Cauchy's integral theorem and formula: Exercises

Definition Standing conventions for this whole page (read first!)

Three ground rules are assumed in every exercise below — they fix the sign and validity of every answer:

Positive orientation. Every loop is traversed counterclockwise (the "positive" sense). Concretely, when we write a circle as $z = e^{i θ}$ we let the angle $θ$ run in the increasing direction from $0$ to $2 π$ . Reversing to clockwise would flip the sign of every integral.
No poles on the contour. Cauchy's Theorem and Formula require the loop to avoid all singularities: a bad point may sit inside or outside, but never on the path itself (there the integrand is undefined and the integral is not even well-posed). We check this silently in each problem — every pole lands strictly inside or strictly outside.
Simply connected region. The region enclosed by the loop must have no holes of its own (it is simply connected — any loop in it shrinks to a point). A disk bounded by a circle always qualifies, which is why we use circles throughout.

Before we begin, one picture fixes the entire mental model for the whole page.

Figure — Cauchy's integral theorem and formula

Here $2 π i$ is just the number $2 π \approx 6.283$ multiplied by $i = - 1$ ; it always appears because a full trip counterclockwise around a circle sweeps an angle of $2 π$ radians.

Level 1 — Recognition

Goal: spot which of the three outcomes applies. No heavy computation.

Recall Solution 1.1

WHAT is the integrand? A polynomial. WHY does that settle it? A polynomial is holomorphic (complex-differentiable) at every point of the plane — it never divides by anything, so it has no bad points anywhere, in particular none inside $∣ z ∣ = 1$ , and none on the circle either. The enclosed disk is simply connected. That is exactly Outcome 1. $\oint_{∣ z ∣ = 1} (3 z^{4} - 5 z + 7) d z = 0.$

Recall Solution 1.2

WHAT is the bad point? The denominator $z$ is zero at $z = 0$ , and $0$ is the centre of the circle — definitely inside (not on the path). So this is not Outcome 1. WHY Outcome 2? Match the pattern $\frac{f ( z )}{z - a}$ with $a = 0$ and $f (z) = 1$ (the constant function, holomorphic). Cauchy's Integral Formula gives $\oint_{∣ z ∣ = 1} \frac{1}{z} d z = 2 π i f (0) = 2 π i \cdot 1 = 2 π i .$ The point of this classic: a single hole makes the answer nonzero.

Recall Solution 1.3

WHAT is the bad point? $z = 3$ . WHERE is it? Its distance from the origin is $3$ , but the loop only reaches distance $1$ . So the bad point is outside the loop (and not on it). WHY that means zero: inside and on $∣ z ∣ = 1$ the integrand $\frac{1}{z - 3}$ is perfectly holomorphic (nothing blows up there), and the enclosed disk is simply connected. Outcome 1. $\oint_{∣ z ∣ = 1} \frac{1}{z - 3} d z = 0.$

Level 2 — Application

Goal: plug into the right formula and evaluate $f (a)$ .

Recall Solution 2.1

Bad point: $z = 1$ , distance $1$ from origin, which is $< 2$ → inside $∣ z ∣ = 2$ (not on it). Outcome 2. Match the pattern: $\frac{f ( z )}{z - a}$ with $a = 1$ , $f (z) = e^{z}$ (holomorphic everywhere). $\oint_{∣ z ∣ = 2} \frac{e ^{z}}{z - 1} d z = 2 π i f (1) = 2 π i e^{1} = 2 π i e .$ Numerically $2 π e \approx 17.08$ , so the value is about $17.08 i$ .

Recall Solution 2.2

Bad point: $z = 0$ , inside. Pattern: $a = 0$ , $f (z) = sin z$ . $\oint_{∣ z ∣ = 1} \frac{s i n z}{z} d z = 2 π i sin (0) = 2 π i \cdot 0 = 0.$ WHY zero even with a hole? The formula returns $2 π i f (a)$ ; here $f (0) = sin 0 = 0$ . The hole is real, but the numerator vanishes there, so the answer collapses to $0$ . (Geometrically: $sin z / z$ actually has a removable singularity — it can be redefined as $1$ at $0$ and becomes fully holomorphic.)

Recall Solution 2.3

Bad point: $z = 2 i$ . Its distance from origin is $∣2 i ∣ = 2 < 3$ → inside (not on the path). Outcome 2. Pattern: $a = 2 i$ , $f (z) = z^{2} + 1$ . $f (2 i) = (2 i)^{2} + 1 = 4 i^{2} + 1 = - 4 + 1 = - 3.$ $\oint_{∣ z ∣ = 3} \frac{z ^{2} + 1}{z - 2 i} d z = 2 π i (- 3) = - 6 π i .$

Level 3 — Analysis

Goal: higher-order poles (repeated bad points) and choosing $n$ correctly.

Intuition Why we may differentiate

through the integral (the honest reason) The claim "differentiate the basic formula $f (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{z - a} d z$ with respect to $a$ " hides a real question: why is it legal to move the derivative $\frac{d}{d a}$ inside the $\oint$ sign? Here is the justification, from the definition of derivative.

The derivative in $a$ is a limit of a difference quotient: $\frac{d}{d a} (\frac{1}{z - a}) = lim_{h \to 0} \frac{1}{h} (\frac{1}{z - ( a + h )} - \frac{1}{z - a}) = \frac{1}{( z - a ) ^{2}}$ (just algebra: combine the two fractions, the $h$ 's cancel, let $h \to 0$ ).
Because the contour $γ$ is a fixed, bounded curve and, for $a$ strictly inside, $z - a$ stays bounded away from zero on $γ$ (the pole never touches the path — rule 2), the fraction $\frac{f ( z )}{( z - a ) ^{2}}$ is uniformly bounded and continuous along $γ$ .
A finite integral over a fixed curve of a uniformly convergent family of functions may have the limit taken outside the integral — this is the standard "interchange of limit and finite integral" theorem. So the limit $h \to 0$ (i.e. the $a$ -derivative) slides right through the $\oint$ .

Applying it once gives $f^{'} (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{( z - a ) ^{2}} d z$ ; repeating it $n$ times gives $\frac{d ^{n}}{d a ^{n}} \frac{1}{z - a} = \frac{n !}{( z - a ) ^{n + 1}}$ , which is exactly where the $n!$ and the exponent $n + 1$ come from.

Recall Solution 3.1

Read the power: denominator is $z^{3} = (z - 0)^{3}$ . Match $(z - a)^{n + 1}$ : so $n + 1 = 3 \Rightarrow n = 2$ , $a = 0$ , $f (z) = cos z$ . Differentiate twice: $f^{'} (z) = - sin z$ , $f^{''} (z) = - cos z$ , so $f^{''} (0) = - cos 0 = - 1$ . $\oint = \frac{2 π i}{2 !} f^{''} (0) = \frac{2 π i}{2} (- 1) = - π i .$

Recall Solution 3.2

Read the power: $z^{2} = (z - 0)^{2} \Rightarrow n + 1 = 2 \Rightarrow n = 1$ , $a = 0$ , $f (z) = e^{2 z}$ . Differentiate once: $f^{'} (z) = 2 e^{2 z}$ , so $f^{'} (0) = 2 e^{0} = 2$ . $\oint = \frac{2 π i}{1 !} f^{'} (0) = 2 π i \cdot 2 = 4 π i .$

Recall Solution 3.3

Bad point $z = 1$ , distance $1 < 2$ → inside. Power: $(z - 1)^{2} \Rightarrow n + 1 = 2 \Rightarrow n = 1$ , $a = 1$ , $f (z) = z$ . Differentiate once: $f^{'} (z) = 1$ , so $f^{'} (1) = 1$ . $\oint = \frac{2 π i}{1 !} f^{'} (1) = 2 π i \cdot 1 = 2 π i .$

Level 4 — Synthesis

Goal: two or more poles inside one loop — split the work.

Common mistake Watch out — repeated linear factors need MORE than one term

The simple "one term per factor" split only works when the denominator's factors are distinct (like $(z - 1) (z - 2)$ ). If a factor is repeated — say $(z - 1)^{2}$ in the denominator — you must allot one term per power: $\frac{g ( z )}{( z - 1 ) ^{2} ( z - 2 )} = \frac{A}{z - 1} + \frac{B}{( z - 1 ) ^{2}} + \frac{C}{z - 2} .$ Why: the root-plugging trick still finds $B$ (multiply by $(z - 1)^{2}$ , set $z = 1$ ) and $C$ , but $A$ needs an extra step (differentiate, or match another coefficient). And crucially: the $\frac{B}{( z - 1 ) ^{2}}$ piece is an order-2 pole, so you must integrate it with the generalized formula ( $n = 1$ ), not the simple one. Missing the $\frac{A}{z - 1}$ term, or treating $(z - 1)^{2}$ as if it were simple, is the classic error. In this page's exercises all factors are distinct, so one term each suffices — but recognise a repeated factor the moment it appears.

Recall Solution 4.1

Where are the poles? $z = 1$ and $z = 2$ ; distances $1$ and $2$ , both $< 3$ → both inside (neither on the path). Split (partial fractions): we want $\frac{1}{( z - 1 ) ( z - 2 )} = \frac{A}{z - 1} + \frac{B}{z - 2}$ . Clearing denominators, $1 = A (z - 2) + B (z - 1)$ . Set $z = 1$ (kills the $B$ term): $1 = A (- 1) \Rightarrow A = - 1$ . Set $z = 2$ (kills the $A$ term): $1 = B (1) \Rightarrow B = 1$ . So the integrand is $\frac{- 1}{z - 1} + \frac{1}{z - 2}$ . Apply the formula to each (each is $\frac{const}{z - a}$ , i.e. $f =$ that constant): $\oint \frac{- 1}{z - 1} = 2 π i (- 1) = - 2 π i, \oint \frac{1}{z - 2} = 2 π i (1) = 2 π i .$ Add: $- 2 π i + 2 π i = 0.$ Sanity comment: the two residues cancelled — a genuine feature, not a mistake.

Recall Solution 4.2

Recheck which poles are inside: the loop has radius $\frac{3}{2} = 1.5$ . Pole $z = 1$ : distance $1 < 1.5$ → inside. Pole $z = 2$ : distance $2 > 1.5$ → outside. (Neither lies on the path.) Only the inside pole contributes. Using the same split $\frac{- 1}{z - 1} + \frac{1}{z - 2}$ : the $\frac{1}{z - 2}$ term is holomorphic inside this smaller loop, so its integral is $0$ (Cauchy's Theorem). Only the $\frac{- 1}{z - 1}$ term survives: $\oint_{∣ z ∣ = 3/2} = 2 π i (- 1) + 0 = - 2 π i .$ Lesson: changing the radius changes the answer, because it changes which poles are enclosed.

Recall Solution 4.3

Poles: $z = 0$ and $z = 1$ , distances $0$ and $1$ , both $< 2$ → both inside. Split the $\frac{1}{z ( z - 1 )}$ part (same root-plugging trick): $\frac{1}{z ( z - 1 )} = \frac{A}{z} + \frac{B}{z - 1}$ gives $1 = A (z - 1) + B z$ . Set $z = 0$ (kills $B$ ): $A = - 1$ ; set $z = 1$ (kills $A$ ): $B = 1$ . So $\frac{e ^{z}}{z ( z - 1 )} = e^{z} (\frac{- 1}{z} + \frac{1}{z - 1}) = \frac{- e ^{z}}{z} + \frac{e ^{z}}{z - 1} .$ Apply formula (now $f (z) = e^{z}$ on each): $\oint \frac{- e ^{z}}{z} = 2 π i (- e^{0}) = - 2 π i, \oint \frac{e ^{z}}{z - 1} = 2 π i e^{1} = 2 π i e .$ Add: $- 2 π i + 2 π i e = 2 π i (e - 1) .$

Level 5 — Mastery

Goal: combine Cauchy machinery with a downstream consequence (bounds, uniqueness, real integrals).

Recall Solution 5.1

First, the modulus of the prefactor. We need $\frac{1}{2 π i}$ . Recall $∣ i ∣ = 1$ (the number $i$ sits on the unit circle, distance $1$ from origin), and the modulus of a quotient is the quotient of moduli, so $\frac{1}{2 π i} = \frac{1}{2 π ∣ i ∣} = \frac{1}{2 π}$ . The $i$ contributes nothing to the size — it only rotates. Tool — the ML inequality: for any contour, $\oint g d z \leq (length) \cdot (max ∣ g ∣)$ . WHY this tool? We cannot compute the integral exactly (unknown $f$ ), but we can bound it. On $∣ z ∣ = R$ with $a = 0$ : length $= 2 π R$ , and $\frac{f ( z )}{z ^{2}} = \frac{∣ f ( z ) ∣}{∣ z ∣ ^{2}} \leq \frac{M}{R ^{2}} .$ So $∣ f^{'} (0) ∣ = \frac{1}{2 π} \oint \frac{f ( z )}{z ^{2}} d z \leq \frac{1}{2 π} \cdot 2 π R \cdot \frac{M}{R ^{2}} = \frac{M}{R} .$ Payoff: this is exactly the estimate behind Liouville's theorem — a bounded entire function ( $M$ fixed while $R \to \infty$ ) has $f^{'} (0) = 0$ , hence is constant.

Recall Solution 5.2

Set up the parametrisation: on $∣ z ∣ = 1$ traced counterclockwise, $z = e^{i θ}$ with $θ$ increasing from $0$ to $2 π$ ; then $d z = i e^{i θ} d θ$ and $z - a = e^{i θ} - 0 = e^{i θ}$ . With $f \equiv 1$ , $\oint_{∣ z ∣ = 1} \frac{1}{z} d z = \int_{0}^{2 π} \frac{1}{e ^{i θ}} i e^{i θ} d θ = i \int_{0}^{2 π} d θ .$ But we also know this integral equals $2 π i f (0) = 2 π i$ (Exercise 1.2). Equate: $i \int_{0}^{2 π} d θ = 2 π i \Rightarrow \int_{0}^{2 π} d θ = 2 π .$ Meaning: the mysterious $2 π i$ in Cauchy's formula is literally the trip's angle $2 π$ , dressed with an $i$ from $d z$ . (Had we run $θ$ the other way, $2 π \to 0$ , we would have gotten $- 2 π i$ — this is why direction is fixed.)

Recall Solution 5.3

Start from the formula on the circle $∣ z - a ∣ = r$ : parametrise $z = a + r e^{i θ}$ with $θ$ increasing $0 \to 2 π$ (counterclockwise), so $z - a = r e^{i θ}$ and $d z = i r e^{i θ} d θ$ . $f (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{z - a} d z = \frac{1}{2 π i} \int_{0}^{2 π} \frac{f ( a + r e ^{i θ} )}{r e ^{i θ}} i r e^{i θ} d θ .$ Cancel $r e^{i θ}$ and the $i$ against $\frac{1}{2 π i}$ : $f (a) = \frac{1}{2 π} \int_{0}^{2 π} f (a + r e^{i θ}) d θ .$ Meaning: the value at the centre equals the average of the values around any circle. This is the seed of the Maximum modulus principle — an interior point can never exceed the boundary, because it is an average of boundary values.

Recall One-line summary of the whole page

Locate every bad point ::: check which are inside the (counterclockwise) loop and that none lies on it; no poles → $0$ ; simple pole → $2 π i f (a)$ ; order- $(n + 1)$ pole → $\frac{2 π i}{n !} f^{(n)} (a)$ ; several poles → split (one term per power for repeated factors) and add.