Chapter 7 · Contour integration and Cauchy–Goursat

The hook

Two paths, one answer — sometimes

Story

Every integral you have ever computed ran along the real line, from $a$ to $b$ — and there was only one way to get from $a$ to $b$ . In the complex plane that is no longer true. To integrate a function you must travel from one point to another along a path, and between any two points there are infinitely many paths. So a brand-new question appears, one ordinary calculus never had to ask: does the answer depend on which path you take?

Take the two points $0$ and $1 + i$ , and two ways between them: the straight diagonal, and the right-angled "elbow" $0 \to 1 \to 1 + i$ . Integrate $f(z) = z^2$ along each — you get the same number both times. Integrate $f(z) = \bar z$ along each — you get different numbers. Same two paths, same two endpoints; for one function the path is irrelevant, for the other it is everything.

Solid = the straight diagonal; dashed = the elbow. For

z^2

the two agree; for

\bar z

they do not. The dividing line is a property you already know how to test.

That dividing line is analyticity — the property you spent four chapters learning to detect. For an integrand analytic everywhere between and around the two paths — no holes, no bad points anywhere in the region they enclose — the path does not matter; only the endpoints do, exactly as in the Fundamental Theorem of Calculus. That caveat about holes is not pedantry; it returns with a vengeance later in this chapter. The theorem that makes this precise, and that says a loop integral of an analytic function is simply zero, is the Cauchy–Goursat theorem — the destination of this chapter. The four labelled numbers above are computed properly in Beat 3; for now, just hold the contrast.

Why this matters

This is the chapter where the subject turns from what analytic functions are to calculus with them, and almost everything left in Unit 1 is built directly on it. The next chapter, on Cauchy's integral formula, takes the small surprise you are about to meet — that $\oint_{|z|=1} \frac{dz}{z} = 2\pi i$ rather than $0$ — and turns it into a formula that reconstructs an analytic function inside a loop from its values on the loop. The chapter after that, on residues, generalises the same surprise into a machine for evaluating hard real integrals. And for the exam this is among the most reliably tested topics in the course: evaluate a contour integral, state and prove Cauchy's theorem, bound an integral with the ML inequality. This chapter works each one before it asks you for it.

Intuition before formula

March along the curve, and let the function twist each step

Before any formula, a picture of what $\displaystyle\int_C f(z)\,dz$ is. Chop the path $C$ into many tiny steps. Each step is a small complex displacement $\Delta z$ — a little arrow pointing the way you are travelling. At each step the integral adds the contribution

f(z)\,\Delta z,

and here is the one idea that makes complex integration different: $f(z)$ is a complex number, and multiplying $\Delta z$ by a complex number rotates and scales it — exactly the Chapter-1 fact that multiplication is rotation-and-stretch. So a contour integral is: march along the curve, and at each step let the function twist and stretch your stride, then add up all the twisted strides.

This gives path (in)dependence an intuitive reading. For an analytic $f$ the little twists are so well-organised that, when you add them along any path between two fixed endpoints — provided no singularity is trapped in the region the paths enclose — everything in the middle cancels and only the endpoints survive, the path washes out. For a non-analytic $f$ like $\bar z$ , the twists do not cooperate, the middle does not cancel, and the sum remembers the whole journey. That is the entire story of the chapter, in one sentence.

The practical recipe is just substitution

You never add up infinitely many tiny arrows by hand. Parametrise the path: write $z = z(t)$ for $a \le t \le b$ . Then the step $\Delta z$ becomes $z'(t)\,dt$ , and the whole contour integral collapses into an ordinary real-variable integral:

\int_C f(z)\,dz = \int_a^b f\bigl(z(t)\bigr)\,z'(t)\,dt.

Nothing mystical — one substitution turns it into an integral of a complex-valued function of the real variable $t$ , which you handle by integrating its real and imaginary parts separately. This single line is the workhorse of the whole chapter.

Formal statement · 1 of 5

Contours, and the substitution that computes them

First the vocabulary, all of it standard one-mark exam material. A smooth arc is a curve $z(t)$ with a continuous, non-vanishing derivative; a contour is several smooth arcs joined end to end (so it may have corners, like the sides of a square). A contour is simple if it never crosses itself, and closed if it ends where it began, $z(a) = z(b)$ . A simple closed contour is a loop that bounds a well-defined inside and outside.

Definition · contour integral

For a contour $C$ parametrised by $z(t)$ , $a \le t \le b$ ,

\int_C f(z)\,dz = \int_a^b f\bigl(z(t)\bigr)\,z'(t)\,dt.

For a closed contour we write $\displaystyle\oint_C$ . The positive orientation of a closed contour is counterclockwise — traversed so the enclosed region stays on your left. Every closed contour in this chapter is positively oriented unless stated otherwise.

Three properties follow straight from the definition, and you may use them without comment:

Linearity: $\int_C \bigl(\alpha f + \beta g\bigr)\,dz = \alpha\int_C f\,dz + \beta\int_C g\,dz$ .
Reversing orientation flips the sign: $\int_{-C} f\,dz = -\int_C f\,dz$ . (Every step arrow $\Delta z$ reverses, so the whole sum negates.)
Additivity: a contour split into pieces integrates piece by piece.

Worked example A · the recipe on a closed contour

Evaluate $\displaystyle\oint_{|z|=1} \bar z\,dz$ , once counterclockwise around the unit circle.

Parametrise the circle: $z = e^{it}$ , $0 \le t \le 2\pi$ . Then $\bar z = e^{-it}$ and $z'(t) = i\,e^{it}$ , so the integrand is $e^{-it}\cdot i\,e^{it} = i$ . Hence

\oint_{|z|=1} \bar z\,dz = \int_0^{2\pi} i\,dt = 2\pi i.

Two lessons, both worth keeping. (i) The recipe in action: parametrise, substitute, integrate. (ii) A closed-contour integral need not be zero — this one is $2\pi i$ . Keep this example in your pocket; it returns in the gauntlet.

Worked example B · the hook, done honestly (path dependence)

Now compute the hook's $\bar z$ integrals from $0$ to $1 + i$ along the two paths.

Straight diagonal: $z(t) = (1 + i)t$ , $0 \le t \le 1$ , so $\bar z = (1 - i)t$ and $z'(t) = 1 + i$ :

\int_C \bar z\,dz = \int_0^1 (1 - i)t\,(1 + i)\,dt = (1 - i)(1 + i)\int_0^1 t\,dt = 2\cdot\tfrac12 = 1.

Elbow $0 \to 1 \to 1 + i$ : the first leg ( $z = t,\ z' = 1$ ) gives $\int_0^1 t\,dt = \tfrac12$ ; the second leg ( $z = 1 + it,\ z' = i$ ) gives $\int_0^1 (1 - it)\,i\,dt = \int_0^1 (i + t)\,dt = \tfrac12 + i$ . Total $= 1 + i$ .

So $\bar z$ gives $1$ on the diagonal and $1 + i$ on the elbow — genuinely path-dependent, because $\bar z$ is not analytic and has no antiderivative (Beat 3B). This is exactly the right-hand panel of the hook.

One more non-analytic integrand — $f(z) = |z|$ . On the same diagonal $z = (1 + i)t$ we have $|z(t)| = |t(1+i)| = t\sqrt2$ , so

\int_C |z|\,dz = \int_0^1 t\sqrt2\,(1 + i)\,dt = \sqrt2\,(1 + i)\int_0^1 t\,dt = \frac{\sqrt2}{2}\,(1 + i).

Notice that $|z|$ simplified to a plain expression in $t$ — it did not vanish or stay constant. (On a circle centred at the origin $|z|$ would be the constant radius; on a segment it never is.) We will need this modulus-handling move in Reveal 1, so it is worth seeing once here, ungated.

Formal statement · 2 of 5

Bounding an integral, and the antiderivative shortcut

The ML inequality

If $|f(z)| \le M$ for every $z$ on the contour $C$ , and $L$ is the length of $C$ , then

\left|\int_C f(z)\,dz\right| \le M L.

The plausibility is immediate from the Riemann sum: each stride contributes at most $M\,|\Delta z|$ , and the $|\Delta z|$ add up to the length $L$ . It never gives the value of an integral — only a ceiling on its size — but that is often all a proof needs.

Worked example C · an ML bound

Bound $\displaystyle\left|\oint_{|z|=4}\frac{dz}{z^2 + 1}\right|$ .

The load-bearing move: to bound the integrand $\dfrac{1}{z^2+1}$ from above, we must bound its denominator $|z^2 + 1|$ from below. On $|z| = 4$ , the reverse triangle inequality gives

|z^2 + 1| \ge |z|^2 - 1 = 16 - 1 = 15 \qquad(\text{equality at } z = \pm 4i),

so $\left|\dfrac{1}{z^2+1}\right| \le \dfrac{1}{15} = M$ . The length is $L = 2\pi\cdot 4 = 8\pi$ , hence

\left|\oint_{|z|=4}\frac{dz}{z^2+1}\right| \le \frac{8\pi}{15} \approx 1.68.

Bounding $|z^2+1|$ the wrong way — from above — is the classic slip; Quiz 3 tests exactly it.

Antiderivatives and path independence

Fundamental theorem for contour integrals

If $F$ is analytic with $F'(z) = f(z)$ throughout a domain containing the contour $C$ from $z_0$ to $z_1$ , then

\int_C f(z)\,dz = F(z_1) - F(z_0),

which depends only on the endpoints — the path drops out. For a closed contour ( $z_1 = z_0$ ) the integral is therefore $0$ .

Worked example D · two methods, one answer

Evaluate $\displaystyle\int_0^{\,1+i} z^2\,dz$ . Because $z^2$ is entire, it has the antiderivative $F(z) = \dfrac{z^3}{3}$ , so by the theorem above the path is irrelevant and

\int_0^{\,1+i} z^2\,dz = \frac{(1+i)^3}{3} = \frac{-2 + 2i}{3}.

Check by parametrising the diagonal $z = (1+i)t$ : $\int_0^1 (1+i)^2 t^2 (1+i)\,dt = (1+i)^3\int_0^1 t^2\,dt = \dfrac{(1+i)^3}{3}$ — the same number. This is the left-hand panel of the hook: $z^2$ is analytic, so diagonal and elbow both land on $-\tfrac23 + \tfrac23 i$ .

Formal statement · 3 of 5

The fundamental integral

One computation underlies the entire rest of Unit 1. Take a circle of radius $r$ about a point $z_0$ , and integrate the power $(z - z_0)^n$ for an integer $n$ . Parametrise $z = z_0 + r\,e^{it}$ , $0 \le t \le 2\pi$ , so $z - z_0 = r\,e^{it}$ and $dz = i r\,e^{it}\,dt$ :

\oint_{|z - z_0| = r} (z - z_0)^n\,dz = \int_0^{2\pi} r^n e^{int}\cdot i r\,e^{it}\,dt = i\,r^{\,n+1}\int_0^{2\pi} e^{i(n+1)t}\,dt.

If $n \ne -1$ , the integrand $e^{i(n+1)t}$ completes a whole number of oscillations over $[0, 2\pi]$ and integrates to $0$ . If $n = -1$ , the integrand is $e^{0} = 1$ and the integral is $2\pi$ . So:

The fundamental integral

\oint_{|z - z_0| = r} (z - z_0)^n\,dz = \begin{cases} 2\pi i, & n = -1, \\[4pt] 0, & n \ne -1, \end{cases} \qquad n \in \mathbb{Z}.

The answer is independent of the radius $r$ and of the centre $z_0$ . The single non-zero case, $n = -1$ (that is, $\dfrac{1}{z - z_0}$ ), is the $2\pi i$ that the rest of the unit is built on.

We will call this result the fundamental integral throughout the course — the later chapters refer back to it by that name. It is the single most-used computation in everything that follows; commit the boxed result to memory. (One honest caveat for the exam is in Beat 6: that name is ours, not an examiner's.)

Formal statement · 4 of 5

Log z, and where the $2\pi i$ lives

Back in the elementary-functions chapter we promised $\operatorname{Log} z$ a deliberate treatment here, because contour integration is where it finally earns its keep. Recall the principal logarithm

\operatorname{Log} z = \ln|z| + i\,\operatorname{Arg} z, \qquad \operatorname{Arg} z \in (-\pi, \pi],

with its branch cut along the negative real axis $(-\infty, 0]$ . Off that cut $\operatorname{Log} z$ is analytic, and there $\dfrac{d}{dz}\operatorname{Log} z = \dfrac{1}{z}$ . So $1/z$ has an antiderivative — which sets up the chapter's best little puzzle.

The puzzle

If $1/z$ has the antiderivative $\operatorname{Log} z$ , then the fundamental theorem of Beat 3B should force $\displaystyle\oint_{|z|=1}\frac{dz}{z} = 0$ around the closed unit circle. But the fundamental integral just told us it is $2\pi i$ . Both cannot be right — so which rule is being broken?

The resolution is the whole point of the branch cut. The principal $\operatorname{Log}$ is not analytic on any domain containing the whole unit circle — the circle meets the branch cut at $z = -1$ . As you travel once around the origin, the argument cannot stay single-valued and continuous; it must jump. Walk the circle once counterclockwise and $\operatorname{Arg} z$ climbs from just above $-\pi$ up to $\pi$ : the "antiderivative" fails to return to its starting value, jumping by exactly $i\pi - (-i\pi) = 2\pi i$ — which is precisely the value of the integral. The branch cut is not pedantry; it is where the $2\pi i$ lives.

The unit circle crosses the cut at

z = -1

; there the argument snaps between

+\pi

and

-\pi

. That snap, accumulated, is the

2\pi i

.

Formal statement · 5 of 5

Cauchy–Goursat, and deforming contours

Cauchy–Goursat theorem

If $f$ is analytic at every point on and inside a simple closed contour $C$ , then

\oint_C f(z)\,dz = 0.

This is the destination promised in Beat 1. The proof is the standard exam answer, and it leans on one tool from Unit 2 that you have not formally met yet.

Stated tool · Green's theorem

For nice functions $P, Q$ on the region $R$ bounded by $C$ ,

\oint_C (P\,dx + Q\,dy) = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.

(In words: a circulation around the boundary equals a double integral over the inside. You will prove and practise this in the Green's-theorem chapter in Unit 2; here we simply use it.)

Proof of Cauchy–Goursat (Cauchy's version, assuming $f'$ continuous). Write $f = u + iv$ and $dz = dx + i\,dy$ , and split into two real line integrals:

\oint_C f\,dz = \oint_C (u\,dx - v\,dy) + i\oint_C (v\,dx + u\,dy).

Apply Green's theorem to each. The first becomes $\iint_R (-v_x - u_y)\,dA$ ; the second becomes $\iint_R (u_x - v_y)\,dA$ . Now invoke the Cauchy–Riemann equations $u_x = v_y$ and $u_y = -v_x$ (the analyticity you carried from the Cauchy–Riemann chapter): the first integrand is $-v_x - u_y = -v_x + v_x = 0$ , and the second is $u_x - v_y = 0$ . Both double integrals vanish, so $\oint_C f\,dz = 0$ . $\blacksquare$

Goursat's contribution was to prove the same conclusion without assuming $f'$ is continuous — analyticity alone is enough. That refinement is why the theorem carries both names; the full Goursat argument is more technical than anything this course needs, so we state it without proof. On an exam, "state Cauchy–Goursat" earns the name, so mention it.

Deformation of contours

Principle of deformation

If $f$ is analytic at every point on $C_1$ and $C_2$ and at every point between them (one contour lying inside the other), then

\oint_{C_1} f\,dz = \oint_{C_2} f\,dz.

You may deform a contour freely through any region where $f$ stays analytic without changing the integral.

The consequence makes the fundamental integral universal: a simple closed contour around a single singularity can be shrunk to a tiny circle around it without crossing anything where $f$ misbehaves — so $\oint_C \frac{dz}{z} = 2\pi i$ for any loop enclosing the origin once, not just a circle. The value depends only on which singularities the loop encloses, never on its shape.

Worked example E · Cauchy–Goursat applied

Evaluate $\displaystyle\oint_{|z|=1}\frac{dz}{z^2 + 4}$ .

The integrand fails to be analytic only where $z^2 + 4 = 0$ , i.e. $z = \pm 2i$ . Both lie at distance $2$ from the origin, so both are outside $|z| = 1$ . Hence $f$ is analytic on and inside $C$ , and by Cauchy–Goursat

\oint_{|z|=1}\frac{dz}{z^2 + 4} = 0.

The two-step ritual that settles almost every contour-integral exam question: (1) locate the singularities; (2) check which are inside the contour. Here, none — so the answer is $0$ with no integration at all.

End of sitting 1 — the definitions and the theorem are now in hand. Next sitting you drive: a playground that turns Cauchy–Goursat into an on/off switch, three quizzes, two scaffolded reveals, and six exam problems.

Interactive playground

The Cauchy–Goursat meter: watch $2\pi i$ switch on and off

Drag the singularity $z_0$ across the circle's edge and watch the integral $\displaystyle\oint_C \frac{dz}{z - z_0}$ snap between $2\pi i$ (singularity inside) and $0$ (singularity outside). That on/off switch is the Cauchy–Goursat theorem. You can also drag the circle's centre, change its radius, and switch the integrand to the entire function $z^2$ (which gives $0$ no matter what, because it is analytic everywhere).

The Cauchy–Goursat meter

Integrand: Radius: 1.00

Drag the circle's centre (open ring) or the singularity (✕). The integral is computed numerically by the trapezoid rule on 720 points — watch it land on $0$ or on $2\pi i \approx 6.283\,i$ .

What to notice

Three things, each a result you have already met:

Preset (a) — $z_0$ at the centre, $r = 1$ : the integral reads $\approx 6.283\,i = 2\pi i$ . This is the fundamental integral, live.
Drag $z_0$ outward across the circle: as it crosses the edge, the readout switches from $2\pi i$ to $0$ . Right at the edge the meter refuses to answer — the contour passes through the singularity and the integral is undefined.
Switch to $z^2$ : the readout is $0$ wherever you put the circle. An entire integrand is analytic everywhere inside, so Cauchy–Goursat gives $0$ every time — there is nothing to switch.

The numeric value never depends on the radius or on where you put the centre, only on whether $z_0$ is enclosed — deformation of contours, made visible.

Retrieval gauntlet

Test yourself before moving on

Each of these follows a method you have now seen worked. Commit to an answer before you read the feedback — recognising the right option once you see it feels like knowing it, and it is not the same thing.

Quiz 1 · when does the theorem apply

All four contours are the positively oriented unit circle $|z| = 1$ . Which integral is guaranteed to be $0$ by the Cauchy–Goursat theorem?

Quiz 2 · inside or outside

Evaluate $\displaystyle\oint_C \frac{dz}{z - 3}$ , where $C$ is the positively oriented circle $|z| = 1$ .

Quiz 3 · the ML bound

Using the ML inequality, what is the best bound it gives for $\displaystyle\left|\oint_C \frac{dz}{z^2 + 1}\right|$ , where $C$ is the positively oriented circle $|z| = 2$ ?

Evaluate $\displaystyle\int_C |z|\,dz$ , where $C$ is the straight segment from $-i$ to $i$ .

Predict the next step before each reveal: how do you parametrise a vertical segment, and what happens to $|z|$ at the origin?

Step 1 — parametrise. The segment runs up the imaginary axis from

-i

to

i

. Write

z(t)

and its

t

-range.

The segment is $z(t) = it$ with $t$ running from $-1$ to $1$ ; then $z'(t) = i$ . (Watch the direction: $t = -1$ is the start $-i$ , $t = +1$ the end $i$ . Running it $1 \to -1$ would flip the sign of the answer.)

Step 2 — assemble the integrand. Find

|z|

and the factor

z'(t)\,dt

. (Two traps live here.)

$|z| = |it| = |t|$ — note $|t|$ , not $t$ ; and $dz = i\,dt$ . Do not drop the $z'(t) = i$ factor. So

\int_C |z|\,dz = \int_{-1}^{1} |t|\,(i)\,dt = i\int_{-1}^{1} |t|\,dt.

Step 3 — handle the modulus. Integrate

\int_{-1}^{1} |t|\,dt

. (You cannot just integrate

t

.)

Split at $t = 0$ , where $|t|$ has its corner:

\int_{-1}^{1} |t|\,dt = \int_{-1}^{0} (-t)\,dt + \int_{0}^{1} t\,dt = \tfrac12 + \tfrac12 = 1.

(Integrating $t$ straight through would give $0$ — the classic slip, and obviously wrong, since $|t| \ge 0$ has positive area.)

Step 4 — finish.

\int_C |z|\,dz = i\cdot 1 = i.

Why no shortcut. $|z|$ is real-valued and analytic nowhere, so it has no antiderivative and no theorem applies — for integrands like $|z|$ , $\bar z$ , or $\operatorname{Re} z$ , direct parametrisation is the only tool.

Evaluate $\displaystyle\oint_C \frac{2z + 1}{z^2 + z}\,dz$ , where $C$ is the positively oriented circle $|z| = 2$ .

Predict the next step before each reveal: where does the integrand misbehave, and which of those points does $C$ actually enclose?

Step 1 — locate the singularities, and check inside/outside.

$z^2 + z = z(z + 1)$ , so the singularities are $z = 0$ and $z = -1$ . Both satisfy $|z| < 2$ , so both are enclosed by $C$ . (Skipping this check is how marks are lost — you must know which singularities are inside before you do anything else.)

Step 2 — partial fractions. Split

\dfrac{2z + 1}{z(z + 1)}

.

\frac{2z + 1}{z(z + 1)} = \frac{1}{z} + \frac{1}{z + 1}.

Check by recombining: $\dfrac{(z+1) + z}{z(z+1)} = \dfrac{2z + 1}{z(z+1)}$ . ✓

Step 3 — integrate each term by deformation. What does each piece contribute?

By linearity, $\oint_C = \oint_C \dfrac{dz}{z} + \oint_C \dfrac{dz}{z + 1}$ . For each term, deform $C$ to a small circle around its singularity (legal: the integrand of that term is analytic everywhere else inside $C$ ); the fundamental integral then gives $2\pi i$ for each. (Do not say Cauchy–Goursat forces $0$ — the singularity is enclosed, so the theorem does not apply.)

Step 4 — add.

\oint_C \frac{2z + 1}{z^2 + z}\,dz = 2\pi i + 2\pi i = 4\pi i.

What you just did — decompose, locate each pole, add the $2\pi i$ 's — is the residue theorem in embryo. A later chapter mechanises exactly this pattern.

Practice set

Six cold problems

Work each before revealing. The two hardest types each open with a faded bridge — a half-finished solution with the key step blank — before the cold problem. Every solution is complete; nothing points outside this page.

Direct parametrisation

Shape · parametrise a non-analytic integrand

P1. Evaluate $\displaystyle\int_C \operatorname{Re}(z)\,dz$ , where $C$ is the straight segment from $0$ to $1 + i$ .

Parametrise $z = (1 + i)t$ , $0 \le t \le 1$ ; then $\operatorname{Re}(z) = t$ and $z' = 1 + i$ :

\int_C \operatorname{Re}(z)\,dz = \int_0^1 t\,(1 + i)\,dt = (1 + i)\cdot\tfrac12 = \frac{1 + i}{2}.

Answer: $\dfrac{1 + i}{2}$ . ( $\operatorname{Re} z$ is not analytic — no antiderivative shortcut; parametrise.)

Inside check, orientation, and shape

Shape · the fundamental integral, three ways

P2. Let $C$ enclose the point $z = 2$ . Evaluate $\displaystyle\oint_C \frac{dz}{z - 2}$ when: (a) $C$ is $|z| = 3$ , positively oriented; (b) the same circle traversed clockwise; (c) $C$ is instead the square with vertices $\pm 3 \pm 3i$ (positively oriented) — explain in one sentence why the answer is unchanged from (a).

(a) $z = 2$ lies inside $|z| = 3$ , and $1/(z-2)$ is the fundamental integrand with $z_0 = 2$ . Enclosed once, counterclockwise: $\oint_C \dfrac{dz}{z - 2} = 2\pi i$ .

(b) Reversing the orientation flips the sign (Beat 3A): $\displaystyle\oint_{\text{cw}} \frac{dz}{z - 2} = -2\pi i$ .

(c) $\dfrac{1}{z - 2}$ is analytic in the region between the square and the circle $|z| = 3$ (and on both contours), so by the deformation principle the two integrals are equal: $2\pi i$ . The shape of the loop never mattered — only that it encloses $z = 2$ once.

Linearity with Cauchy–Goursat

Shape · split the integrand, judge each piece

P3. Evaluate $\displaystyle\oint_{|z|=1} \left(z^2 + \frac{1}{z - 2}\right) dz$ .

By linearity, integrate each term. $z^2$ is entire, so Cauchy–Goursat gives $\oint z^2\,dz = 0$ . The term $\dfrac{1}{z - 2}$ has its only singularity at $z = 2$ , which is outside $|z| = 1$ ; analytic on and inside $C$ , so that integral is $0$ too. Total:

\oint_{|z|=1} \left(z^2 + \frac{1}{z - 2}\right) dz = 0 + 0 = 0.

Antiderivative method

Shape · antiderivative on an open path

P4. Evaluate $\displaystyle\int_0^{\,\pi i} e^z\,dz$ .

$e^z$ is entire with antiderivative $e^z$ , so the path is irrelevant:

\int_0^{\,\pi i} e^z\,dz = e^{\pi i} - e^{0} = -1 - 1 = -2,

using Euler's $e^{\pi i} = -1$ from Chapter 1.

ML inequality

Faded bridge · find

M

first

Bound $\displaystyle\left|\oint_{|z|=3}\frac{dz}{z^2 - 1}\right|$ . Fill the blank, then reveal: the bridge to everything is one question — what is the smallest $|z^2 - 1|$ can be when $|z| = 3$ ? Answer: ___. From it, $M =$ ___ and $L =$ ___.

On $|z| = 3$ : $|z^2 - 1| \ge |z|^2 - 1 = 9 - 1 = 8$ (equality at $z = \pm 3$ ). So the smallest value is $8$ , giving $M = \tfrac18$ , and $L = 2\pi\cdot 3 = 6\pi$ .

Shape · ML with the reverse triangle inequality

P5. Show that $\displaystyle\left|\oint_{|z|=3}\frac{dz}{z^2 - 1}\right| \le \frac{3\pi}{4}$ .

From the bridge, $M = \tfrac18$ and $L = 6\pi$ , so

\left|\oint_{|z|=3}\frac{dz}{z^2 - 1}\right| \le M L = \frac{1}{8}\cdot 6\pi = \frac{6\pi}{8} = \frac{3\pi}{4}. \qquad\blacksquare

Split poles — one in, one out

Faded bridge · check each pole against THIS circle

Set up $\displaystyle\oint_{|z - 1| = 1}\frac{dz}{z^2 - 1}$ . Fill the blanks, then reveal: decompose first — $\dfrac{1}{z^2 - 1} = \tfrac12\!\left(\dfrac{1}{z - 1} - \dfrac{1}{z + 1}\right)$ . The contour is the circle of radius $1$ centred at $z = 1$ . Pole $z = 1$ is ___ (inside/outside); pole $z = -1$ is ___.

$z = 1$ is the centre of the circle — inside. $z = -1$ is at distance $|-1 - 1| = 2 > 1$ from the centre — outside. Check each pole against this circle, not against $|z| = 2$ .

Shape · partial fractions, split poles

P6. Evaluate $\displaystyle\oint_{|z - 1| = 1}\frac{dz}{z^2 - 1}$ .

Decompose $\dfrac{1}{z^2 - 1} = \tfrac12\!\left(\dfrac{1}{z - 1} - \dfrac{1}{z + 1}\right)$ . The headline is the in/out asymmetry: on the circle $|z - 1| = 1$ , the pole $z = 1$ is enclosed (it is the centre), while $z = -1$ is outside (distance $2$ ). So the first term contributes $\tfrac12\cdot 2\pi i$ and the second contributes $\tfrac12\cdot 0$ :

\oint_{|z - 1| = 1}\frac{dz}{z^2 - 1} = \tfrac12(2\pi i) - \tfrac12(0) = \pi i.

This one-in-one-out twist on Reveal 2's pattern is an exam favourite.

Exam translator

How this appears on exams

Exam shapes

In standard Indian engineering math exams, and likely in MCC201A unless your class notes differ, contour-integration questions come in five recognisable shapes — each one you have now seen worked:

"State and prove Cauchy's theorem" — the Beat 3E Green's-theorem proof; name Goursat for the extra mark.
"Evaluate $\oint \frac{dz}{z - a}$ over a given circle" — the two-step ritual (locate, then inside/outside). Both the enclosed ( $2\pi i$ ) and not-enclosed ( $0$ ) variants appear. One caution: "the fundamental integral" is this course's internal name — no examiner will recognise it. On paper, write "by direct evaluation of $\oint \frac{dz}{z - z_0}$ over a circle" or show the two-line parametrisation.
"Evaluate by parametrisation" (integrands like $\bar z$ , $\operatorname{Re} z$ , $|z|$ ) — Reveal 1's recipe; no shortcuts exist.
"Use the ML inequality to bound…" — Quiz 3 and P5: find $M$ by bounding the denominator below, multiply by the length $L$ .
Partial-fraction contour integrals — Reveal 2 and P6: decompose, locate each pole, add a $2\pi i$ for each enclosed one.

What costs marks

Skipping the inside/outside check — then applying Cauchy–Goursat to a loop that encloses a singularity, or the fundamental integral to one that does not (Quiz 2, P6).
Dropping the $z'(t)$ factor when parametrising (Reveal 1).
Bounding the denominator the wrong way in ML — you must bound $|z^2+1|$ below to bound the fraction above (Quiz 3b).
Writing $2\pi$ for $2\pi i$ — the factor of $i$ is not optional (Quiz 2d).
Confusing "the theorem does not apply" with "the integral is nonzero" (Quiz 1d) — and its converse, invoking Cauchy–Goursat when the integrand is not analytic inside.
Forgetting the orientation sign on a clockwise contour (P2b), and treating $\operatorname{Log} z$ as a global antiderivative across its branch cut (Beat 3D).

Seven things to carry out of this chapter

The recipe: $\displaystyle\int_C f\,dz = \int_a^b f(z(t))\,z'(t)\,dt$ . Every evaluation starts by writing $z(t)$ ; reversing orientation negates the integral.
ML inequality: $\left|\int_C f\,dz\right| \le ML$ , with $M$ any number satisfying $|f| \le M$ on $C$ and $L = \operatorname{length}(C)$ — a ceiling, never the value.
Antiderivative ⇒ path independence: if $F' = f$ with $F$ analytic, $\int_C f\,dz = F(z_1) - F(z_0)$ (endpoints only); $\oint_C f\,dz = 0$ round a closed loop.
The fundamental integral: $\oint_{|z - z_0| = r} (z - z_0)^n\,dz = 2\pi i$ if $n = -1$ , else $0$ — independent of $r$ and $z_0$ . The seed of residues.
Cauchy–Goursat: $f$ analytic on and inside a simple closed contour $C$ $\Rightarrow \oint_C f\,dz = 0$ . It is Cauchy–Riemann integrated (Green's-theorem proof); Goursat removed the " $f'$ continuous" assumption.
Deformation of contours: deform $C$ freely through any region where $f$ stays analytic; the value depends only on which singularities are enclosed.
$\operatorname{Log} z$ and its cut: $1/z$ has the antiderivative $\operatorname{Log} z$ , but only off the branch cut $(-\infty, 0]$ ; a loop around the origin crosses it, and the argument's $2\pi$ jump is exactly the $2\pi i$ .

Before you move on

Come back to two of the six problems tomorrow. Re-doing P1–P6 from a blank page after a night's sleep — not re-reading them, re-doing them — is worth more than another hour now, and it is the kind of spaced practice that pays off at revision time.

What comes next. The next chapter, on Cauchy's integral formula, upgrades today's on/off switch ( $0$ or $2\pi i$ ) into a machine that reads off the value $f(z_0)$ itself from a loop around $z_0$ — and from it spring four famous consequences (Morera, Liouville, the maximum-modulus principle, and the Fundamental Theorem of Algebra). Everything there stands on the contour integral and the deformation principle you built here.

Same material, another voice

If a different explanation would help, this one is worth your time — free, from MIT OpenCourseWare:

Read: Orloff, MIT 18.04, Topic 3 — Line integrals and Cauchy's theorem — the same path-integral / Cauchy's-theorem material from another angle, with the ML estimate and path independence laid out cleanly.

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.

Two paths, one answer — sometimes

March along the curve, and let the function twist each step

Contours, and the substitution that computes them

Bounding an integral, and the antiderivative shortcut

The ML inequality

Antiderivatives and path independence

The fundamental integral

Log z, and where the 2πi2\pi i2πi lives

Cauchy–Goursat, and deforming contours

Deformation of contours

The Cauchy–Goursat meter: watch 2πi2\pi i2πi switch on and off

Test yourself before moving on

Six cold problems

Direct parametrisation

Inside check, orientation, and shape

Linearity with Cauchy–Goursat

Antiderivative method

ML inequality

Split poles — one in, one out

How this appears on exams

Seven things to carry out of this chapter

Same material, another voice

Log z, and where the $2\pi i$ lives

The Cauchy–Goursat meter: watch $2\pi i$ switch on and off