Complex integration — contour integrals: Exercises

Before we start, one shared picture of the playground. A contour is a curve we walk along in the flat plane of complex numbers; a pole is a point where the function blows up (a "spike"). The whole game is: is the spike inside my loop or not?

Figure 1 — read it like this. The black circle is the contour $∣ z ∣ = 2$ with a counterclockwise arrow (the positive direction). The red cross at $z = 1 + \frac{1}{2} i$ sits inside: a pole there contributes $2 π i Res$ . The black cross at $z = 3$ sits outside: it contributes $0$ . This single "inside vs outside" test drives every problem below.

Figure — Complex integration — contour integrals

L1 — Recognition

Goal: name the tool. No heavy computation.

Recall Solution 1.1

What tool? Tool 2, the $z^{n}$ -loop lemma: $\oint_{∣ z ∣ = 1} z^{n} d z = 2 π i$ only when $n = - 1$ , else $0$ .

$z^{4}$ has $n = 4 \neq = - 1$ , so $\oint z^{4} d z = 0$ .
$z^{- 1}$ has $n = - 1$ , so $\oint z^{- 1} d z = 2 π i$ .

Why: on the unit circle $z = e^{i t}$ , the integrand becomes $e^{i (n + 1) t}$ . A full loop averages any wobbling exponential to zero — unless $n + 1 = 0$ makes it the constant $1$ , which survives.

Recall Solution 1.2

What tool? Tool 3, Cauchy's Integral Theorem. The integrand is a polynomial, which is analytic everywhere — no spikes anywhere, let alone inside the loop. $\oint_{∣ z ∣ = 1} (3 z^{2} - 5 z + 7) d z = 0.$ Why, spelled out via Green's theorem. Writing $f = u + i v$ , $z = x + i y$ , the real and imaginary parts of $\oint f d z$ become the two line integrals $\oint (u d x - v d y)$ and $\oint (v d x + u d y)$ . Green's theorem turns each closed line integral into an area integral over the enclosed region $D$ : the "Green's-theorem integrands" are $(- v_{x} - u_{y})$ and $(u_{x} - v_{y})$ . For an analytic $f$ the Cauchy–Riemann equations $u_{x} = v_{y}, u_{y} = - v_{x}$ make both of these exactly zero, so both area integrals vanish and the loop is $0$ .

Recall Solution 1.3

What tool? "Is the spike inside?" The pole sits at $z = 3$ . The contour is the circle of radius $2$ . Since $∣3∣ = 3 > 2$ , the pole is outside (like the black cross in Figure 1). Inside the loop the function is analytic, so $\oint_{∣ z ∣ = 2} \frac{d z}{z - 3} = 0.$

L2 — Application

Goal: run one tool cleanly, start to finish.

Recall Solution 2.1

Parametrise (Tool 1): a straight line is linear, $z (t) = t (2 + i)$ for $t \in [0, 1]$ , so $z^{'} (t) = 2 + i$ and $d z = (2 + i) d t$ . $\int_{0}^{1} t (2 + i) \cdot (2 + i) d t = (2 + i)^{2} \int_{0}^{1} t d t = (2 + i)^{2} \cdot \frac{1}{2} .$ Now $(2 + i)^{2} = 4 + 4 i + i^{2} = 3 + 4 i$ , so the answer is $\frac{1}{2} (3 + 4 i) = \frac{3}{2} + 2 i$ . Sanity check: $z$ is analytic, so the answer should equal the antiderivative $\frac{z ^{2}}{2}$ evaluated at the endpoints: $\frac{( 2 + i ) ^{2}}{2} - 0 = \frac{3 + 4 i}{2} = \frac{3}{2} + 2 i$ . ✔

Recall Solution 2.2

What tool? Tool 3, CIF: $f (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{z - a} d z$ . Match: $f (z) = e^{z}$ , $a = 2$ . The pole $z = 2$ satisfies $∣2∣ = 2 < 3$ , so it is inside $∣ z ∣ = 3$ , and $e^{z}$ is analytic there. $\oint_{∣ z ∣ = 3} \frac{e ^{z}}{z - 2} d z = 2 π i f (2) = 2 π i e^{2} .$

Recall Solution 2.3

Factor first: $z^{2} - 1 = (z - 1) (z + 1)$ , giving simple poles at $z = 1$ and $z = - 1$ ; both have modulus $1 < 2$ , so both are inside. Cover-up method (see the definition above): for $z = 1$ cover $(z - 1)$ , leaving $\frac{1}{z + 1}$ at $z = 1$ ; for $z = - 1$ cover $(z + 1)$ , leaving $\frac{1}{z - 1}$ at $z = - 1$ : $Res_{z = 1} = \frac{1}{1 + 1} = \frac{1}{2}, Res_{z = - 1} = \frac{1}{- 1 - 1} = - \frac{1}{2} .$ Sum $= \frac{1}{2} - \frac{1}{2} = 0$ , so by the Residue Theorem $\oint_{∣ z ∣ = 2} \frac{d z}{z ^{2} - 1} = 2 π i \cdot 0 = 0.$

L3 — Analysis

Goal: pick the right tool and justify why the others fail.

Recall Solution 3.1

Locate poles: $z = 1$ (modulus $1 < 2$ , inside) and $z = 3$ (modulus $3 > 2$ , outside). Only $z = 1$ counts (exactly the "inside vs outside" test of Figure 1). Cover-up at the simple pole $z = 1$ : cover the factor $(z - 1)$ , substitute $z = 1$ into the rest: $Res_{z = 1} = \frac{z}{z - 3}_{z = 1} = \frac{1}{1 - 3} = - \frac{1}{2} .$ $\oint_{∣ z ∣ = 2} \frac{z d z}{( z - 1 ) ( z - 3 )} = 2 π i (- \frac{1}{2}) = - π i .$

Recall Solution 3.2

Why not residues? $\overset{z}{ˉ}$ (the conjugate) is not analytic — it fails the Cauchy–Riemann equations — so no Cauchy/residue machinery applies. We must parametrise by hand (Tool 1). Parametrise: $z (t) = e^{i t}$ for $t \in [0, π]$ ; then $\overset{z}{ˉ} = e^{- i t}$ and $z^{'} (t) = i e^{i t}$ . $\int_{0}^{π} e^{- i t} \cdot i e^{i t} d t = i \int_{0}^{π} d t = iπ .$ Path-dependence check: along the straight line from $1$ to $- 1$ , take $z (t) = 1 - 2 t$ ( $t \in [0, 1]$ ), which is real so $\overset{z}{ˉ} = z = 1 - 2 t$ and $z^{'} = - 2$ : $\int_{0}^{1} (1 - 2 t) (- 2) d t = - 2 [t - t^{2}]_{0}^{1} = - 2 (0) = 0.$ Two paths, same endpoints, different answers ( $iπ$ vs $0$ ) — path-dependence confirmed, exactly because $\overset{z}{ˉ}$ is not analytic.

Figure 2 — read it like this. Both paths start at $z = 1$ and end at $z = - 1$ . The red upper semicircle gives $iπ$ ; the black straight segment along the real axis gives $0$ . Same endpoints, different answers — the visual proof that $\int \overset{z}{ˉ} d z$ is path-dependent because $\overset{z}{ˉ}$ is not analytic. (Contrast an analytic $f$ , where Figure 1's logic would force the two paths to agree.)

L4 — Synthesis

Goal: combine contour tricks to crack a real integral.

Definition When does the big arc vanish? (Jordan's lemma & the degree test)

Closing the real line with a semicircle only works if the arc's contribution dies as its radius $R \to \infty$ . Two clean sufficient conditions:

Degree test (rational functions). If $f = \frac{P ( z )}{Q ( z )}$ with $de g Q \geq de g P + 2$ , then on the arc $∣ f ∣ \leq \frac{c}{R ^{2}}$ while the arc length is $π R$ , so the arc integral is at most $\frac{c π R}{R ^{2}} = \frac{c π}{R} \to 0$ . This is the " $∣ f ∣ \sim 1/ R^{2}$ " heuristic made precise: you need the denominator to outrun the numerator by at least two degrees.
Jordan's lemma (oscillatory factor). For integrals carrying $e^{im z}$ with $m > 0$ , the arc in the upper half-plane vanishes as long as $f (z) \to 0$ uniformly on the arc — even if only $de g Q \geq de g P + 1$ — because $∣ e^{im z} ∣ = e^{- m Im z}$ decays on the upper arc.

Figure 3 shows the two ingredients: the shrinking bound $c π / R$ and the growing arc length $π R$ , whose product still $\to 0$ .

Recall Solution 4.1

Plan: replace $x$ by complex $z$ , integrate $\frac{1}{z ^{2} + 4}$ around the real axis closed by a big semicircle of radius $R$ in the upper half-plane (counterclockwise, the positive orientation), let $R \to \infty$ . Arc check (degree test): here $P = 1$ , $Q = z^{2} + 4$ , and $de g Q = 2 \geq de g P + 2 = 2$ , so the arc vanishes. Thus the closed loop equals the real-line integral. Poles: $z^{2} + 4 = 0 \Rightarrow z = \pm 2 i$ . Only $z = 2 i$ is in the upper half-plane. Cover-up / derivative residue at $z = 2 i$ ( $q = z^{2} + 4$ , $q^{'} = 2 z$ ): $Res_{z = 2 i} = \frac{1}{2 z}_{z = 2 i} = \frac{1}{4 i} .$ $\int_{- \infty}^{\infty} \frac{d x}{x ^{2} + 4} = 2 π i \cdot \frac{1}{4 i} = \frac{π}{2} .$ Check: $\frac{1}{a} arctan \frac{x}{a}$ with $a = 2$ gives $\frac{1}{2} \cdot π = \frac{π}{2}$ . ✔

Edge case — closing in the LOWER half-plane. You may instead close with a semicircle below the real axis. That loop is traced clockwise, so it equals $- 2 π i \sum (lower residues)$ . The only lower pole is $z = - 2 i$ with $Res_{- 2 i} = \frac{1}{2 ( - 2 i )} = - \frac{1}{4 i}$ , giving $- 2 π i \cdot (- \frac{1}{4 i}) = \frac{π}{2}$ — the same answer, as it must be. Two orientations, one truth: the $- 1$ from reversing orientation cancels the sign flip of the other residue.

Recall Solution 4.2

Poles: roots of $z^{4} = - 1$ are $z = e^{iπ /4}, e^{i 3 π /4}, e^{i 5 π /4}, e^{i 7 π /4}$ . The two in the upper half-plane are $z_{1} = e^{iπ /4}$ and $z_{2} = e^{i 3 π /4}$ . Arc check (degree test): $P = z^{2}$ , $Q = z^{4} + 1$ , $de g Q = 4 \geq de g P + 2 = 4$ , so the arc contribution $\to 0$ . Cover-up / derivative residues of $\frac{z ^{2}}{z ^{4} + 1}$ , with $q^{'} = 4 z^{3}$ , so $Res = \frac{z ^{2}}{4 z ^{3}} = \frac{1}{4 z}$ :

\operatorname{Res}_{z_2}=\frac{1}{4e^{i3\pi/4}}=\tfrac14 e^{-i3\pi/4}.$$ Sum $=\tfrac14\big(e^{-i\pi/4}+e^{-i3\pi/4}\big)=\tfrac14\big(\cos\tfrac\pi4 - i\sin\tfrac\pi4 - \cos\tfrac\pi4 - i\sin\tfrac\pi4\big)=\tfrac14\big(-2i\sin\tfrac\pi4\big)=\tfrac14\left(-2i\cdot\tfrac{\sqrt2}{2}\right)=-\tfrac{i\sqrt2}{4}.$ $$\int_{-\infty}^{\infty}\frac{x^2}{x^4+1}\,dx = 2\pi i\left(-\frac{i\sqrt2}{4}\right)=\frac{\pi\sqrt2}{2}=\frac{\pi}{\sqrt2}.$$

L5 — Mastery

Goal: higher-order poles, Laurent series, and full case-handling.

Definition The Laurent series and the order-

$m$ residue formula Near an isolated pole $z_{0}$ , a function that is not analytic there can still be written as a Laurent series — an ordinary power series plus a finite tail of negative powers:

+\ c_0 + c_1(z-z_0)+c_2(z-z_0)^2+\cdots$$ The pole has **order $m$** if $c_{-m}\neq 0$ but all deeper coefficients vanish. The **residue** is always the single coefficient $c_{-1}$ — the only term that survives a loop (Tool 2). - **How to build the series in practice:** take a known power series (like $e^z=1+z+\tfrac{z^2}{2}+\cdots$) and divide by the pole factor. Dividing $e^z$ by $z^2$ shifts every power down by two, instantly exposing the $c_{-1}$ term. - **Order-$m$ residue formula (when the series is awkward):** $$\operatorname{Res}_{z_0}f=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{\,m-1}}{dz^{\,m-1}}\Big[(z-z_0)^m f(z)\Big].$$ For a **simple** pole ($m=1$) this reduces to the cover-up rule (no derivative, just substitute). For a **double** pole ($m=2$) it says: multiply by $(z-z_0)^2$, differentiate **once**, then take the limit.

Recall Solution 5.1

Method A — Laurent series (read off $c_{- 1}$ ). Expand $e^{z} = 1 + z + \frac{z ^{2}}{2} + \dots$ and divide by $z^{2}$ (every power drops by two): $\frac{e ^{z}}{z ^{2}} = \frac{1}{z ^{2}} + \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \dots$ The coefficient of $\frac{1}{z}$ is $c_{- 1} = 1$ . That single term is the only survivor of a loop (Tool 2), so $\oint_{∣ z ∣ = 1} \frac{e ^{z}}{z ^{2}} d z = 2 π i c_{- 1} = 2 π i \cdot 1 = 2 π i .$ Method B — order- $m$ formula (cross-check). Here $m = 2$ , $z_{0} = 0$ . Multiply by $z^{2}$ : $(z - 0)^{2} f = e^{z}$ . Differentiate once: $\frac{d}{d z} e^{z} = e^{z}$ . Take the limit $z \to 0$ and divide by $(m - 1)! = 1! = 1$ : $Res_{0} \frac{e ^{z}}{z ^{2}} = \frac{1}{1 !} lim_{z \to 0} \frac{d}{d z} [e^{z}] = e^{0} = 1.$ Both methods agree: $c_{- 1} = 1$ , integral $= 2 π i$ . ✔

Recall Solution 5.2

Poles: $z = 0$ ( $∣0∣ = 0 < 3$ , inside), $z = 1$ ( $< 3$ , inside), $z = 4$ ( $> 3$ , outside). Keep $0$ and $1$ . Cover-up method (cover the factor at each pole, plug in the rest): $Res_{0} = \frac{2 z + 1}{( z - 1 ) ( z - 4 )}_{z = 0} = \frac{1}{( - 1 ) ( - 4 )} = \frac{1}{4},$ $Res_{1} = \frac{2 z + 1}{z ( z - 4 )}_{z = 1} = \frac{3}{( 1 ) ( - 3 )} = - 1.$ Sum $= \frac{1}{4} - 1 = - \frac{3}{4}$ , so by the Residue Theorem $\oint_{∣ z ∣ = 3} \frac{( 2 z + 1 ) d z}{z ( z - 1 ) ( z - 4 )} = 2 π i (- \frac{3}{4}) = - \frac{3 π i}{2} .$

Recall Solution 5.3

The substitution trick: set $z = e^{i θ}$ so as $θ$ runs $0 \to 2 π$ , $z$ traverses the unit circle once counterclockwise (the positive orientation). Then $cos θ = \frac{z + z ^{- 1}}{2}, d θ = \frac{d z}{i z} .$ Why: $d z = i e^{i θ} d θ = i z d θ$ , so $d θ = d z / (i z)$ ; this converts the trig integral into a rational contour integral. Full algebra of the substitution (shown step by step): $5 + 4 cos θ = 5 + 4 \cdot \frac{z + z ^{- 1}}{2} = 5 + 2 z + 2 z^{- 1} .$ So the integrand times $d θ$ is

= \frac{1}{iz\,(5+2z+2z^{-1})}\,dz = \frac{1}{i\,(5z+2z^{2}+2)}\,dz,$$ where the last step multiplied the $z$ from $iz$ into the bracket: $z(5+2z+2z^{-1})=5z+2z^2+2$. Hence $$\int_0^{2\pi}\frac{d\theta}{5+4\cos\theta}=\oint_{|z|=1}\frac{dz}{i\,(2z^2+5z+2)}.$$ **Factor the denominator:** $2z^2+5z+2=(2z+1)(z+2)$, so poles at $z=-\tfrac12$ (inside, $|{-\tfrac12}|<1$) and $z=-2$ (outside). Only $z=-\tfrac12$ counts. **Cover-up residue** of $\dfrac{1}{i(2z+1)(z+2)}$ at $z=-\tfrac12$ (write $2z+1=2(z+\tfrac12)$, cover $(z+\tfrac12)$): $$\operatorname{Res}_{-1/2}=\frac{1}{i\cdot 2\,(z+2)}\bigg|_{z=-1/2}=\frac{1}{i\cdot 2\cdot \tfrac32}=\frac{1}{3i}.$$ $$\int_0^{2\pi}\frac{d\theta}{5+4\cos\theta}=2\pi i\cdot\frac{1}{3i}=\frac{2\pi}{3}.$$

Recall Master check — one-line reasons

Why did $z = 3$ not contribute in Ex 3.1? ::: It lies outside $∣ z ∣ = 2$ , so it is not enclosed. Which tool handles $\int \overset{z}{ˉ} d z$ and why? ::: Direct parametrisation, because $\overset{z}{ˉ}$ is not analytic (no residues). How do you turn $\int_{0}^{2 π} (\dots cos θ) d θ$ into a contour integral? ::: Set $z = e^{i θ}$ , $cos θ = \frac{z + z ^{- 1}}{2}$ , $d θ = \frac{d z}{i z}$ . Residue of $e^{z} / z^{2}$ at $0$ ? ::: $c_{- 1} = 1$ , since $e^{z} / z^{2} = z^{- 2} + z^{- 1} + \frac{1}{2} + \dots$ . When does the order- $m$ formula reduce to plain cover-up? ::: When $m = 1$ (simple pole): no derivative, just substitute. What does reversing a contour's orientation do? ::: Multiplies the integral by $- 1$ (clockwise = negative).

Connections

Residue theorem applications — L4 and L5 are exactly these techniques.
Laurent series — the double-pole residue in Ex 5.1 is a coefficient read-off.
Cauchy–Riemann equations — why Ex 3.2's $\overset{z}{ˉ}$ is path-dependent.
Analytic functions — the dividing line between "loop $= 0$ " and "loop counts spikes".
Green's theorem — the engine that made Cauchy's theorem true in the first place.

Concept Map

The flowchart below is the decision tree you should run on every contour integral: first ask whether $f$ is analytic inside; if yes the loop is $0$ ; if not, find the enclosed poles, compute each residue (simple $\Rightarrow$ cover-up; higher-order $\Rightarrow$ Laurent $c_{- 1}$ or order- $m$ formula), and sum them times $2 π i$ . Real integrals feed in by closing a semicircle or substituting $z = e^{i θ}$ .