Worked examples — Cauchy's integral theorem and formula

Two tools are the whole toolbox. Let $\gamma$ be a closed loop, $f$ a function, $a$ a point.

The reader who forgets what "holomorphic" or "$\oint$" mean should reread §1 of the parent first.

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The scenario matrix

Every contour integral you meet falls into exactly one of these cells. Each row is one thing that can go wrong or be special. The examples below are labelled by cell.

Cell	Description	Which weapon	Example
A	No pole inside $\gamma$	Theorem $\to 0$	Ex 1
B	One simple pole inside ($n=0$)	Formula, $f(a)$	Ex 2
C	Higher-order pole ($n\ge 1$)	Formula, $f^{(n)}(a)$	Ex 3
D	Pole outside $\gamma$ (degenerate: nothing enclosed)	Theorem $\to 0$	Ex 4
E	Two poles inside → split by partial fractions	Formula twice	Ex 5
F	Pole on the contour (illegal input)	Neither — principal value	Ex 6
G	Limiting case: radius $\to$ swallowing/excluding the pole	Boundary of A vs B	Ex 7
H	Real-world word problem (physics/AC circuit flavour)	Formula in disguise	Ex 8
I	Exam twist: same integral, two contours, different answers	A vs B side by side	Ex 9
J	Contour reversed (clockwise)	Sign flip	Ex 10
K	Non-pole singularities: branch points and essential singularities	Beyond these two weapons	Ex 11

Let me set one convention used everywhere: $|z|=R$ means the circle of radius $R$ centred at the origin, traversed counter-clockwise (the positive direction). Look at the figure — counter-clockwise is the arrow direction that keeps the enclosed disk on your left.

(Figure s01 — a blue unit circle with yellow arrows all pointing counter-clockwise around it; the shaded disk sits on the left of each arrow. If the image does not load: picture a clock face with the hands running backwards, and the whole clock face shaded — that shaded face is the region the loop encloses.)

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Example 1 — Cell A: nothing inside, so zero

Steps.

1. Identify singularities of $f(z)=z^3-4z+7$. Why this step? The theorem only fires if $f$ is holomorphic on the entire enclosed disk. A polynomial is differentiable at every point of $\mathbb{C}$ — no denominators, no roots to divide by — so there are no singularities anywhere.
2. All of them (there are none) lie... nowhere, hence none inside $|z|=3$. Cell A confirmed. Why this step? We must be certain the "no holes" condition holds; here it holds trivially.
3. Apply the Theorem: ${\oint }_{|z|=3}fdz=0.$

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Example 2 — Cell B: one simple pole, use $f(a)$

Steps.

1. Match the shape $\frac{f(z)}{z-a}$: here $f(z)=\sin z$ and $a=\frac{\pi }{2}$ (so $n=0$). Why this step? The formula needs the denominator written exactly as $z-a$ so we can read off $a$.
2. Is $a=\pi /2\approx 1.571$ inside $|z|=2$? Yes, since $|\pi /2|<2$. And $f(z)=\sin z$ is holomorphic everywhere. Cell B confirmed. Why this step? If $a$ were outside we'd be in Cell D and the answer would be $0$ instead.
3. Apply $\oint =2\pi if(a)=2\pi i\sin (\frac{\pi }{2})=2\pi i\cdot 1=2\pi i.$

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Example 3 — Cell C: a higher-order pole needs a derivative

Steps.

1. Write $z^4=(z-0{)}^{n+1}$. So $n+1=4\Rightarrow n=3$, with $a=0$ and $f(z)=e^{2z}$. Why this step? The generalized formula reads the order of the pole off the exponent; matching $n+1$ to the power tells us how many derivatives to take.
2. Confirm $a=0$ is inside $|z|=1$ (yes) and $e^{2z}$ is holomorphic (yes). Cell C confirmed.
3. Compute $f^{(3)}(z)$. Each derivative of $e^{2z}$ multiplies by $2$: $f^{\prime }(z)=2e^{2z}$, $f^{\prime \prime }=4e^{2z}$, $f^{\prime \prime \prime }=8e^{2z}$. So $f^{(3)}(0)=8$. Why this step? The formula needs $f^{(n)}(a)=f^{(3)}(0)$, nothing else.
4. Apply $\oint =\frac{2\pi i}{n!}{f}^{(n)}(a)=\frac{2\pi i}{3!}\cdot 8=\frac{2\pi i}{6}\cdot 8=\frac{8\pi i}{3}.$

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Example 4 — Cell D: pole outside → the theorem still rules, answer is zero

Steps.

1. Singularity of the integrand: denominator zero at $z=5$. Why this step? We must locate the pole before deciding which weapon fires.
2. Is $z=5$ inside $|z|=1$? No — $|5|=5>1$. So inside the loop, the whole integrand $\frac{e^z}{z-5}$ is holomorphic (nowhere does the denominator vanish there). Cell D confirmed. Why this step? This is the crucial contrast with Ex 2: same shape of integrand, but the pole is excluded, so it is invisible to the loop.
3. Apply the Theorem (not the formula): ${\oint }_{|z|=1}\frac{e^z}{z-5}dz=0.$

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Example 5 — Cell E: two poles inside → split them apart

Steps.

1. Factor and split: $\frac{1}{z^2-1}=\frac{1}{(z-1)(z+1)}=\frac{1/2}{z-1}-\frac{1/2}{z+1}.$ Why this step? The formula handles one enclosed pole at a time. Partial fractions turn a two-pole integrand into two Cell-B integrals we already know how to do. To be sure the split is correct, put the two pieces over a common denominator: $\frac{1/2(z+1)-1/2(z-1)}{(z-1)(z+1)}=\frac{1}{(z-1)(z+1)}$, which is exactly what we started with — so the split is valid.
2. Both $z=1$ and $z=-1$ lie inside $|z|=3$. Cell E confirmed. Why this step? Only enclosed poles contribute; we verify both are grabbed by this large circle before summing them.
3. Integrate each piece with $f=$ constant $\frac{1}{2}$, $n=0$: $\oint \frac{1/2}{z-1}dz=2\pi i\cdot \frac{1}{2}=\pi i,\oint \frac{1/2}{z+1}dz=2\pi i\cdot \frac{1}{2}=\pi i.$ Why this step? Each is exactly Cauchy's formula with $f(z)=\frac{1}{2}$ evaluated at the pole.
4. Combine with the minus sign from the split: $\oint =\pi i-\pi i=0.$ Why this step? The contour integral is linear: the integral of a sum (or difference) of pieces equals the sum (or difference) of their integrals, $\oint (g-h)=\oint g-\oint h$. So the algebraic minus sign between the two partial fractions carries straight through to a minus sign between the two contour integrals. We are not re-deriving anything — just using that integration distributes over $+$ and $-$.

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Example 6 — Cell F: a pole sitting ON the contour (degenerate/illegal)

Before we start, one honest warning. When the trouble-spot sits on the path, the integral does not have a single value; different sensible ways of handling it give different answers. The standard repair is the principal value: cut a tiny symmetric arc of angular half-width $\delta$ out of the contour, centred exactly on the pole, integrate over what remains, then let $\delta \to 0$. "Symmetric" means you remove equal angle on each side of the pole, so the blow-ups on the two sides are given a fair chance to cancel. Keep that picture in mind for step 3.

Steps.

1. Locate the pole: $z=1$. Is it inside, outside, or on $|z|=1$? Exactly on it, since $|1|=1$. Why this step? The formula's hypothesis "$a$ inside $\gamma$" is now violated, and the theorem's hypothesis "$f$ holomorphic on the contour" is also violated.
2. Neither weapon legally applies. As written, the integrand is unbounded on the path and the integral does not converge as an ordinary contour integral. Why this step? We must not blindly plug $z=1$ into either formula — that would give a fake number.
3. The symmetric detour: replace the removed arc by a tiny semicircle of radius $\epsilon$ that bulges around the pole. A full small circle around a simple pole contributes $2\pi i$ (that is Ex 2 in miniature); a half circle — exactly $\pi$ radians of arc — therefore contributes half of that, $\pi i$. Because the principal-value cut is symmetric, the detour is exactly this half-circle, so the principal value is $\pi i$. If instead you push the semicircle so the pole ends up strictly outside, you keep none of the residue and get $0$; push it so the pole ends up strictly inside, you keep all of it and get $2\pi i$. Why this step? This makes the mechanism explicit: the answer is literally the fraction of the little circle you swing around the pole, and $\pi$ out of $2\pi$ radians is exactly one half — that is where $\pi i$ comes from.

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Example 7 — Cell G: the limiting knife-edge (radius crossing the pole)

Steps.

1. Pole at $z=2$. Compare with radius $R$. Why this step? Whether the pole is enclosed is the only thing that changes as $R$ varies.
2. Case $R<2$ (pole outside): Cell D → Theorem → $I(R)=0.$ Why? Integrand holomorphic on the whole disk of radius $R<2$.
3. Case $R>2$ (pole inside): Cell B with $f=1$, $a=2$ → $I(R)=2\pi i\cdot 1=2\pi i.$ Why? Now the pole is enclosed and the formula fires.
4. Case $R=2$: Cell F — the pole lies on the contour, integral undefined (see Ex 6).

(Figure s02 — a step graph: the value of $I(R)$ is a flat green line at $0$ for $R<2$, then jumps up a dashed yellow arrow of height $2\pi i$ to a flat red line for $R>2$; a blue dotted vertical line marks the pole at $R=2$. If the image does not load: picture a light switch that is OFF for every radius below $2$ and suddenly ON at value $2\pi i$ for every radius above $2$.)

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Example 8 — Cell H: word problem (an AC-circuit / averaging flavour)

Steps.

1. Recognise the mean-value property. Start from Cauchy's formula on $|z|=1$ with $a=0$: $f(0)=\frac{1}{2\pi i}{\oint }_{|z|=1}\frac{f(z)}{z}dz.$ Why this step? The word "average around the ring" is Cauchy's formula in disguise; we convert the given angular average into a contour integral.
2. Parametrise $z=e^{i\theta }$, $dz=i{e}^{i\theta }d\theta$, so $\frac{dz}{z}=id\theta$: $f(0)=\frac{1}{2\pi i}{\int }_0^{2\pi }f(e^{i\theta })id\theta =\frac{1}{2\pi }{\int }_0^{2\pi }f(e^{i\theta })d\theta =A.$ Why this step? This shows the designer's $A$ is exactly $f(0)$ — the mean-value property for holomorphic functions.
3. Evaluate the centre: $f(0)=0^2+3\cdot 0+5=5.$ Why this step? The whole ring average collapses to a single number, $f(a)$ at the centre.
4. So $A=5.$

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Example 9 — Cell I: exam twist, one integrand, two contours

Steps.

1. Factor the denominator: $z^2+4=(z-2i)(z+2i)$; poles at $z=2i$ and $z=-2i$. Why this step? Knowing pole locations is the whole game once the contours differ; each circle will enclose a different subset of these two points.
2. Contour (a) $|z-2i|=1$ — the circle of radius $1$ centred at $2i$. Distance from centre $2i$ to the pole $2i$ is $0<1$, so $2i$ is inside; distance to $-2i$ is $|2i-(-2i)|=4>1$, so $-2i$ is outside. This is Cell B: exactly one enclosed simple pole. Why this step? We must confirm which single pole is grabbed before applying the one-pole formula; the other pole must be safely outside or the method changes.
3. Fold the far factor into $f$. Write the integrand as $\frac{f(z)}{z-2i}$ with $f(z)=\frac{z}{z+2i}$, so $a=2i$. Why this step? Cauchy's formula needs the enclosed pole isolated as a bare $(z-a)$; the factor $(z+2i)$ from the outside pole never vanishes inside ${\gamma }_1$, so it is holomorphic there and legally becomes part of $f$.
4. Apply the formula: $J_{{\gamma }_1}=2\pi if(2i)=2\pi i\cdot \frac{2i}{2i+2i}=2\pi i\cdot \frac{2i}{4i}=2\pi i\cdot \frac{1}{2}=\pi i.$ Why this step? With $f$ holomorphic on and inside ${\gamma }_1$ and one simple pole at $a=2i$, the formula gives the value directly.
5. Contour (b) $|z|=1$ — radius-$1$ circle about the origin. Distance to each pole is $|\pm 2i|=2>1$, so neither pole is enclosed. This is Cell A. Why this step? If no singularity is enclosed, the integrand is holomorphic on the whole disk and the theorem — not the formula — applies.
6. Apply the theorem: $J_{{\gamma }_2}=0.$ Why this step? No enclosed pole ⇒ loop integral of a holomorphic function ⇒ zero, no matter how complicated the integrand looks.

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Example 10 — Cell J: the same loop walked backwards (orientation)

Steps.

1. Name the clockwise loop $-\gamma$, where $\gamma$ is the usual counter-clockwise unit circle. Why this step? We only know how to evaluate counter-clockwise loops; we must express the clockwise one in terms of a counter-clockwise one.
2. Apply the reversal rule ${\oint }_{-\gamma }fdz=-{\oint }_{\gamma }fdz$. Why this step? Reversing direction negates every $\Delta z$ step, hence negates the whole sum — this is the sign law stated in the orientation definition, no new machinery needed.
3. Substitute the known value: ${\oint }_{-\gamma }\frac{dz}{z}=-{\oint }_{\gamma }\frac{dz}{z}=-2\pi i.$

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Example 11 — Cell K: when the singularity is NOT a pole

Our two weapons assume the only bad points are poles — spots where $f$ blows up like $1/(z-a{)}^m$ for some finite whole number $m$. Two other creatures live in contour calculus, and it is vital to recognise that they are outside the reach of Cauchy's formula as stated.

Steps.

1. (a) Branch point. $\sqrt{z}$ has a branch point at $0$: going once around the origin sends $\sqrt{z}\to -\sqrt{z}$, so it is not a single-valued holomorphic function on the punctured disk. Cauchy's formula requires a single-valued holomorphic $f$; that hypothesis fails. Why this step? The formula's very first word is "holomorphic" — and holomorphic includes single-valued. A multivalued expression must first be tamed with a branch cut (a curve you forbid the contour to cross), which changes the problem entirely. So we do not get to write "$2\pi if(a)$" here.
2. (b) Essential singularity — use the Laurent series. Expand $e^{1/z}={\sum }_{k=0}^{\infty }\frac{1}{k!}{z}^{-k}=1+\frac{1}{z}+\frac{1}{2!z^2}+\cdots$. Why this step? The loop integral of $z^n$ over $|z|=1$ is $0$ for every integer $n\ne -1$ and equals $2\pi i$ for $n=-1$ (that is the $\oint dz/z$ fact). So the only term that survives is the coefficient of $z^{-1}$.
3. Read off the coefficient of $z^{-1}$ (the $k=1$ term): it is $\frac{1}{1!}=1$. Why this step? This coefficient is called the residue; the whole integral is $2\pi i$ times it.
4. Therefore ${\oint }_{|z|=1}{e}^{1/z}dz=2\pi i\cdot 1=2\pi i.$

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Master checklist (run this on any contour integral)

Recall Quick self-test

${\oint }_{|z|=1}\frac{\cos z}{z^2}dz=?$ ::: $n=1$, $f=\cos z$, $f^{\prime }(0)=-\sin 0=0$, so $\frac{2\pi i}{1!}\cdot 0=0$. ${\oint }_{|z|=1}\frac{dz}{z-3}=?$ ::: pole at $3$ is outside; theorem ⇒ $0$. ${\oint }_{|z|=4}\frac{dz}{z-3}=?$ ::: pole at $3$ now inside; $2\pi i\cdot 1=2\pi i$. ${\oint }_{|z|=1}\frac{dz}{z}$ walked clockwise $=?$ ::: $-2\pi i$ (reverse orientation flips the sign).

Related deeper tools: enclosing many poles at once is the Residue theorem; the mean-value idea in Ex 8 leads to the Maximum modulus principle and Liouville's theorem; expanding an integrand near a pole or essential singularity uses a Laurent series; the derivation itself rests on the Cauchy-Riemann equations and Green's theorem.