Chapter 8 · Cauchy's integral formula and its four children

The hook

The boundary knows the interior

Story

Here is something that should not be allowed. Pick an analytic function — say $f(z) = z^2$ — and a point inside a loop, say $z_0 = i$ inside the circle $|z| = 2$ . I am going to tell you the value $f(i)$ without ever evaluating $f$ at $i$ . I will only walk around the boundary circle, sampling $f$ at points that are nowhere near $i$ , add those boundary values up in the right way — and out drops $f(i)$ exactly.

That is not a trick of this one example. For any function analytic on and inside a loop, the values on the boundary completely determine every value inside. The inside has no freedom of its own; it is the boundary's shadow. This is Cauchy's integral formula, and it has no analogue anywhere in real calculus.

Walk the boundary, sample

f

on it, combine — and the value at the interior point

z_0

falls out. The boundary determines the inside.

The concrete claim this chapter opens with, computed properly in Beat 3:

\oint_{|z|=2} \frac{z^2}{z - i}\,dz = 2\pi i\cdot f(i) = 2\pi i\cdot i^2 = -2\pi i,

where $f(z) = z^2$ , so $f(i) = i^2 = -1$ . A loop integral of boundary values reproduced $f(i)$ — multiplied by the now-familiar $2\pi i$ from Chapter 7.

Why this matters

Cauchy's integral formula is the hinge of the whole unit. From it, in this one chapter, fall four classical theorems — Morera, Liouville, the maximum-modulus principle, and the Fundamental Theorem of Algebra — results that resisted other methods for centuries and here drop out in a few lines each. Its derivative form is a reliable exam earner ("evaluate $\oint f(z)/(z-z_0)^n\,dz$ "), and its proof skeletons are exactly the "state and prove" theory questions the course rewards. The next chapter turns the multi-pole version you will meet here into a machine for the hard real integrals ordinary calculus can't touch.

Intuition before formula

Shrink the loop until the function can't vary

The formula looks like magic, but it is built from exactly two things you already own from Chapter 7: the fundamental integral $\displaystyle\oint_{|z-z_0|=r}\frac{dz}{z - z_0} = 2\pi i$ , and the principle of deformation — a contour may be deformed freely through any region where the integrand stays analytic, without changing the integral.

We want $\displaystyle\oint_C \frac{f(z)}{z - z_0}\,dz$ . The integrand misbehaves only at the single point $z_0$ (where $f$ is analytic). So deform $C$ down to a tiny circle $C_\rho$ of radius $\rho$ centred at $z_0$ — legal, because nothing between $C$ and $C_\rho$ is singular. On that tiny circle, $f$ is continuous, so it barely changes: $f(z) \approx f(z_0)$ everywhere on $C_\rho$ . Pull that near-constant out of the integral:

\oint_{C_\rho} \frac{f(z)}{z - z_0}\,dz \;\approx\; f(z_0)\oint_{C_\rho} \frac{dz}{z - z_0} \;=\; f(z_0)\cdot 2\pi i.

This paragraph is the proof idea — not yet the proof

Shrink the loop to a point; the function freezes to $f(z_0)$ ; the bare $\displaystyle\oint \frac{dz}{z - z_0} = 2\pi i$ does the rest. That is the whole mechanism, and it is worth holding in one breath. But the step " $f(z) \approx f(z_0)$ " is a heuristic: it hides an error term, and a proof has to show that error vanishes as $\rho \to 0$ . We do exactly that in Beat 6 (Reveal 1 — the full "state and prove" answer), where the approximation $\approx$ is upgraded to an equality $=$ with one line of the ML inequality. For now, carry the picture: the boundary value at $z_0$ is what a vanishing loop around $z_0$ reads off.

Formal statement · 1 of 4

Cauchy's integral formula

Let $C$ be a positively oriented simple closed contour, and let $f$ be analytic on and inside $C$ . Then for every point $z_0$ inside $C$ ,

f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}\,dz, \qquad\text{equivalently}\qquad \oint_C \frac{f(z)}{z - z_0}\,dz = 2\pi i\,f(z_0).

State the hypotheses every time: $C$ positively oriented, simple, closed; $f$ analytic on and inside $C$ ; $z_0$ inside $C$ . Drop any one and the formula need not hold.

Step 0 of every evaluation — locate the pole. Before reaching for any formula, place the point $z_0$ (the root of the denominator) relative to $C$ : inside, outside, or on it. Everything downstream depends on that one check, and it is the single most common place marks are lost.

Worked example 1 · the hook, done properly

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

Step 0 — locate. The denominator vanishes at $z_0 = i$ , and $|i| = 1 < 2$ , so $z_0$ is inside $C$ . Identify $f$ . What multiplies $\dfrac{1}{z - z_0}$ is $f(z) = z^2$ , analytic everywhere — so CIF applies. Apply.

\oint_{|z|=2} \frac{z^2}{z - i}\,dz = 2\pi i\,f(i) = 2\pi i\cdot i^2 = 2\pi i\cdot(-1) = -2\pi i.

The boundary integral reconstructed $f(i) = -1$ without the integrand ever being evaluated at $i$ . Keep this in your pocket — it is the hook, and it returns in the gauntlet.

Worked example 2 · an exponential numerator

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

Step 0 — locate. Pole at $z_0 = 1$ ; $|1| = 1 < 2$ , inside. Identify $f$ : $f(z) = e^z$ , entire. Apply:

\oint_{|z|=2} \frac{e^z}{z - 1}\,dz = 2\pi i\,f(1) = 2\pi i\,e^1 = 2\pi i\,e.

Same three moves every time: locate, identify $f$ , evaluate $f$ at the pole, multiply by $2\pi i$ .

Formal statement · 2 of 4

When the pole is outside: nothing to apply

Step 0 is not a formality — half the value of CIF questions is knowing when not to use it. If the pole lies outside $C$ , the integrand is analytic on and inside $C$ , and you are back in Chapter 7: Cauchy–Goursat gives $0$ , with no formula at all.

Worked example 3 · the examiner's favourite trap

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

Step 0 — locate. Pole at $z_0 = 2$ ; $|2| = 2 > 1$ , so it is outside $C$ . Then $\dfrac{z^2 + 1}{z - 2}$ is analytic on and inside $|z| = 1$ — no singularity is enclosed — and by Cauchy–Goursat (Chapter 7)

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

CIF never enters. The trap is to see the $\dfrac{1}{z - z_0}$ shape, reach for $2\pi i\,f(z_0)$ , and forget to check where $z_0$ actually is. Step 0 saves you every time.

The decision so far

For $\displaystyle\oint_C \frac{f(z)}{z - z_0}\,dz$ with $f$ analytic on and inside $C$ :

$z_0$ inside $C$ $\;\Rightarrow\;$ $2\pi i\,f(z_0)$ (Cauchy's integral formula).
$z_0$ outside $C$ $\;\Rightarrow\;$ $0$ (Cauchy–Goursat — integrand analytic inside).
$z_0$ on $C$ $\;\Rightarrow\;$ the integral is improper / undefined; exam contours are arranged so this does not happen.

Formal statement · 3 of 4

Higher powers in the denominator: the derivative formula

What if the denominator is $(z - z_0)^2$ , or $(z - z_0)^3$ ? Differentiating Cauchy's formula under the integral sign — legal here — produces a whole family:

Cauchy's integral formula for derivatives

With the same hypotheses ( $C$ positively oriented simple closed, $f$ analytic on and inside $C$ , $z_0$ inside), for every integer $n \ge 0$ ,

f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz, \qquad\text{i.e.}\qquad \oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz = \frac{2\pi i}{n!}\,f^{(n)}(z_0).

The off-by-one trap — read this twice

The denominator exponent is $n + 1$ ; the derivative order is $n$ . A cube $(z - z_0)^3$ means $n + 1 = 3$ , so $n = 2$ — the second derivative, not the third. A bare $(z - z_0)$ is $n + 1 = 1$ , so $n = 0$ : plain CIF, $f^{(0)} = f$ . Read the exponent, subtract one — that is the derivative order. This single off-by-one is the most common slip in the whole topic.

A corollary with no real-calculus analogue

Look at what the derivative formula quietly says: if $f$ is analytic (merely once differentiable in the complex sense) on a region, then $f^{(n)}(z_0)$ exists for every $n$ — analytic once forces analytic infinitely often. Nothing like this happens on the real line: $f(x) = |x|\cdot x$ is differentiable once and never again. In $\mathbb{C}$ , one derivative buys you all of them. We will spend this corollary as hard currency in the next sitting (it is the engine inside Morera's theorem).

Worked example 4 · the derivative formula in action

Evaluate $\displaystyle\oint_{|z|=2} \frac{e^{2z}}{(z - 1)^2}\,dz$ .

Step 0 — locate. Pole at $z_0 = 1$ , $|1| < 2$ , inside. Read the exponent: denominator is $(z-1)^2$ , so $n + 1 = 2$ , giving $n = 1$ — the first derivative. Identify and differentiate $f$ : $f(z) = e^{2z}$ , and by the chain rule

f'(z) = 2e^{2z}, \qquad f'(1) = 2e^{2}.

The chain-rule factor of $2$ is its own slip point — $(e^{2z})' = 2e^{2z}$ , not $e^{2z}$ . Assemble:

\oint_{|z|=2} \frac{e^{2z}}{(z - 1)^2}\,dz = \frac{2\pi i}{1!}\,f'(1) = 2\pi i\cdot 2e^{2} = 4\pi i\,e^{2}.

One composed move, shown once — an exponential at an imaginary point

You will need this exactly once, in the capstone problem P6, so meet it here in the calm. Evaluating $e^{iz}$ at an imaginary point collapses to a real number:

e^{iz}\Big|_{z = i} = e^{i\cdot i} = e^{i^2} = e^{-1}, \qquad e^{iz}\Big|_{z = -i} = e^{i\cdot(-i)} = e^{-i^2} = e^{+1}.

Both ingredients are already yours: evaluating an exponential at a point, and $i^2 = -1$ . The only new thing is watching them compose — the imaginary exponent $i\cdot i$ becoming the real exponent $-1$ . Hold it for P6.

Formal statement · 4 of 4

Two poles inside: split, then apply CIF to each

When the denominator carries two different roots inside $C$ , CIF does not apply directly — its statement has a single pole $z_0$ . The fix is purely algebraic: partial fractions break the integrand into one-pole pieces, and CIF (or Cauchy–Goursat) handles each piece on its own.

Worked example 5 · two poles, both enclosed

Evaluate $\displaystyle\oint_{|z|=3} \frac{z}{(z - 1)(z - 2)}\,dz$ .

Step 0 — locate. Poles at $z = 1$ and $z = 2$ ; both satisfy $|z| < 3$ , so both are enclosed. Split. Write $\dfrac{z}{(z-1)(z-2)} = \dfrac{A}{z-1} + \dfrac{B}{z-2}$ ; cover-up gives $A = \dfrac{z}{z-2}\big|_{z=1} = \dfrac{1}{-1} = -1$ and $B = \dfrac{z}{z-1}\big|_{z=2} = \dfrac{2}{1} = 2$ :

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

Apply CIF to each enclosed pole (here $f =$ the constant numerators, evaluated trivially): $\displaystyle\oint \frac{-1}{z-1}\,dz = 2\pi i(-1)$ and $\displaystyle\oint \frac{2}{z-2}\,dz = 2\pi i(2)$ . Sum:

\oint_{|z|=3} \frac{z}{(z - 1)(z - 2)}\,dz = 2\pi i(-1) + 2\pi i(2) = 2\pi i.

The general recipe for several poles

Locate every root of the denominator (Step 0), and mark which lie inside $C$ .
Split by partial fractions into one-pole terms.
Apply CIF (or the derivative formula) to each term whose pole is inside; a term whose pole is outside contributes $0$ .
Sum.

You may notice this is doing real work mechanically. The next chapter gives it a name and turns it into a single formula — for now, the split-and-apply recipe is all you need, and it is exactly what the exam rewards.

Interactive playground

The boundary oracle: watch the loop reconstruct $f(z_0)$

Drag the point $z_0$ around inside and outside a fixed circle, and watch two numbers. LOOP is $\dfrac{1}{2\pi i}\displaystyle\oint_C \dfrac{f(z)}{z - z_0}\,dz$ , computed numerically by walking the boundary circle (the trapezoid rule on 720 points — nothing here knows $f(z_0)$ directly). DIRECT is $f(z_0)$ , evaluated at the point. Inside the circle the two agree to the decimal — the loop has reconstructed $f(z_0)$ from the boundary alone. Outside, the loop reads $0$ : it cannot see what it does not enclose.

The boundary oracle

Function f(z): Radius r: 2.00

The circle is fixed at the origin; drag the point $z_0$ (✕) and slide the radius. Compare LOOP (the boundary integral, divided by $2\pi i$ ) against DIRECT ( $f(z_0)$ itself).

What to notice

z₀ inside (presets a, b): LOOP and DIRECT print the same complex number to three decimals. The loop reconstructed $f(z_0)$ without ever touching $z_0$ — Cauchy's integral formula, live.
Drag z₀ outward across the edge: LOOP snaps to $\approx 0$ while DIRECT keeps reporting $f(z_0)$ . The loop cannot see outside itself — outside, the integrand is analytic inside $C$ , so Cauchy–Goursat gives $0$ , and CIF never fires.
On the edge: the readout refuses to answer — $z_0$ sits on the contour and the integral is undefined.
Change $f$ or the radius: inside, LOOP still tracks DIRECT exactly. The value depends only on $f$ and on $z_0$ being enclosed — never on how big the loop is.

End of sitting 1 — you can now compute with Cauchy's formula: locate the pole, pick CIF or the derivative formula or a partial-fraction split, and assemble. After the break: four giants that fall straight out of it.

Consequences of CIF · the proof block

Four children of Cauchy's formula — and the first, Morera

Everything in this sitting is "state and prove" material — the theory questions the course rewards most. Cauchy's integral formula has four famous consequences; stated in the order they appear on cards and the formula sheet, they are Morera, Liouville, the maximum-modulus principle, and the Fundamental Theorem of Algebra. We will prove them in a slightly different order — Morera, Liouville, FTA, max-modulus — so that Liouville and FTA stay next to each other, because the two-line jump from one to the other is the best story in the chapter. Call them "consequences," not "direct corollaries": some reach CIF through the infinite-differentiability corollary rather than from the formula itself.

1 · Morera's theorem (the converse of Cauchy–Goursat)

Assumed $f$ is continuous on a domain $D$ , and $\displaystyle\oint_C f(z)\,dz = 0$ for every closed contour $C$ in $D$ .

Proved $f$ is analytic in $D$ .

Key idea zero loops give an antiderivative; an antiderivative is analytic; an analytic function is infinitely differentiable — so its derivative $f$ is analytic too.

Because $\oint_C f\,dz = 0$ around every loop, the integral of $f$ between two points is path-independent in $D$ (Chapter 7).
Fix a basepoint $z_*$ and define $F(z) = \displaystyle\int_{z_*}^{z} f(w)\,dw$ . Path-independence makes this well-defined, and the Chapter-7 antiderivative theorem gives $F'(z) = f(z)$ — so $F$ is analytic on $D$ .
Now spend the corollary from Beat 3C: an analytic function is infinitely differentiable. So $F''$ exists everywhere in $D$ .
But $F'' = (F')' = f'$ . For $f'$ to exist throughout $D$ is exactly for $f$ to be analytic. $\blacksquare$

Honest framing. This is a sketch resting on one cited theorem, not a self-contained five-step proof. The load-bearing move — " $\oint f = 0$ over every loop $\Rightarrow$ an analytic antiderivative $F$ exists" — is the path-independence / antiderivative theorem from Chapter 7, used here, not re-proved. Given that theorem, the rest is short. The infinite-differentiability corollary (Beat 3C) is the spark that turns $F$ analytic into $f$ analytic; name it when you write this out.

Consequences of CIF · the proof block

Liouville, the Fundamental Theorem of Algebra, and max-modulus

These three are the heart of the sitting. Liouville comes from a one-line size estimate; FTA falls out of Liouville in four lines; the maximum-modulus principle comes from reading Cauchy's formula as an average.

2 · Cauchy's inequality, and Liouville's theorem

Cauchy's inequality if $f$ is analytic on and inside the circle $|z - z_0| = R$ and $|f| \le M_R$ on that circle, then

\bigl|f^{(n)}(z_0)\bigr| \le \frac{n!\,M_R}{R^{\,n}}.

Proof apply the derivative formula and the ML inequality (Chapter 7). On $|z - z_0| = R$ the integrand $\dfrac{f(z)}{(z - z_0)^{n+1}}$ has size at most $\dfrac{M_R}{R^{\,n+1}}$ , and the contour has length $2\pi R$ , so

\bigl|f^{(n)}(z_0)\bigr| = \frac{n!}{2\pi}\left|\oint \frac{f(z)}{(z-z_0)^{n+1}}\,dz\right| \le \frac{n!}{2\pi}\cdot\frac{M_R}{R^{\,n+1}}\cdot 2\pi R = \frac{n!\,M_R}{R^{\,n}}.

Liouville's theorem if $f$ is entire (analytic on all of $\mathbb{C}$ ) and bounded ( $|f| \le M$ everywhere), then $f$ is constant.

Take $n = 1$ in Cauchy's inequality. Since $f$ is entire we may use a circle of any radius $R$ about any point $z_0$ , and $M_R \le M$ : $\;|f'(z_0)| \le \dfrac{M}{R}$ .
This holds for every $R$ . Let $R \to \infty$ : the right-hand side $\to 0$ , forcing $f'(z_0) = 0$ .
$z_0$ was arbitrary, so $f' \equiv 0$ on $\mathbb{C}$ — hence $f$ is constant. $\blacksquare$

Both hypotheses are load-bearing. "Entire" is what lets $R \to \infty$ ; "bounded" is what makes the bound die. Drop either and the conclusion fails — $e^z$ is entire but unbounded; $z$ on a disk is bounded but not entire.

3 · The Fundamental Theorem of Algebra

Claim every non-constant polynomial $p(z)$ with complex coefficients has at least one root in $\mathbb{C}$ .

Proof (by contradiction, via Liouville)

Suppose $p$ is non-constant and has no root. Then $g(z) = \dfrac{1}{p(z)}$ is defined everywhere — and, being a ratio of analytic functions with non-vanishing denominator, $g$ is entire.
As $|z| \to \infty$ , $|p(z)| \to \infty$ (the top-degree term dominates), so $|g(z)| \to 0$ — in particular $|g(z)| \le 1$ outside some disk $|z| \le R$ . On that closed disk $g = 1/p$ is continuous ( $p$ has no zeros), hence bounded there; therefore $g$ is bounded on all of $\mathbb{C}$ .
Entire and bounded $\Rightarrow$ by Liouville, $g$ is constant — so $p$ is constant, contradicting "non-constant."
The contradiction kills the assumption: $p$ must have a root. $\blacksquare$ (Dividing out that root and repeating gives all $n$ roots of a degree- $n$ polynomial.)

The story beat. A theorem about polynomials — pure algebra, which resisted purely algebraic proof for centuries — falls in four lines the moment Liouville exists. That is why we kept Liouville and FTA adjacent.

4 · The maximum-modulus principle (state and apply)

Statement if $f$ is analytic and non-constant on a domain, then $|f|$ attains no maximum at an interior point; on a closed bounded region, the maximum of $|f|$ is therefore attained on the boundary.

Light sketch set $z = z_0 + re^{i\theta}$ in Cauchy's formula. The $dz = ire^{i\theta}\,d\theta$ cancels the $(z - z_0) = re^{i\theta}$ downstairs, leaving the mean-value property:

f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f\bigl(z_0 + re^{i\theta}\bigr)\,d\theta.

The centre value is the average of the values on any circle around it. Take the modulus of both sides and use $\bigl|\int g\bigr| \le \int |g|$ :

|f(z_0)| \le \frac{1}{2\pi}\int_0^{2\pi}\bigl|f(z_0 + re^{i\theta})\bigr|\,d\theta \le \max_{|z - z_0| = r} |f(z)|.

So the modulus at the centre cannot exceed the largest modulus on the surrounding circle — a value that is an average of its neighbours cannot exceed the largest of them. (Keep the wording at "cannot exceed," never "cannot beat": at the constant case the centre ties the maximum, and the strict-maximum / forced-constant refinement is deliberately left out — the exam form here is state-and-apply.)

Worked application · max of

|e^z|

on the closed disk

|z| \le 1

Since $|e^z| = |e^{x + iy}| = e^{x} = e^{\operatorname{Re} z}$ , maximising $|e^z|$ means maximising $\operatorname{Re} z$ on the disk. On $|z| \le 1$ , $\operatorname{Re} z$ is largest at $z = 1$ , where it equals $1$ . So

\max_{|z| \le 1} |e^z| = e^{1} = e, \quad\text{attained at } z = 1 \text{ — on the boundary, as the principle demands.}

Theory done — four classical theorems, all standing on Cauchy's one formula. Last stretch is all practice: the patterns the exam actually asks for.

Retrieval gauntlet

Test yourself before moving on

Each follows a method you have now seen worked. Commit to an answer before reading the feedback — recognising the right option once you see it feels like knowing it, and it is not the same thing. Step 0 (locate the pole) wins or loses most of these.

Quiz 1 · Step 0 first

Evaluate $\displaystyle\oint_{|z|=2} \frac{\cos z}{z - \pi}\,dz$ .

Quiz 2 · inside, so apply CIF

Evaluate $\displaystyle\oint_{|z|=2} \frac{\sin z}{z - \pi/2}\,dz$ .

Quiz 3 · the derivative formula, off-by-one and chain rule

Evaluate $\displaystyle\oint_{|z|=1} \frac{e^{2z}}{z^3}\,dz$ .

Quiz 4 · two poles, and they cancel

Evaluate $\displaystyle\oint_{|z|=3} \frac{dz}{z^2 - 1}$ .

Quiz 5 · which hypotheses does Liouville need

Liouville's theorem forces $f$ to be constant when $f$ is:

Quiz 6 · max-modulus, applied

The maximum of $|e^z|$ on the closed disk $|z| \le 1$ is:

Practice set

Six cold problems

Work each before revealing. The last two open with a faded bridge — part of the work done for you, the key step left blank — before the full solution. Every solution is complete; nothing points outside this page.

Plain CIF, with a sign trap

Shape · locate, identify f, evaluate, assemble

P1. Evaluate $\displaystyle\oint_{|z|=3} \frac{z^2}{z + 2i}\,dz$ .

Step 0 — locate. Write $z + 2i = z - (-2i)$ , so the pole is $z_0 = -2i$ ; $|-2i| = 2 < 3$ , inside. Identify $f$ : $f(z) = z^2$ . Evaluate (the sign is the slip point):

f(-2i) = (-2i)^2 = 4i^2 = -4.

Assemble: $\displaystyle\oint_{|z|=3} \frac{z^2}{z + 2i}\,dz = 2\pi i\,f(-2i) = 2\pi i(-4) = -8\pi i.$

Answer: $-8\pi i$ .

Looks fearsome, dies at Step 0

Shape · both poles outside ⇒ Cauchy–Goursat

P2. Evaluate $\displaystyle\oint_{|z|=1} \frac{\cos z}{z^2 - 6z + 8}\,dz$ .

Step 0 — locate. Factor the denominator: $z^2 - 6z + 8 = (z - 2)(z - 4)$ . The poles are $z = 2$ and $z = 4$ ; both have modulus $> 1$ , so both are outside $|z| = 1$ . The integrand is analytic on and inside $C$ , so by Cauchy–Goursat

\oint_{|z|=1} \frac{\cos z}{z^2 - 6z + 8}\,dz = 0.

A fearsome-looking integrand that dies at Step 0 with no integration at all — that is the lesson. Always factor and locate before anything else.

Derivative formula with a trig second derivative

Shape · read the exponent, differentiate, divide by n!

P3. Evaluate $\displaystyle\oint_{|z|=2} \frac{\cos z}{z^3}\,dz$ .

Step 0 — locate. Pole $z_0 = 0$ , inside $|z| = 2$ . Read the exponent: $z^3$ means $n + 1 = 3$ , so $n = 2$ . Differentiate $f(z) = \cos z$ twice: $(\cos z)'' = -\cos z$ , so $f''(0) = -\cos 0 = -1$ . Assemble:

\oint_{|z|=2} \frac{\cos z}{z^3}\,dz = \frac{2\pi i}{2!}\,f''(0) = \frac{2\pi i}{2}(-1) = -\pi i.

Answer: $-\pi i$ .

Faded bridge · same integrand, two different contours

Shape · the answer depends on the contour

P4. Let $I(C) = \displaystyle\oint_C \frac{z + 1}{(z - 1)(z - 2)}\,dz$ . Evaluate it for (a) $C : |z| = 3$ , and (b) $C : |z| = \tfrac32$ .

Bridge: the partial-fraction split is $\dfrac{z + 1}{(z - 1)(z - 2)} = \dfrac{-2}{z - 1} + \dfrac{3}{z - 2}$ (given for (a)). Before revealing (b), predict: which of the poles $z = 1, z = 2$ does the smaller circle $|z| = \tfrac32$ actually enclose?

(a) $C : |z| = 3$ . Both poles $z = 1$ and $z = 2$ are inside ( $1, 2 < 3$ ). Apply CIF to each PF term:

I = 2\pi i(-2) + 2\pi i(3) = 2\pi i(-2 + 3) = 2\pi i.

(b) $C : |z| = \tfrac32$ . Now $z = 1$ is inside ( $1 < 1.5$ ) but $z = 2$ is outside ( $2 > 1.5$ ). Two equivalent routes:

Via the split: keep only the enclosed term. $\dfrac{-2}{z - 1}$ contributes $2\pi i(-2)$ ; the $\dfrac{3}{z - 2}$ term has its pole outside, so it contributes $0$ . Total $-4\pi i$ .
Direct CIF: with one pole enclosed, write the integrand as $\dfrac{g(z)}{z - 1}$ with $g(z) = \dfrac{z + 1}{z - 2}$ , analytic on and inside $C$ . Then $g(1) = \dfrac{2}{-1} = -2$ , so $I = 2\pi i\,g(1) = -4\pi i$ .

Both routes agree: $I = -4\pi i$ . The teaching point: the same integrand gave $2\pi i$ on one contour and $-4\pi i$ on another — the value of a contour integral depends on which singularities the loop encloses, not on the integrand alone.

Theory · reproduce the Liouville proof

P5. Show that an entire function with $|f(z)| \le 10$ for all $z$ is constant.

Predict the next step before each reveal: this is Liouville with $M = 10$ — what is the one inequality you start from?

Step 1 — the starting inequality. Which estimate, and at what

n

?

Cauchy's inequality at $n = 1$ . Since $f$ is entire, for any point $z_0$ and any radius $R$ we may take $M_R \le 10$ , giving

|f'(z_0)| \le \frac{1!\cdot 10}{R} = \frac{10}{R}.

Step 2 — exploit "for every $R$ ." The bound holds for all

R

; what do you do with that?

Let $R \to \infty$ . The right-hand side $\dfrac{10}{R} \to 0$ , so we are forced to $|f'(z_0)| \le 0$ , i.e. $f'(z_0) = 0$ . (This step needs $f$ entire — only then is every $R$ allowed.)

Step 3 — conclude.

$z_0$ was arbitrary, so $f'(z) = 0$ for every $z \in \mathbb{C}$ . A function with zero derivative everywhere on $\mathbb{C}$ is constant. $\blacksquare$ (The number $10$ never mattered — any finite bound dies the same way. That is the content of Liouville.)

Faded bridge · capstone — a loop integral that returns a real number

Shape · two complex poles, collapsing to a real answer

P6. Evaluate $\displaystyle\oint_{|z|=2} \frac{e^{iz}}{z^2 + 1}\,dz$ .

Bridge: $z^2 + 1 = (z - i)(z + i)$ . Before revealing, predict two things — which poles lie inside $|z| = 2$ , and (recalling the Beat 3C aside) what $e^{iz}$ becomes at $z = i$ and at $z = -i$ .

Step 0 — locate. Poles $z = \pm i$ ; both have modulus $1 < 2$ , so both are inside. Split $\dfrac{e^{iz}}{(z - i)(z + i)}$ and apply CIF at each pole.

At $z = i$ : write the integrand as $\dfrac{g(z)}{z - i}$ with $g(z) = \dfrac{e^{iz}}{z + i}$ . The exponent $i\cdot i = -1$ , so $e^{iz}|_{z=i} = e^{-1}$ , and $g(i) = \dfrac{e^{-1}}{2i}$ .

At $z = -i$ : write it as $\dfrac{h(z)}{z + i}$ with $h(z) = \dfrac{e^{iz}}{z - i}$ . The exponent $i\cdot(-i) = +1$ , so $e^{iz}|_{z=-i} = e^{1}$ , and $h(-i) = \dfrac{e^{1}}{-2i}$ .

Sum the two $2\pi i$ contributions:

\oint_{|z|=2}\frac{e^{iz}}{z^2 + 1}\,dz = 2\pi i\!\left(\frac{e^{-1}}{2i} - \frac{e^{1}}{2i}\right) = \pi\bigl(e^{-1} - e\bigr) = \pi\!\left(\frac1e - e\right) = -2\pi\sinh 1 \approx -7.384.

Answer: $\pi\!\left(\tfrac1e - e\right) = -2\pi\sinh 1$ . A loop integral of a complex function produced a real number — the two imaginary poles' contributions combined into something with no $i$ left. Chapter 9 turns exactly this into a machine for evaluating real integrals that defeat ordinary calculus.

Exam translator

How this appears on exams

Exam shapes

In standard Indian engineering math exams, and likely in MCC201A unless your class notes differ, this chapter shows up in four recognisable costumes — each one you have now seen worked:

"Evaluate $\oint f(z)/(z - z_0)^n\,dz"$ — run the decision tree below. The single most valuable habit is Step 0.
"State and prove Cauchy's integral formula" — Reveal 1, a full proof (the Beat 2 idea, made rigorous).
"State and prove Liouville's theorem; hence prove the Fundamental Theorem of Algebra" — Reveal 2. "Hence" means use Liouville, not algebra.
"State the maximum-modulus principle and find $\max|f|$ on …" — state it, then the $|e^z| = e^{\operatorname{Re} z}$ pattern from the four-children section (Q6).

Costume 1 · the evaluation decision tree

For $\displaystyle\oint_C \frac{f(z)}{(z - z_0)^{\,k}}\,dz$ with $f$ analytic on and inside $C$ , every problem is one walk down this list:

Step 0 — locate. Find every root of the denominator and mark which lie inside $C$ .
No pole inside? Integrand analytic inside $\Rightarrow$ Cauchy–Goursat: the integral is $0$ . Stop.
One pole inside, $k = 1$ ? Plain CIF: $2\pi i\,f(z_0)$ .
One pole inside, $k \ge 2$ ? Derivative formula with $n = k - 1$ : $\dfrac{2\pi i}{(k-1)!}\,f^{(k-1)}(z_0)$ . (Read the exponent, subtract one.)
Several poles inside? Partial-fraction split, apply the rule above to each enclosed term, and add.

Reveal 1 — "State and prove Cauchy's integral formula." This is Beat 2's idea turned into a proof: the difference is the one line that bounds the error and sends it to zero.

Predict the next step before each reveal: how do you isolate the $2\pi i\,f(z_0)$ term, and how do you kill what is left over?

Step 1 — deform and split.

Deform $C$ to a small circle $C_\rho$ of radius $\rho$ about $z_0$ (legal: the integrand is analytic between $C$ and $C_\rho$ ). On $C_\rho$ write $f(z) = f(z_0) + \bigl[f(z) - f(z_0)\bigr]$ , so

\oint_{C_\rho}\frac{f(z)}{z - z_0}\,dz = f(z_0)\oint_{C_\rho}\frac{dz}{z - z_0} + \underbrace{\oint_{C_\rho}\frac{f(z) - f(z_0)}{z - z_0}\,dz}_{R(\rho)} = 2\pi i\,f(z_0) + R(\rho),

using the fundamental integral $\oint_{C_\rho}\frac{dz}{z - z_0} = 2\pi i$ .

Step 2 — bound the remainder (the load-bearing line).

By the ML inequality on $C_\rho$ , where $|z - z_0| = \rho$ and the length is $2\pi\rho$ ,

|R(\rho)| \le \frac{\max_{C_\rho}\bigl|f(z) - f(z_0)\bigr|}{\rho}\cdot 2\pi\rho = 2\pi\,\max_{C_\rho}\bigl|f(z) - f(z_0)\bigr|.

As $\rho \to 0$ , continuity of $f$ at $z_0$ sends $\max_{C_\rho}|f(z) - f(z_0)| \to 0$ , so $R(\rho) \to 0$ . This is the step the examiner is marking — without it, the argument is a sketch, not a proof.

Step 3 — conclude.

By deformation, the left-hand side $\oint_C \frac{f(z)}{z - z_0}\,dz$ is independent of $\rho$ . So $2\pi i\,f(z_0) + R(\rho)$ is a constant in $\rho$ , which means $R(\rho)$ is itself constant — and a constant that tends to $0$ is $0$ . Hence $R(\rho) \equiv 0$ and

\oint_C \frac{f(z)}{z - z_0}\,dz = 2\pi i\,f(z_0), \qquad\text{i.e.}\qquad f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}\,dz. \quad\blacksquare

Reveal 2 — "State and prove Liouville's theorem; hence prove the Fundamental Theorem of Algebra." The exam-answer shape of the four-children section. The word "hence" is an instruction: build FTA on Liouville.

Predict the next step: after Liouville is proved, what entire, bounded function do you build from a rootless polynomial?

Part 1 — state and prove Liouville.

Statement. A bounded entire function is constant.

Proof. Let $|f| \le M$ on $\mathbb{C}$ . Cauchy's inequality at $n = 1$ , on a circle of radius $R$ about any $z_0$ , gives $|f'(z_0)| \le M/R$ . Valid for every $R$ because $f$ is entire; let $R \to \infty$ to get $f'(z_0) = 0$ . As $z_0$ was arbitrary, $f' \equiv 0$ , so $f$ is constant. $\blacksquare$

Part 2 — hence FTA.

Let $p$ be a non-constant polynomial and suppose, for contradiction, it has no root. Then $g = 1/p$ is entire (denominator never zero). Since $|p(z)| \to \infty$ as $|z| \to \infty$ , $g(z) \to 0$ there — so $g$ is bounded outside a large disk, and bounded on that closed disk by continuity, hence bounded on all of $\mathbb{C}$ . By Liouville, $g$ is constant — hence $p$ is constant, contradicting "non-constant." So $p$ has a root; factor it out and repeat for all $n$ roots. $\blacksquare$

What costs marks

Skipping Step 0 — applying CIF to a pole that is outside, or Cauchy–Goursat to a loop that encloses one (Q1, Q2, P2).
The $n + 1$ off-by-one — reading $(z - z_0)^3$ as the third derivative instead of the second (Q3, P3).
Dropping the chain-rule factor — $(e^{2z})' = 2e^{2z}$ , not $e^{2z}$ (Q3, WE-4).
A partial-fraction sign error — the coefficients in Q4 are $+\tfrac12$ and $-\tfrac12$ and cancel; a sign slip turns $0$ into $2\pi i$ (Q4, P4).
For "prove CIF": dropping the error bound — stopping at " $f(z) \approx f(z_0)$ " without showing $R(\rho) \to 0$ (Reveal 1). This is the single most common way a "prove Cauchy's formula" answer loses marks.

Six things to carry out of this chapter

Cauchy's integral formula: $f(z_0) = \dfrac{1}{2\pi i}\oint_C \dfrac{f(z)}{z - z_0}\,dz$ — boundary values determine the inside; $z_0$ must be enclosed.
Derivative formula: $\oint_C \dfrac{f(z)}{(z - z_0)^{n+1}}\,dz = \dfrac{2\pi i}{n!}\,f^{(n)}(z_0)$ — denominator exponent $n + 1$ , derivative order $n$ . Read the exponent, subtract one.
Step 0 always: locate every pole and check inside / outside / on the contour before any formula.
Several poles: partial-fraction split, CIF on each enclosed term, add.
The four consequences: Morera (zero loops $\Rightarrow$ analytic), Liouville (bounded entire $\Rightarrow$ constant), the maximum-modulus principle (max on the boundary), and the Fundamental Theorem of Algebra (Liouville $\Rightarrow$ every non-constant polynomial has a root).
Cauchy's inequality: $|f^{(n)}(z_0)| \le \dfrac{n!\,M_R}{R^{\,n}}$ — the size estimate behind Liouville, and behind FTA in turn.

Before you move on

Re-do two of P1–P6 from a blank page tomorrow — not re-read, re-do. Spaced retrieval of the locate-then-apply pattern is worth more than another hour now.

What comes next. The two-pole partial-fraction trick you used in 3D, P4 and P6 is doing real work by hand. The next chapter — Taylor and Laurent series, singularities — turns that hand-work into a single machine, and turns the "real number from a complex loop" surprise of P6 into a tool for hard real integrals that ordinary calculus can't touch.

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 4 — Cauchy's integral formula — the same formula and its consequences from another angle, with the Cauchy-inequality and Liouville arguments laid out cleanly.

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