Cauchy's theorem and integral formula

In one sentence

On a simply connected domain every closed contour integral of a holomorphic function vanishes, the single exception $\oint dz/z = 2\pi i$ is where all the content lives, and the integral formula turns that exception into a machine for evaluating functions and their derivatives.

Why the exam asks it

Because "simply connected" is the hypothesis candidates skip, and because almost every contour integral in the paper reduces to one computation — $\oint z^n\,dz$ round a circle — which is $0$ for every integer $n$ except $-1$. A candidate who knows that one fact and where the singularities are can do most of this topic without integrating anything.

The idea

For a contour $\gamma$ and continuous $f$, $\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\gamma'(t)\,dt$.

The ML inequality bounds it: $\left|\int_\gamma f\right| \le M \cdot L$ where $M$ bounds $|f|$ on $\gamma$ and $L$ is the length. Crude, and it proves more of this topic than anything else.

The one computation

Round $|z| = r$, parametrise $z = re^{it}$, so $dz = ire^{it}dt$ and

$$\oint_{|z|=r} z^{n}\,dz = \int_0^{2\pi} i r^{n+1}e^{i(n+1)t}\,dt = \begin{cases} 2\pi i & n = -1\\ 0 & \text{otherwise.}\end{cases}$$

Note what is special: not that $n$ is negative — $z^{-2}$ integrates to $0$ as well — but that $n = -1$ is the one power whose antiderivative, $\log z$, cannot be defined single-valuedly around the origin. Everything else has a single-valued antiderivative and so integrates to zero round any loop.

Cauchy's theorem

Cauchy–Goursat. If $f$ is holomorphic on a simply connected domain $D$, then $\oint_\gamma f(z)\,dz = 0$ for every closed contour $\gamma$ in $D$.

The hypothesis is about the domain, not the function. $1/z$ is holomorphic on $\mathbb{C}\setminus\{0\}$, perfectly well behaved everywhere it is defined, and $\oint_{|z|=1} dz/z = 2\pi i \ne 0$. Nothing is wrong with $f$; the domain has a hole and the loop cannot be shrunk past it.

The practical consequence is deformation: a contour may be moved freely as long as it does not cross a singularity. So any loop winding once around the origin gives $2\pi i$, whatever its shape, and this is what lets you replace an awkward contour with a circle.

The integral formula

If $f$ is holomorphic on and inside a positively oriented simple closed contour $\gamma$, and $a$ is inside $\gamma$, then

$$f(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z - a}\,dz .$$

The values on the contour determine every value inside — a rigidity with no real-variable analogue. If $a$ lies outside, the integrand is holomorphic inside and the integral is $0$ instead. Which of the two applies is decided entirely by where $a$ sits, and that is the commonest way these questions are made.

Differentiating under the integral gives

$$f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\,dz,$$

and with it the fact that a holomorphic function is infinitely differentiable — one derivative buys all of them, which is false in every real-variable sense.

Cauchy's estimates

From the formula and ML, if $|f| \le M$ on $|z - a| = r$ then

$$|f^{(n)}(a)| \le \frac{n!\,M}{r^{n}} .$$

Everything in the next chapter — Liouville, the fundamental theorem of algebra, the characterisation of polynomials by growth — comes out of letting $r \to \infty$ here.

Several singularities

With more than one bad point inside, split by partial fractions. For $\dfrac{1}{z(z-1)} = \dfrac{-1}{z} + \dfrac{1}{z-1}$ round $|z| = 2$: each term contributes $2\pi i$ times its coefficient, giving $-2\pi i + 2\pi i = 0$. This is the residue theorem arriving early, and partial fractions will answer most questions before you need it formally.

Where intuition breaks

"$f$ is holomorphic on the domain, so $\oint f = 0$." $1/z$ on $\mathbb{C}\setminus\{0\}$. The missing hypothesis is simple connectedness of the domain, and it is missing in the option rather than in the function. This is the topic's central hypothesis dropped.

"$\oint_\gamma f = 0$ for some closed contour, so $f$ is holomorphic." $\oint_{|z|=1} \overline{z}\,dz = \oint z^{-1}\cdot|z|^2 dz$, and on the unit circle $\overline z = 1/z$, giving $2\pi i$ — but take $f(z) = \overline z$ round a contour not enclosing the origin and you get $0$ from a nowhere-holomorphic function. One vanishing integral proves nothing; Morera needs it for every triangle, plus continuity.

"The integral formula gives $f(a)$ wherever $a$ is." Only for $a$ inside. Outside, the answer is $0$. Half the options in this topic are the same integral with the point moved.

"$\oint z^{-2}dz = 2\pi i$ because the exponent is negative." It is $0$. Only $n = -1$ survives, because only $1/z$ lacks a single-valued antiderivative.

"$f$ holomorphic inside $\gamma$ is enough." The theorem asks for holomorphic on and inside. A singularity sitting on the contour itself invalidates it, and that case is set deliberately.

"Deformation lets me move the contour anywhere." Anywhere it does not cross a singularity. Crossing one changes the answer, which is precisely what the winding number counts.

The exam's angle

1. Locate the singularities, then ask which are inside. That question answers most contour integrals before any calculation.
2. Reduce to $\oint z^n dz$. Partial fractions, or a series expansion, turns nearly everything into a sum of these, and only the $n = -1$ terms survive.
3. Check whether the domain is simply connected. If it is, and $f$ is holomorphic there, the answer is $0$ and you are done.
4. For $f^{(n)}$, match the power in the denominator. An integrand with $(z-a)^{n+1}$ underneath is the derivative formula wearing a disguise, and the factor is $n!$.
5. Reach for ML when asked to bound rather than evaluate. Estimate $|f|$ on the contour, multiply by the length, and that is usually the intended route.

The night before

• $\oint_{|z|=r} z^n dz = 2\pi i$ if $n = -1$, else $0$. The whole topic rests on this.
• Cauchy–Goursat: holomorphic on a simply connected domain $\Rightarrow$ every closed contour integral is $0$. The hypothesis is on the domain.
• $\oint_{|z|=1} dz/z = 2\pi i$: nothing is wrong with $1/z$, the domain has a hole.
• Deformation: move the contour freely, never across a singularity.
• Integral formula: $f(a) = \frac{1}{2\pi i}\oint \frac{f(z)}{z-a}dz$ for $a$ inside; $0$ for $a$ outside.
• $f^{(n)}(a) = \frac{n!}{2\pi i}\oint \frac{f(z)}{(z-a)^{n+1}}dz$. Holomorphic $\Rightarrow$ infinitely differentiable.
• Cauchy estimates: $|f^{(n)}(a)| \le n!M/r^n$.
• ML: $|\int_\gamma f| \le ML$.
• Several poles: partial fractions, then add the $n=-1$ pieces.