Complex integration — contour integrals: Visual walkthrough

Step 0 — What are we even integrating along?

Because $z$ is a point on a 2D sheet, "moving a little" — the thing we write $d z$ — is a tiny arrow with a direction, not just a length. That single fact is why complex integration is different from the real line, where $d x$ can only point left or right.

Figure — Complex integration — contour integrals

In the picture: the burnt-orange path is our contour $γ$ . At each point a little teal arrow $d z$ shows which way we are stepping. To add up $f (z)$ along the curve, we need a way to walk the curve in order — that is the parametrisation below.

Step 1 — Turning the walk into a clock: the parametrisation

WHY this works: the left side is a mysterious 2D sum; the right side is an ordinary real integral in the single real variable $t$ . We have converted a walk on a sheet into a clock reading.

Look at the teal velocity arrow $z^{'} (t)$ : it is tangent to the curve and its length is the walking speed. Multiplying the collected value $f (z (t))$ by this arrow is what makes the sum keep track of direction. See Analytic functions for the class of $f$ where the answer stops caring about the path at all.

Step 2 — The seed experiment: walk once around, collecting $1/ z$

Take the simplest interesting loop: the unit circle, $z (t) = e^{i t}$ , $t \in [0, 2 π]$ , going counterclockwise. Here $e^{i t} = cos t + i sin t$ is exactly "the point on the circle at angle $t$ ".

Its velocity: $z^{'} (t) = i e^{i t}$ . The factor $i$ rotates the position arrow by $9 0^{\circ}$ — which is precisely the direction you must face to move around a circle. That is the geometry, not algebra.

$\oint_{∣ z ∣ = 1} \frac{d z}{z} = \int_{0}^{2 π} f (z) = 1/ z \frac{1}{e ^{i t}} z^{'} (t) i e^{i t} d t = \int_{0}^{2 π} i d t = 2 π i .$

$\frac{1}{e ^{i t}}$ — the value we collect at each point.
$i e^{i t}$ — the step direction; the $e^{i t}$ cancels the $1/ e^{i t}$ .
$\int_{0}^{2 π} i d t = 2 π i$ — a full lap contributes a clean $2 π i$ .

WHY it isn't zero: the answer $2 π i \neq = 0$ is the entire source of residue theory. Notice $2 π$ is exactly the total angle swept — going around once. The $i$ marks it as a pure rotation.

The plum arrow tracks the accumulated value: it winds around and lands at $2 π i$ after one lap.

Step 3 — Every other power dies: $z^{n}$ with $n \neq = - 1$

Same circle, now collect $z^{n}$ :

$\oint_{∣ z ∣ = 1} z^{n} d z = \int_{0}^{2 π} z^{n} e^{in t} z^{'} i e^{i t} d t = i \int_{0}^{2 π} e^{i (n + 1) t} d t .$

$e^{in t} = (e^{i t})^{n} = z^{n}$ on the circle.
$e^{i (n + 1) t}$ — a point spinning around at whole-number speed $n + 1$ .

WHY it vanishes: for any integer $n \neq = - 1$ , the exponent $n + 1$ is a nonzero whole number, so $e^{i (n + 1) t}$ makes $∣ n + 1∣$ complete loops as $t$ goes $0 \to 2 π$ . A vector that spins around a whole number of full circles averages to nothing — its integral is $0$ . Only $n = - 1$ makes the exponent $0$ , so the arrow stands still and just accumulates.

$\oint_{∣ z ∣ = 1} z^{n} d z = {2 π i, 0, n = - 1 n \in Z, n \neq = - 1$

Left panel ( $n \neq = - 1$ ): the arrow closes back on the origin — net zero. Right panel ( $n = - 1$ ): it never closes, spiralling to $2 π i$ . This is the whole content of Laurent series residues: only the $c_{- 1} / z$ term survives a loop.

Step 4 — Smooth everywhere inside ⇒ the loop gives zero (Cauchy)

Now split $f$ into its real and imaginary parts, $f = u + i v$ , with $z = x + i y$ and $d z = d x + i d y$ :

$\oint_{γ} f d z = \oint_{γ} (u d x - v d y) + i \oint_{γ} (v d x + u d y) .$

Each piece is a real line integral around a loop, so Green's theorem turns each loop into an area integral over the region $D$ the loop encloses:

+i\iint_D\!\Big(\tfrac{\partial u}{\partial x}-\tfrac{\partial v}{\partial y}\Big)dA.$$ - $\tfrac{\partial u}{\partial x}$ etc. — how fast each part changes as you nudge in each direction. - The two brackets are the two "swirl" densities Green's theorem measures over the whole disc $D$. **WHY both vanish for analytic $f$:** an [[Analytic functions|analytic]] function obeys the [[Cauchy–Riemann equations]] $u_x = v_y$ and $u_y = -v_x$. Substitute: $-v_x - u_y = -v_x - (-v_x) = 0$ and $u_x - v_y = u_x - u_x = 0$. Both integrands are **exactly zero at every interior point** — no swirl anywhere — so the whole loop integral collapses to $0$. > [!formula] Cauchy's Integral Theorem > If $f$ is analytic on and inside a closed contour $\gamma$, then $\displaystyle\oint_\gamma f\,dz = 0$. ![[deepdives/dd-maths-4.10.02-d2-s05.png]] Every little cell of $D$ has zero swirl (blank cells), so summing them gives nothing. Compare Step 2: there the "spike" at $z=0$ sat inside, and $1/z$ is **not** analytic there — that is the loophole. --- ## Step 5 — Shrink the loop onto the spike (Cauchy's Integral Formula) Take an analytic $f$ and a point $a$ **inside** $\gamma$. Look at $\dfrac{f(z)}{z-a}$: it is smooth everywhere except at the single spike $z=a$, where the denominator vanishes. **Key move — deform the loop.** In the ring between $\gamma$ and a tiny circle around $a$, the function $f/(z-a)$ is analytic, so by Step 4 the loop integral over that ring is $0$. That means we may **slide $\gamma$ inward** onto a tiny circle $z = a + \varepsilon e^{it}$ without changing the integral. $$\oint_\gamma\frac{f(z)}{z-a}\,dz=\int_0^{2\pi}\frac{f(a+\varepsilon e^{it})}{\varepsilon e^{it}}\,\underbrace{i\varepsilon e^{it}}_{z'(t)}\,dt = i\int_0^{2\pi}f(a+\varepsilon e^{it})\,dt.$$ - $\varepsilon e^{it}$ — the tiny step away from $a$ at angle $t$. - The $\varepsilon e^{it}$ in $z'$ **cancels** the one in the denominator (just like Step 2!). **WHY the limit is clean:** as $\varepsilon \to 0$ the tiny circle collapses onto $a$, so $f(a+\varepsilon e^{it}) \to f(a)$, a constant. Then $i\int_0^{2\pi} f(a)\,dt = 2\pi i\,f(a)$. > [!formula] Cauchy Integral Formula & Residue Theorem > $$f(a)=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-a}\,dz,\qquad \oint_\gamma f\,dz = 2\pi i\sum_k \operatorname{Res}_{z=z_k} f.$$ > A **residue** is the $c_{-1}$ coefficient of the [[Laurent series]] — the only surviving term of Step 3. ![[deepdives/dd-maths-4.10.02-d2-s06.png]] The big orange loop shrinks (dashed teal) onto the plum micro-circle around the spike $a$; the value never changes because the ring between them is swirl-free. See [[Residue theorem applications]] for using this on real integrals. --- ## Step 6 — Edge cases you must never trip on > [!mistake] "Closed loop ⇒ 0, always." > Only if $f$ is analytic **inside**. A spike inside (like $1/z$) breaks it — you get $2\pi i\sum$Res. > [!mistake] "Path doesn't matter." > True only for analytic $f$. For $\bar z$ (not analytic), swapping the straight line for an arc > changes the answer. Check analyticity first. - **Spike ON the contour** (not inside): the integral is not defined by this theorem — the loop passes through the singularity. Nudge the contour off it (principal-value techniques). - **Clockwise orientation:** reverses the walk, so every $dz \to -dz$ — the answer flips sign: clockwise gives $-2\pi i\sum$Res. - **No spike at all inside** ($n\neq -1$ case, analytic case): the accumulated arrow closes up — $0$. ![[deepdives/dd-maths-4.10.02-d2-s07.png]] Three loops: (1) spike inside → $2\pi i$; (2) spike outside → the loop encloses only smooth ground → $0$; (3) same spike but clockwise → $-2\pi i$. --- ## The one-picture summary ![[deepdives/dd-maths-4.10.02-d2-s08.png]] One frame compresses the whole story: **parametrise** the walk ($dz = z'\,dt$) → on a loop **only the $1/z$ term survives** ($2\pi i$, everything else $0$) → **smooth interior kills the loop** (Cauchy) → **a spike contributes $2\pi i\times$ its residue** (residue theorem). Hard real integrals become *counting the spikes inside*. > [!recall]- Feynman retelling — the whole walkthrough in plain words > You stroll around a flat field, and at every step someone hands you a little arrow (that's > $f(z)\,dz$). To keep it honest you walk by a clock, $t$ — that's the parametrisation. On a perfectly > smooth field, walking a full loop, all the arrows cancel and you finish empty-handed: **zero**. > But bury a magic spike under the field. Now every single lap around the spike hands you the **exact > same** bundle, $2\pi i$ times the spike's strength (its residue), no matter how you wiggle the loop — > because the smooth ring between two loops always cancels, so you can shrink any loop down onto the > spike. That's why a fearsome integral turns into a schoolyard game: **count the spikes inside your > loop, add up their strengths, multiply by $2\pi i$.** Done. > [!mnemonic] > **"Only $1/z$ survives the loop; smooth dies to zero; a spike pays $2\pi i\times$Res."** --- ## Connections - [[Cauchy–Riemann equations]] — the equations that zero out both swirl integrands (Step 4). - [[Green's theorem]] — the real-analysis engine turning loop into area (Step 4). - [[Laurent series]] — residue = the surviving $c_{-1}$ term (Steps 3, 5). - [[Analytic functions]] — the class for which paths don't matter (Steps 1, 4). - [[Residue theorem applications]] — cashing the residue theorem into real integrals.