0;
    enclosed.forEach(p => { sumRe+=p.res.re; sumIm+=p.res.im; });
    const integral2piRe = -2*Math.PI*sumIm;
    const integral2piIm = 2*Math.PI*sumRe;

    const infoDiv=document.createElement("div");
    infoDiv.style.cssText="margin-top:0.5rem; font-family:sans-serif; font-size:0.85em; background:#f0f9ff; border-radius:4px; padding:0.5rem;";
    const encStr=enclosed.length===0?"none":enclosed.map(p=>p.label).join(", ");
    const resStr=enclosed.length===0?"0":(`${sumRe.toFixed(4)}${sumIm>=0?"+":""}${sumIm.toFixed(4)}i`);
    const intStr=enclosed.length===0?"0":`${integral2piRe.toFixed(4)} + ${integral2piIm.toFixed(4)}i`;
    const realStr=(fd.realValue!==null&&enclosed.some(p=>p.im>0))
      ?`<div style="color:#059669">Real integral = 2\u03c0 \xd7 Im(upper-half sum) = ${(2*Math.PI*sumRe).toFixed(4)} [predicted: ${fd.realIntegral}]</div>`
      :"";
    infoDiv.innerHTML=`
      <div><strong>Enclosed poles:</strong> ${encStr}</div>
      <div><strong>Sum of residues:</strong> ${resStr}</div>
      <div><strong>\u222e<sub>C</sub> f\u202fdz = 2\u03c0i \xd7 \u03a3Res = </strong><span style="color:#1e40af">${intStr}</span></div>
      ${realStr}
    `;

    const wrap=document.createElement("div");
    wrap.appendChild(canvas);
    wrap.appendChild(infoDiv);
    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="Residue theorem: sweeping poles with the contour";
  container.appendChild(title);
  container.appendChild(funcSelector);
  container.appendChild(radiusSlider);

  const vizDiv3=document.createElement("div");
  container.appendChild(vizDiv3);

  const note=document.createElement("div");
  note.style.cssText="margin-top:0.5rem; font-size:0.82em; color:#6b7280; font-style:italic;";
  note.textContent="Domain colouring shows f(z): poles appear as white blooms. Gold \u00d7 marks = poles inside contour. As R increases past each pole, the running residue sum updates and the integral changes. Each pole contributes 2\u03c0i times its residue.";
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The residue theorem and Laurent series complete the core tools of complex integration. The next chapter turns to a different application of analyticity: conformal mapping, which uses analytic functions to transform complicated geometries into simple ones where Laplace’s equation can be solved by inspection.

50.7 Exercises

Code

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50.7.1 Exercise 1: Laurent series around a pole

Find the Laurent series of \(f(z) = \dfrac{1}{z(z-1)}\) valid for \(0 < |z| < 1\), and classify the singularity at \(z = 0\).

Code

makeStepperHTML([
  {
    op: "Partial fractions",
    eq: "\\frac{1}{z(z-1)} = \\frac{A}{z} + \\frac{B}{z-1} \\implies A = -1,\\; B = 1",
    note: "Multiply both sides by z: at z=0, A = 1/(0−1) = −1. Multiply by (z−1): at z=1, B = 1/1 = 1."
  },
  {
    op: "Expand 1/(z−1) using geometric series for |z| < 1",
    eq: "\\frac{1}{z-1} = \\frac{-1}{1-z} = -\\sum_{n=0}^{\\infty}z^n"
  },
  {
    op: "Combine",
    eq: "\\frac{1}{z(z-1)} = -\\frac{1}{z} - 1 - z - z^2 - z^3 - \\cdots \\quad (0 < |z| < 1)"
  },
  {
    op: "Classify the singularity at z = 0",
    eq: "\\text{Principal part: } -1/z \\text{ (one term)} \\implies \\text{simple pole of order 1}",
    note: "The residue at z = 0 is the coefficient c_{−1} = −1."
  }
])

50.7.2 Exercise 2: Classifying \(\sin z / z^2\) at \(z = 0\)

Code

makeStepperHTML([
  {
    op: "Expand sin z in Taylor series",
    eq: "\\sin z = z - \\frac{z^3}{6} + \\frac{z^5}{120} - \\cdots"
  },
  {
    op: "Divide by z²",
    eq: "\\frac{\\sin z}{z^2} = \\frac{1}{z} - \\frac{z}{6} + \\frac{z^3}{120} - \\cdots"
  },
  {
    op: "Principal part has one term: 1/z",
    eq: "\\text{Simple pole at } z = 0, \\quad \\text{Res}\\!\\left[\\frac{\\sin z}{z^2},\\, 0\\right] = 1"
  },
  {
    op: "Compare with sin(z)/z and sin(z)/z³",
    eq: "\\frac{\\sin z}{z} \\to 1 \\text{ (removable)}, \\quad \\frac{\\sin z}{z^3} \\text{ has principal part } 1/z^2 \\text{ (order-2 pole)}",
    note: "The order of the pole equals (order of zero of denominator) − (order of zero of numerator): z² has a zero of order 2 at 0; sin z has a zero of order 1; difference = 1."
  }
])

50.7.3 Exercise 3: Residue at a simple pole

Find \(\text{Res}\!\left[\dfrac{z^2+1}{(z-2)(z+3)},\; z=2\right]\).

Code

makeStepperHTML([
  {
    op: "Apply the simple-pole formula",
    eq: "\\text{Res}[f, 2] = \\lim_{z\\to 2}(z-2)f(z) = \\lim_{z\\to 2}\\frac{z^2+1}{z+3}"
  },
  {
    op: "Evaluate at z = 2",
    eq: "= \\frac{4+1}{2+3} = \\frac{5}{5} = 1"
  },
  {
    op: "Alternatively, use p/q′ formula with q = (z−2)(z+3)",
    eq: "q'(z) = 2z+1, \\quad q'(2) = 5, \\quad p(2) = 5 \\implies \\text{Res} = 5/5 = 1 \\checkmark"
  }
])

50.7.4 Exercise 4: Residue at a double pole

Find \(\text{Res}\!\left[\dfrac{e^z}{(z-1)^2},\; z=1\right]\).

Code

makeStepperHTML([
  {
    op: "Apply the order-m formula with m = 2",
    eq: "\\text{Res}[f,1] = \\frac{1}{(2-1)!}\\lim_{z\\to 1}\\frac{d}{dz}\\left[(z-1)^2\\cdot\\frac{e^z}{(z-1)^2}\\right] = \\lim_{z\\to 1}\\frac{d}{dz}e^z"
  },
  {
    op: "Evaluate",
    eq: "\\frac{d}{dz}e^z\\bigg|_{z=1} = e^1 = e"
  },
  {
    op: "Verify via Laurent expansion",
    eq: "\\frac{e^z}{(z-1)^2} = \\frac{e\\,e^{z-1}}{(z-1)^2} = \\frac{e}{(z-1)^2}\\left[1 + (z-1) + \\frac{(z-1)^2}{2} + \\cdots\\right] = \\frac{e}{(z-1)^2} + \\frac{e}{z-1} + \\cdots",
    note: "Coefficient of (z−1)^{−1} is e. ✓"
  }
])

50.7.5 Exercise 5: Trigonometric integral

Evaluate \(\displaystyle\int_0^{2\pi}\frac{d\theta}{2 + \cos\theta}\).

Code

makeStepperHTML([
  {
    op: "Substitute z = e^{iθ}: cos θ = (z+1/z)/2, dθ = dz/(iz)",
    eq: "\\int_0^{2\\pi}\\frac{d\\theta}{2+\\cos\\theta} = \\oint_{|z|=1}\\frac{1}{2+\\frac{z+z^{-1}}{2}}\\cdot\\frac{dz}{iz} = \\oint_{|z|=1}\\frac{2\\,dz}{i(z^2+4z+1)}"
  },
  {
    op: "Find roots of z²+4z+1 = 0",
    eq: "z = \\frac{-4\\pm\\sqrt{12}}{2} = -2\\pm\\sqrt{3}",
    note: "z₁ = −2+√3 ≈ −0.27 (inside |z|=1); z₂ = −2−√3 ≈ −3.73 (outside)."
  },
  {
    op: "Compute Res[2/(i(z²+4z+1)), z₁]",
    eq: "\\text{Res} = \\frac{2}{i(2z_1+4)} = \\frac{2}{i\\cdot 2\\sqrt{3}} = \\frac{1}{i\\sqrt{3}} = \\frac{-i}{\\sqrt{3}}"
  },
  {
    op: "Apply the residue theorem",
    eq: "\\int_0^{2\\pi}\\frac{d\\theta}{2+\\cos\\theta} = 2\\pi i\\cdot\\frac{-i}{\\sqrt{3}} = \\frac{2\\pi}{\\sqrt{3}}"
  }
])

50.7.6 Exercise 6: Improper real integral via residues

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

Code

makeStepperHTML([
  {
    op: "Poles of 1/(1+z²)² at z = ±i, each of order 2",
    eq: "\\frac{1}{(1+z^2)^2} = \\frac{1}{(z-i)^2(z+i)^2}"
  },
  {
    op: "Close contour with upper semicircle; only z = i is in upper half-plane",
    eq: "\\text{Res}\\left[\\frac{1}{(z-i)^2(z+i)^2},\\,i\\right] = \\lim_{z\\to i}\\frac{d}{dz}\\frac{1}{(z+i)^2}"
  },
  {
    op: "Differentiate",
    eq: "\\frac{d}{dz}(z+i)^{-2} = -2(z+i)^{-3} \\implies \\text{Res} = \\frac{-2}{(2i)^3} = \\frac{-2}{-8i} = \\frac{1}{4i} = \\frac{-i}{4}"
  },
  {
    op: "Apply residue theorem (semicircle integral → 0 as R → ∞)",
    eq: "\\int_{-\\infty}^{\\infty}\\frac{dx}{(1+x^2)^2} = 2\\pi i\\cdot\\frac{-i}{4} = \\frac{2\\pi}{4} = \\frac{\\pi}{2}"
  }
])