Complex integration — contour integrals: Worked examples

Before anything: three facts, one convention, and one piece of notation we will lean on constantly. So you never have to chase the parent note, here they are, restated in full.

Recall The three parent facts, restated here (so you never leave this page)

Parent fact 1 — the power lemma. On the unit circle $z = e^{i t}$ , every integer power obeys $\oint_{∣ z ∣ = 1} z^{n} d z = {2 π i, 0, n = - 1 n \in Z, n \neq = - 1.$ Why: substituting $z = e^{i t}$ turns the integral into $i \int_{0}^{2 π} e^{i (n + 1) t} d t$ ; that exponential completes whole loops and cancels to $0$ unless $n + 1 = 0$ , where it is the constant $i$ , giving $2 π i$ . Only the $\frac{1}{z}$ term survives a loop — this is the seed of everything.

Parent fact 2 — Cauchy's theorem. If $f$ is analytic on and inside a closed loop $γ$ , then $\oint_{γ} f d z = 0$ . Why: writing $f = u + i v$ and applying Green's theorem turns the loop integral into area integrals whose integrands are exactly the Cauchy–Riemann equations $u_{x} = v_{y}, u_{y} = - v_{x}$ — both vanish.

Parent fact 3 — the deform-to-a-tiny-circle argument (the "inside" test). If $f$ is analytic in the ring between a big loop and a small circle around a pole $a$ , the two loop integrals are equal (the ring encloses no singularity, so by fact 2 the difference is $0$ ). Shrinking the small circle $z = a + ε e^{i t}$ gives $\oint_{γ} \frac{f ( z )}{z - a} d z = 2 π i f (a)$ . Consequence: a pole outside your loop can be pushed off to nothing and contributes $0$ ; a pole inside always contributes $2 π i$ times its residue.

The scenario matrix

Every contour problem lives in one of these cells. The examples below are labelled by cell.

Cell	What makes it distinct	Which tool wins
A — non-analytic integrand	$f$ like $\overset{z}{ˉ}$ ; Cauchy fails	brute parametrise $d z = z^{'} d t$
B — analytic, no pole inside	$f$ smooth everywhere in the loop	Cauchy's theorem $\Rightarrow 0$
C — one simple pole inside	one spike	$2 π i \cdot \frac{p}{q ^{'}}$
D — several poles, some in some out	selection by "is it inside?"	sum only the enclosed residues
E — higher-order (repeated) pole	spike counted twice	derivative residue formula
F — pole exactly ON the contour	degenerate/limiting	undefined (or principal value)
G — real integral $\int_{- \infty}^{\infty}$	close with a big arc	arc $\to 0$ , then residues
H — real word/exam twist	phrased in disguise	translate, then use C–G
I — essential singularity	$e^{1/ z}$ -type, infinitely sharp	read $c_{- 1}$ off the Laurent series
J — branch point / branch cut	$z, lo g z, x^{s}$	keyhole contour, cannot encircle freely

We now cover cells A → J. Watch the sign, the "inside?" test, and the degenerate cases.

Cell A — Non-analytic: brute force is the only way

Worked example A1 —

$\int_{γ} \overset{z}{ˉ}^{2} d z$ along $z (t) = e^{i t}$ , $t : 0 \to π$ (upper half of unit circle, counterclockwise) Forecast: $\overset{z}{ˉ}$ is not analytic, so path matters and the answer need not be $0$ or a multiple of $2 π i$ . Guess: some plain complex number. Write your guess down.

Parametrise. $z = e^{i t} \Rightarrow \overset{z}{ˉ} = e^{- i t}$ , so $\overset{z}{ˉ}^{2} = e^{- 2 i t}$ , and $z^{'} (t) = i e^{i t}$ . Why this step? Cauchy's theorem needs analyticity; $\overset{z}{ˉ}$ has $u_{x} \neq = v_{y}$ , so we cannot shortcut — we must literally walk the path with clock $t$ .
Assemble the integrand. $\overset{z}{ˉ}^{2} z^{'} = e^{- 2 i t} \cdot i e^{i t} = i e^{- i t}$ . Why? This is just $f (z (t)) z^{'} (t)$ from the definition.
Integrate in $t$ . $\int_{0}^{π} i e^{- i t} d t = i \cdot \frac{e ^{- i t}}{- i}_{0}^{π} = - e^{- i t}_{0}^{π} = - (e^{- iπ} - e^{0}) = - (- 1 - 1) = 2.$

Verify: the answer $2$ is a plain real number — exactly what "non-analytic, path matters" predicted. It is not $0$ , and not a multiple of $2 π i$ . If you got $0$ you probably treated $\overset{z}{ˉ}$ as analytic — see the Analytic functions warning.

Cell B — Analytic, no pole: the answer is $0$

Worked example B1 —

$\oint_{∣ z ∣ = 1} cos z d z$ Forecast: $cos z$ is analytic on the whole plane (an entire function — no spikes anywhere). Predict the loop value before computing.

Check analyticity inside the loop. $cos z$ satisfies the Cauchy–Riemann equations everywhere; there is no denominator to vanish, so no pole is enclosed. Why this step? Cauchy's theorem's only hypothesis is "analytic on and inside $γ$ " (parent fact 2).
Apply Cauchy's Integral Theorem. $\oint_{∣ z ∣ = 1} cos z d z = 0$ . Why this step? An analytic function on and inside the loop has a single-valued antiderivative there ( $sin z$ for $cos z$ ); a closed loop returns to its starting point, so antiderivative at end minus antiderivative at start $= 0$ . This is exactly Cauchy's theorem, and it is why "analytic inside" forces the loop to vanish.

Verify (brute check that it really is $0$ ). Parametrise $z = e^{i t}$ : $\int_{0}^{2 π} cos (e^{i t}) i e^{i t} d t$ . The antiderivative of $cos z$ is $sin z$ , and $sin z$ returns to its start after a full loop ( $z$ ends where it began), so the integral is $sin z$ evaluated at the same point minus itself $= 0$ . ✔

Cell C — One simple pole inside

Cell D — Several poles: only the enclosed ones count

Figure — Complex integration — contour integrals

Figure s01 — the inside/outside test. The lavender loop is the counterclockwise unit circle $∣ z ∣ = 1$ . A coral star at $z = \frac{1}{2}$ sits inside the loop (its residue counts); a grey star at $z = 2$ sits outside (its residue is ignored — parent fact 3 pushes it off to nothing). The small lavender arrow on the circle reminds you the walk is counterclockwise. The whole trick of Cell D is reading this picture: circle, then sum only the stars inside it.

Worked example D1 —

$\oint_{∣ z ∣ = 1} \frac{1}{( z - \frac{1}{2} ) ( z - 2 )} d z$ Forecast: two poles, $z = \frac{1}{2}$ and $z = 2$ . Only $z = \frac{1}{2}$ is inside. Predict: a single residue survives.

Inside/outside test. $∣ \frac{1}{2} ∣ = \frac{1}{2} < 1$ → inside. $∣2∣ = 2 > 1$ → outside, ignored. Why? Deforming the contour past a pole you don't enclose leaves it out of the count entirely.
Residue at $z = \frac{1}{2}$ . Simple pole; cover the $(z - \frac{1}{2})$ factor and plug in the rest: $Res_{1/2} = \frac{1}{( \frac{1}{2} - 2 )} = \frac{1}{- \frac{3}{2}} = - \frac{2}{3}$ . Why cover-up? $Res = lim_{z \to 1/2} (z - \frac{1}{2}) f (z)$ removes the singular factor, leaving the finite value of the rest.
Sum only enclosed residues. $\oint = 2 π i (- \frac{2}{3}) = - \frac{4 π i}{3}$ .

Verify: partial fractions $\frac{1}{( z - \frac{1}{2} ) ( z - 2 )} = \frac{- 2/3}{z - \frac{1}{2}} + \frac{2/3}{z - 2}$ . The second term is analytic inside $∣ z ∣ = 1$ (its pole is outside) → its loop integral is $0$ . The first gives $- \frac{2}{3} \cdot 2 π i = - \frac{4 π i}{3}$ . ✔ Only the inside pole spoke.

Cell E — Higher-order (repeated) pole

Cell F — Pole exactly ON the contour (the degenerate case)

Worked example F1 —

$\oint_{∣ z ∣ = 1} \frac{d z}{z - 1}$ Forecast: pole at $z = 1$ . But $∣1∣ = 1$ — it lies on the circle. Predict: trouble.

Inside/outside test fails. $z = 1$ is neither inside nor outside; it is on $γ$ . Why this matters? The deformation argument (parent fact 3) assumes clear space around the pole; here there is none.
Diagnose. Near $t = 0$ the integrand $\frac{1}{e ^{i t} - 1}$ behaves like $\frac{1}{i t}$ , whose integral over the singularity diverges. So the ordinary contour integral does not exist.
The repair (using the indentation rule above): notch a tiny semicircle around $z = 1$ . That detour contributes $π i \cdot Res_{z = 1} \frac{1}{z - 1} = π i \cdot 1 = π i$ — half a full loop — and the remaining principal-value arc supplies the other half's worth of the analysis. See Residue theorem applications for the full principal-value bookkeeping.

Verify (why "no value"): compare with $z = 1$ moved slightly inside ( $\oint = 2 π i$ ) versus slightly outside ( $\oint = 0$ ). The two limits disagree ( $2 π i \neq = 0$ ), so the on-contour case genuinely has no single value. ✔ The disagreement is the proof it is degenerate — and note the half-residue $π i$ sits exactly midway between $0$ and $2 π i$ , as the indentation rule predicts.

Cell G — A real integral, closed with a big arc

Figure s02 — the semicircular (Jordan) contour. The mint segment along the real axis is the integral we actually want, $[- R, R]$ ; the lavender arc of radius $R$ closes the loop through the upper half-plane. Coral stars mark the poles the loop encloses (upper half); grey stars mark the ones it leaves out (lower half). The two arrows show the walk: rightward along the real axis, then counterclockwise back along the arc. As $R \to \infty$ the arc's contribution shrinks to $0$ , so the real integral equals $2 π i$ times the coral residues only.

Worked example G1 —

$\int_{- \infty}^{\infty} \frac{x ^{2}}{x ^{4} + 1} d x$ Forecast: the integrand is positive and decays like $1/ x^{2}$ , so the answer is a positive real number. Guess a value near $2$ (spoiler: it is $\frac{π}{2} \approx 2.22$ ).

Choose the contour. Real axis $[- R, R]$ + upper semicircular arc $z = R e^{i θ}$ , $θ : 0 \to π$ . Call the loop $γ_{R}$ . Why upper? We need the arc to vanish so that the closed loop equals the real-line integral in the limit.
Show the arc contribution $\to 0$ (the key WHY). On the arc, $∣ z ∣ = R$ , so $z^{4} + 1 \geq R^{4} - 1$ and $∣ z^{2} ∣ = R^{2}$ . Hence the integrand is bounded by $\frac{R ^{2}}{R ^{4} - 1}$ . By the ML-estimate (integral $\leq$ max-modulus $M$ times arc-length $L$ ), with $L = π R$ , $=\frac{\pi R^3}{R^4-1}\xrightarrow{R\to\infty}0.$$ *Why this bound?* $R^3/R^4=1/R\to0$: the denominator's degree beats the numerator by two powers *plus* the extra $R$ from the arc length, so the arc genuinely dies. (This is the elementary version of Jordan's lemma.)$
Find enclosed poles. $z^{4} + 1 = 0 \Rightarrow z = e^{iπ /4}, e^{i 3 π /4}, e^{i 5 π /4}, e^{i 7 π /4}$ . The two in the upper half-plane are $z_{1} = e^{iπ /4}$ and $z_{2} = e^{i 3 π /4}$ . Why only upper? Our loop only encloses the top; the bottom two are outside.
Residues via $p / q^{'}$ . Name top $p (z) = z^{2}$ and bottom $q (z) = z^{4} + 1$ , so $q^{'} (z) = 4 z^{3}$ and $Res = \frac{z ^{2}}{4 z ^{3}} = \frac{1}{4 z}$ .
- At $z_{1} = e^{iπ /4}$ : $\frac{1}{4} e^{- iπ /4}$ .
- At $z_{2} = e^{i 3 π /4}$ : $\frac{1}{4} e^{- i 3 π /4}$ .
Sum, multiply, take $R \to \infty$ . Sum $= \frac{1}{4} (e^{- iπ /4} + e^{- i 3 π /4}) = \frac{1}{4} (\frac{2}{2} - i \frac{2}{2} - \frac{2}{2} - i \frac{2}{2}) = \frac{1}{4} (- i 2) = - \frac{i}{2 2} .$ So $\oint_{γ_{R}} = 2 π i \cdot (- \frac{i}{2 2}) = \frac{π}{2}$ . As $R \to \infty$ the arc drops out (Step 2), leaving $\int_{- \infty}^{\infty} \frac{x ^{2}}{x ^{4} + 1} d x = \frac{π}{2} .$

Verify: numerically $\frac{π}{2} \approx 2.2214$ , a positive real — matching the forecast and the fact that the answer of a real integral of a real function must be real (the $i$ 's cancelled, as they must). ✔

Cell H — Real-world / exam twist (translate first)

Worked example H1 — A signal-processing average:

$\frac{1}{2 π} \int_{0}^{2 π} \frac{d θ}{5 + 4 cos θ}$ Forecast: this is a real trig integral, but the $2 π$ and $cos θ$ scream "unit circle in disguise." Guess a small positive number (it is an average of $\frac{1}{5 + 4 c o s θ}$ , which ranges from $\frac{1}{9}$ to $1$ ).

Substitute $z = e^{i θ}$ . Then $cos θ = \frac{1}{2} (z + \frac{1}{z})$ and $d θ = \frac{d z}{i z}$ , and as $θ : 0 \to 2 π$ , $z$ traces $∣ z ∣ = 1$ once counterclockwise. Why this step? It converts the trig integral into a contour integral we can attack with residues — the whole reason contour methods matter to "real" problems.
Rewrite. $5 + 4 cos θ = 5 + 2 (z + \frac{1}{z}) = \frac{2 z ^{2} + 5 z + 2}{z}$ . So $=\frac{1}{2\pi i}\oint_{|z|=1}\frac{dz}{2z^2+5z+2}.$$$
Poles inside. $2 z^{2} + 5 z + 2 = (2 z + 1) (z + 2) = 0 \Rightarrow z = - \frac{1}{2}$ (inside) or $z = - 2$ (outside). Why check? Only enclosed spikes count (Cell D logic).
Residue at $z = - \frac{1}{2}$ . Name top $p (z) = 1$ and bottom $q (z) = 2 z^{2} + 5 z + 2$ , so $q^{'} (z) = 4 z + 5$ ; at $z = - \frac{1}{2}$ , $q^{'} = - 2 + 5 = 3$ , giving $Res = \frac{1}{3}$ .
Assemble. $\frac{1}{2 π i} \cdot 2 π i \cdot \frac{1}{3} = \frac{1}{3}$ .

Verify: $\frac{1}{3} \approx 0.333$ sits between $\frac{1}{9} \approx 0.111$ and $1$ — a legitimate average of $\frac{1}{5 + 4 c o s θ}$ . Units: dimensionless average, matches. ✔

Cell I — Essential singularity (no finite pole order)

Worked example I1 —

$\oint_{∣ z ∣ = 1} e^{1/ z} d z$ Forecast: infinitely sharp spike at $0$ . You cannot use $p / q^{'}$ or the derivative formula. Guess: the answer is still $2 π i \times (the 1/ z coefficient)$ .

Expand the Laurent series. $e^{1/ z} = \sum_{k \geq 0} \frac{1}{k ! z ^{k}} = 1 + \frac{1}{z} + \frac{1}{2 ! z ^{2}} + \dots$ . Why series? No pole order exists, so the only handle is the coefficient list itself (parent fact 1: only $\frac{1}{z}$ survives).
Read off $c_{- 1}$ . The coefficient of $\frac{1}{z}$ is the $k = 1$ term: $\frac{1}{1 !} = 1$ . So $Res_{0} e^{1/ z} = 1$ .
Residue theorem. $\oint = 2 π i \cdot 1 = 2 π i$ .

Verify: integrate term by term. Every term $\frac{1}{k !} z^{- k}$ is a power $z^{n}$ with $n = - k$ ; by parent fact 1 all give $0$ except $n = - 1$ (i.e. $k = 1$ ), which gives $2 π i \cdot 1$ . So $\oint = 2 π i$ . ✔ The infinitely many other terms integrate to $0$ — the spike's shape is irrelevant, only its $c_{- 1}$ matters.

Cell J — Branch point / branch cut (you cannot encircle freely)

Figure s03 — the keyhole contour. Functions like $z$ or $x^{s}$ are multivalued: walk once around $z = 0$ and the value changes (e.g. $z$ flips sign). A branch point at $0$ forces a branch cut (the coral ray along the positive real axis) that the contour is forbidden to cross. The lavender keyhole hugs the cut: a big circle out, back just above the cut, a tiny circle around $0$ , and back just below. The gap between "above" and "below" the cut is exactly what makes the method work. This is why you cannot naively drop a residue at a branch point — there is no single-valued $c_{- 1}$ to read.

Worked example J1 —

$\int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x$ via a keyhole Forecast: $x^{- 1/2}$ has a branch point at $0$ ; the standard result is $π$ . Guess a value near $3$ .

Set up the keyhole. Use $f (z) = \frac{z ^{- 1/2}}{z + 1}$ with the branch cut along the positive real axis, and define $z^{- 1/2} = e^{- \frac{1}{2} l o g z}$ with $ar g z \in [0, 2 π)$ . Why a keyhole? A plain loop around $0$ is illegal — the value jumps across the cut; the keyhole runs along both sides of the cut instead of crossing it.
Two edges disagree by a known factor. Above the cut ( $ar g z = 0$ ) the integrand is $\frac{x ^{- 1/2}}{x + 1}$ . Below the cut ( $ar g z = 2 π$ ), $z^{- 1/2} = e^{- \frac{1}{2} (2 π i)} x^{- 1/2} = e^{- iπ} x^{- 1/2} = - x^{- 1/2}$ . Walking back the other way flips the direction too, so the two straight edges add to $(1 - e^{- iπ}) \int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x = 2 \int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x$ . Why this step? The whole trick is that the multivaluedness turns the same integral into a nonzero difference we can solve for.
Residue at the enclosed pole $z = - 1$ . In our branch $z = - 1 = e^{iπ}$ , so $z^{- 1/2} = e^{- iπ /2} = - i$ ; residue of $\frac{z ^{- 1/2}}{z + 1}$ at the simple pole $z = - 1$ is $z^{- 1/2}_{z = - 1} = - i$ . The small and large circles contribute $0$ as radii $\to 0, \infty$ .
Assemble. $2 \int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x = 2 π i \cdot (- i) = 2 π$ , so the integral is $π$ .

Verify: substitute $x = u^{2}$ : $\int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x = \int_{0}^{\infty} \frac{2 d u}{u ^{2} + 1} = 2 \cdot \frac{π}{2} = π$ . ✔ Matches the keyhole answer — and shows why a branch point demands its own contour rather than a residue drop.

Sign & orientation sweep — the full clockwise case

Worked example H2 — Same integrand, reversed loop:

$\oint_{∣ z ∣ = 1, clockwise} \frac{d z}{z}$ Forecast: counterclockwise gave $+ 2 π i$ (parent fact 1 with $n = - 1$ ). Reversing the walk should flip the sign to $- 2 π i$ . Predict that before computing.

Reverse the parametrisation. Clockwise means the angle decreases, so take $z = e^{- i t}$ , $t : 0 \to 2 π$ . Then $z^{'} (t) = - i e^{- i t}$ . Why? The sign of the exponent's derivative is what encodes "which way round" — the $- i$ carries the flip.
Assemble the integrand. $\frac{1}{z} z^{'} = \frac{1}{e ^{- i t}} \cdot (- i e^{- i t}) = - i$ . Why? Same $f (z (t)) z^{'} (t)$ recipe; the $e^{- i t}$ factors cancel, leaving a constant.
Integrate in $t$ . $\int_{0}^{2 π} (- i) d t = - i \cdot 2 π = - 2 π i .$

Verify: this is exactly $- 1$ times the counterclockwise value $2 π i$ . ✔ Orientation is a single global sign, nothing more. General rule (from our convention): counterclockwise is $+$ , clockwise is $-$ ; reversing $γ$ multiplies the integral by $- 1$ . Every earlier example assumed the $+$ convention; running one clockwise just negates the result.

Recall checkpoints

Recall Which cell, and what's the first move?

$\oint_{∣ z ∣ = 1} \overset{z}{ˉ} d z$ ::: Cell A (non-analytic) — brute parametrise, cannot use residues. $\oint_{∣ z ∣ = 1} sin z d z$ ::: Cell B (analytic, no pole) — equals $0$ by Cauchy's theorem. $\oint_{∣ z ∣ = 1} \frac{e ^{z}}{z ^{3}} d z$ ::: Cell E (order-3 pole) — need two derivatives; answer $2 π i \cdot \frac{1}{2} = π i$ . A pole sits at $z = 1$ on $∣ z ∣ = 1$ ::: Cell F — integral undefined; needs principal value / indentation. $\oint_{∣ z ∣ = 1} e^{1/ z} d z$ ::: Cell I (essential singularity) — read $c_{- 1} = 1$ off the series; answer $2 π i$ . $\int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x$ ::: Cell J (branch point) — keyhole contour, not a plain loop. Reverse a loop to clockwise ::: Multiply the whole integral by $- 1$ .

Connections

Analytic functions — Cell A vs B: analyticity decides whether paths matter.
Cauchy–Riemann equations — the test behind "is it analytic inside?"
Laurent series — reading residues off $c_{- 1}$ (Cells E, I, and the origin of it all).
Green's theorem — the engine that makes Cell B's answer $0$ .
Residue theorem applications — Cells G, H, J and the principal-value repair of Cell F.

Worked examples — Complex integration — contour integrals