Lesson 5: Consequences of Cauchy's Formula
Lesson 4 ended with the Cauchy Integral Formula: boundary values determine interior values, completely. This lesson is where that single fact cascades into a sequence of results that would each be hard-won in real analysis and here fall out almost for free, culminating in a proof of the Fundamental Theorem of Algebra that has nothing to do with algebra at all.
Cauchy's differentiation formulas
CIF holds for every point $z_0$ inside a contour $\gamma$, so it can be differentiated with respect to $z_0$ under the integral sign, one derivative at a time. This might look suspicious — real calculus teaches derivatives and integrals as things that cancel via the Fundamental Theorem of Calculus. But FTC concerns differentiating with respect to a moving bound of integration; here the contour $\gamma$ is fixed, and $z_0$ is a parameter sitting inside the integrand, never touching the path. Differentiating under a fixed integral sign is a different, separate result (Leibniz's rule), a swap of two limiting processes, legitimate here because $f$ is holomorphic.
Differentiating $f(z_0)=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-z_0}\,dz$ once: $$\frac{d}{dz_0}(z-z_0)^{-1} = (z-z_0)^{-2}$$ via $u=z-z_0$, $du/dz_0=-1$, and the two negative signs canceling. Differentiating again gives a $(z-z_0)^{-3}$ term, and the pattern that emerges by extrapolation is easy to get slightly wrong in two specific ways: the constant out front is not the newest factor alone but the running product $1\times2\times3=\dots=n!$, compounding multiplicatively across derivatives; and the denominator power is $(z-z_0)^{n+1}$, not $(z-z_0)^n$; checking directly against the worked cases (powers $2,3,4$ for $n=1,2,3$) confirms the $+1$. The general formula: $$f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\,dz$$
The headline consequence: holomorphic (complex-differentiable once) automatically implies infinitely differentiable, no extra hypotheses needed. This has no real-variable analogue at all: a real function's derivative can exist without its second derivative existing.
Key takeaway: Cauchy's differentiation formula: $f^{(n)}(z_0)=\dfrac{n!}{2\pi i}\displaystyle\oint_\gamma \dfrac{f(z)}{(z-z_0)^{n+1}}\,dz$; holomorphic $\Rightarrow$ infinitely differentiable, automatically.
Cauchy's estimates
This step needs one more tool first: the ML-estimate, $$\left|\int_\gamma h(z)\,dz\right| \le M\cdot L,$$ where $M$ bounds $|h|$ on $\gamma$ and $L$ is $\gamma$'s length. It's the integral version of the vector triangle inequality: the magnitude of a sum (or integral) is at most the sum (or integral) of magnitudes. A quick sanity check: $\left|\oint_{|z|=1}z^2\,dz\right|\le 2\pi$ (with $M=1$, $L=2\pi$), consistent with the true value, $0$, from Lesson 4.
Now apply it to the differentiation formula on a circle $C_R$ of radius $R$ centered at $z_0$, with $|f|\le M$ on $C_R$. On $C_R$, $|z-z_0|=R$ exactly (not just a bound), so the integrand satisfies $\left|\frac{f(z)}{(z-z_0)^{n+1}}\right|\le \frac{M}{R^{n+1}}$: exact in the denominator, a bound in the numerator. Multiplying by the circle's circumference $2\pi R$ and then by $\frac{n!}{2\pi}$: $$\frac{M}{R^{n+1}}\cdot2\pi R = \frac{2\pi M}{R^n}, \qquad \frac{n!}{2\pi}\cdot\frac{2\pi M}{R^n} = \frac{n!\,M}{R^n}$$ (Writing the multiplication out as a single fraction before simplifying is worth doing explicitly — it's easy to lose track of a $2\pi/2\pi$ cancellation otherwise.) This gives Cauchy's estimate: $$|f^{(n)}(z_0)| \le \frac{n!\,M}{R^n}$$ For a bounded entire function with, say, $M=10$: at $R=100$, the bound on $|f'(z_0)|$ is $1/10$; at $R=10{,}000$, it's $1/1000$. The bound keeps shrinking as $R$ grows, exactly the setup Liouville's theorem needs next.
Note the hypothesis carefully: Cauchy's estimate requires $f$ holomorphic on the entire closed disk bounded by $C_R$, not merely on the circle itself, inherited directly from CIF, which needs the same thing.
Key takeaway: ML-estimate: $\left|\displaystyle\int_\gamma h(z)\,dz\right| \le M\cdot L$. Cauchy's estimate: $|f^{(n)}(z_0)| \le \dfrac{n!\,M}{R^n}$, where $M$ bounds $|f|$ on the circle $C_R$ of radius $R$ centered at $z_0$.
Liouville's theorem
Cauchy's estimate does all the work here: the proof is five short moves, and the whole trick is watching which quantities stay fixed and which are free. Each quantity keeps its color from step to step below (hover or tap one to trace it through the argument).
Proof, step by step
Liouville's theorem. A bounded entire function is constant.
- 1 · the hypotheses$$\htmlClass{pf-f}{f} \text{ entire}, \qquad \bigl|\htmlClass{pf-f}{f}(z)\bigr| \le \htmlClass{pf-M}{M} \ \text{ for all } z \in \mathbb{C}$$
Two assumptions, and the second is global: one fixed $M$ works at every point of the plane at once. That single number is the lever the whole proof pulls.
- 2 · Cauchy's estimate with n = 1$$\bigl|\htmlClass{pf-fp}{f'(z_0)}\bigr| \le \frac{1!\,\htmlClass{pf-M}{M}}{\htmlClass{pf-R}{R}} = \frac{\htmlClass{pf-M}{M}}{\htmlClass{pf-R}{R}}$$
Pick any point $z_0$ and any circle of radius $R$ around it. Entire means holomorphic on every closed disk, so the estimate from earlier in this lesson applies at every radius — the hypothesis CIF needs comes free.
- 3 · let R → ∞$$\frac{\htmlClass{pf-M}{M}}{\htmlClass{pf-R}{R}} \longrightarrow 0 \quad \text{as } \htmlClass{pf-R}{R} \to \infty$$
Boundedness earns its keep: $M$ is pinned while $R$ is ours, so the bound can be driven as small as we like. For an unbounded function, $M$ would grow along with $R$ and this lever wouldn't exist.
- 4 · the squeeze$$0 \le \bigl|\htmlClass{pf-fp}{f'(z_0)}\bigr| \le \frac{\htmlClass{pf-M}{M}}{\htmlClass{pf-R}{R}} \ \text{ for every } \htmlClass{pf-R}{R} \implies \bigl|\htmlClass{pf-fp}{f'(z_0)}\bigr| = 0$$
$f'(z_0)$ is a single fixed number sitting below an arbitrarily small bound. The only nonnegative number smaller than every $M/R$ is $0$.
- 5 · z₀ was arbitrary$$\htmlClass{pf-fp}{f'} \equiv 0 \ \text{ on } \mathbb{C} \implies \htmlClass{pf-f}{f} \text{ is constant}$$
Nothing above chose $z_0$ specially, so the derivative vanishes at every point. $f$ is constant — Liouville's theorem.
Why does this specifically need $R\to\infty$, and not just any circle? Test it on $f(z)=z^2$ (entire, unbounded). At $z_0=0$: $M(R)=R^2$, so the bound $R^2/R=R\to0$ as $R\to0$ too, but this is a coincidence, since $f(0)=0$ exactly. At a generic point $z_0=1$: $M(R)=(1+R)^2$ (the farthest point on the circle from the origin), so the bound $(1+R)^2/R\to\infty$ as $R\to0$: no shrinkage at all. The real distinction: for a bounded entire function, $M$ is one fixed number that works for a circle of any size, so growing $R$ is the only lever available, and it works. For an unbounded function, $M$ grows right along with $R$, so that lever doesn't exist. Small circles never help either way.
The result is genuinely strange when checked against familiar functions: $\sin z$ and $\cos z$ are bounded and non-constant on $\mathbb{R}$, but neither is bounded on all of $\mathbb{C}$: for instance $|\sin(iy)|=\sinh(y)\to\infty$. Liouville's theorem is exactly why that has to be true: a bounded, non-constant entire function simply cannot exist.
Key takeaway: Liouville's theorem: a bounded entire function is constant.
The Fundamental Theorem of Algebra
A proof by contradiction: the thing to watch is which moves are mere setup and which single move is the payoff. Steps 1–5 below just take the assumed-rootless polynomial, build $f=1/p$ out of it, and check that function's passport: entire, and bounded. None of them is clever. The flagged final step is the only one that closes a loop: Liouville detonates the contradiction, and the degree $n\ge1$ from step 1 comes back to bite.
Proof, step by step
The Fundamental Theorem of Algebra. Every nonconstant polynomial over $\mathbb{C}$ has at least one root.
- 1 · the setup, and the assumption to kill$$\htmlClass{pf-p}{p}(z) = a_{\htmlClass{pf-n}{n}} z^{\htmlClass{pf-n}{n}} + \dots + a_0, \quad \htmlClass{pf-n}{n} \ge 1, \ a_{\htmlClass{pf-n}{n}} \ne 0; \qquad \text{suppose } \htmlClass{pf-p}{p}(z) \ne 0 \ \text{ for every } z \in \mathbb{C}$$
Degree $n \ge 1$ with a nonzero leading coefficient is exactly what "nonconstant" means — file that away, it returns at the very end. Everything after this line is a consequence of the assumption no root anywhere, held only so it can destroy itself.
- 2 · the assumption builds a function$$\htmlClass{pf-f}{f}(z) = \frac{1}{\htmlClass{pf-p}{p}(z)} \ \text{ is entire}$$
Pure setup, no cleverness: $p$ is holomorphic everywhere and — by assumption — never zero, so the quotient $f$ is defined and holomorphic on all of $\mathbb{C}$.
- 3 · far away: the leading term takes over$$|\htmlClass{pf-p}{p}(z)| \to \infty \ \text{ as } |z| \to \infty \implies |\htmlClass{pf-f}{f}(z)| \to 0$$
For huge $|z|$ the term $a_nz^n$ dwarfs every other term of $p$ — this is the one place the polynomial's shape gets used. So outside some large disk, $f$ is not just bounded but tiny.
- 4 · close in: continuity does the rest$$\htmlClass{pf-f}{f} \ \text{ continuous on the closed disk} \implies |\htmlClass{pf-f}{f}| \le M \ \text{ there}$$
Still setup. There is no singularity hiding in the disk — $p$ never vanishes, by assumption — and a continuous function on a closed disk is bounded. (This is the step that feels skippable and isn't: $|f| \to 0$ far away says nothing about the middle.)
- 5 · combine the two regions$$\htmlClass{pf-f}{f} \ \text{ is entire and bounded on all of } \mathbb{C}$$
Small outside the disk, at most $M$ inside it: one bound covers the whole plane. Notice what just happened — the assumption "$p$ has no root" has manufactured exactly the hypotheses of Liouville's theorem.
- the contradiction6 · Liouville fires$$\htmlClass{pf-f}{f} \ \text{constant} \implies \htmlClass{pf-p}{p} = \frac{1}{\htmlClass{pf-f}{f}} \ \text{constant} \implies \text{contradicts } \htmlClass{pf-n}{n} \ge 1$$
Everything before this was scaffolding; this is the only move that closes a loop. Liouville (previous section) forces the bounded entire $f$ to be constant, so $p$ is constant — but step 1 declared $n \ge 1$. The assumption was false: $p$ must have a root.
That's the Fundamental Theorem of Algebra (every nonconstant polynomial over $\mathbb{C}$ has at least one root), proved with no algebra at all, only Liouville's theorem.
(Worked example, arbitrary; the technique is the point: $p(z)=z^3+1$ has roots $z=e^{i\pi/3}$, $e^{i\pi}=-1$, $e^{i5\pi/3}$, by the same roots-of-unity work from Lesson 1.)
Key takeaway: Fundamental Theorem of Algebra: every nonconstant polynomial over $\mathbb{C}$ has at least one root. Proof: assume not; $1/p$ is entire and bounded; Liouville forces it constant, contradicting the polynomial's degree.
The whole chain, replayed
Everything above was one continuous argument: CIF $\to$ differentiation formulas $\to$ Cauchy's estimates $\to$ Liouville $\to$ FTA, each link using only the one before it. The recap below replays that chain end to end against live pictures: as you scroll, the pinned figure moves through the four layers, and every one stays interactive the whole way (drag $z_0$, drag the sliders — the animations rerun if you scroll back).
The purple shading is the integrand's magnitude $|f(z)/(z-z_0)^{n+1}|$ for $f(z)=e^z$, and at $n=0$ that is the CIF integrand itself. The formula only ever reads $f$ on the circle $\gamma$, yet it pins down $f$ at $z_0$. Drag $z_0$; the numerically computed integral tracks the exact value $e^{z_0}$ wherever it goes.
Step $n$ up to $3$: each derivative pulls one more factor down and adds one power to the denominator, and the integrand's weight clamps ever tighter around $z_0$. Higher derivatives are more local information, and the integrand concentrates to match. Nothing about higher derivatives was ever assumed, so $f^{(n)}(z_0)$ exists for every $n$: holomorphic $\Rightarrow$ infinitely differentiable.
Take the contour to be a circle $C_R$: on it, $|z-z_0| = R$ exactly, which is what makes circles the right contour for estimates. The ticks sample $|f|$ around the circle; the red dot marks the largest sample: that value is $M$. The readouts assemble Cauchy's estimate from its three raw parts: $M$, the length $2\pi R$, and the finished bound $M/R$.
Slide $R$ outward and the bound does not shrink here: for $f=e^z$, the max $M(R)=e^R$ grows far faster than $R$. That's the tell. Cauchy's estimate only becomes a weapon when $M$ stays pinned while $R$ grows, which is exactly what bounded is about to buy.
Now let $f$ be entire and bounded: $|f| \le M = 10$, one fixed $M$, everywhere. Entire means no singularity ever stops the circle: $C_R$ is legal at every radius. Bounded means that same $M$ works on all of them at once. Both hypotheses are about to do real work: entirety frees $R$, boundedness pins $M$.
Let $R \to \infty$ (running now — the circle creeps toward the edge while the true $R$ runs off to infinity). The gauge is the live bound $M/R$ draining to zero, and $|f'(z_0)|$ is trapped beneath it the whole way down. A fixed nonnegative number smaller than every $M/R$ has no choice: $|f'(z_0)| = 0$.
Nothing in that squeeze used where $z_0$ sits — so the identical argument runs at every point of the plane (stamping now). $f' \equiv 0$ on all of $\mathbb{C}$, so $f$ is constant: Liouville's theorem, the same five moves as the stepper earlier on this page, now drawn instead of written.
Domain coloring of $p(z)=z^3+1$: hue is $\arg p$, dark is $|p|$ near $0$ — the three dark points are the roots FTA promises. Assume, for contradiction, $p$ has no root: then $f = 1/p$ is entire. The dashed ring is $|z|=r$, and the readout tracks $\max |1/p|$ on it.
Push $r$ outward (running): the leading term dominates, $|p| \to \infty$, so $\max|1/p| \to 0$, combined with continuity on the inner disk, $f=1/p$ is bounded on all of $\mathbb{C}$. Entire and bounded forces $f$ constant by step 7, so $p$ is constant, contradicting $n \ge 1$. Every nonconstant polynomial has a root, and the whole chain (CIF $\to$ derivatives $\to$ estimates $\to$ Liouville $\to$ FTA) ran on one formula.
(The standalone version keeps the full four-section walkthrough with its proof steps and explore cards, if you want the long-form version back.)
The maximum modulus principle
Plugging $z=z_0+re^{i\theta}$ into CIF gives the mean-value property in one line: $$f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta$$ This says the value at the center of a circle is the average of the values on the circle. For that average to also be a strict local maximum, every neighboring value on the circle would already have to equal it: an average can't exceed all the things it's averaging unless they're all equal to it. That reasoning propagates: constant on one circle forces constant everywhere nearby (a preview of the identity theorem), and the conclusion is the maximum modulus principle: for nonconstant holomorphic $f$, $|f|$ has no interior local maximum — only on the boundary of a region.
(Practice check: if $|f|\le5$ on the boundary circle $|z|=1$, then $|f|\le5$ everywhere inside too, an immediate consequence.)
Morera's theorem
The converse of Cauchy's theorem: if $f$ is continuous and $\oint_\gamma f\,dz=0$ for every closed contour $\gamma$ in a domain, then $f$ is holomorphic there. The proof builds the antiderivative directly, $F(z)=\int_{z_0}^z f\,dw$ (well-defined precisely because the vanishing contour integrals make it path-independent), giving $F'=f$; then Cauchy's differentiation formula from earlier in this lesson (holomorphic implies infinitely differentiable) gives $F''=f'$ exists too. Morera's theorem is best thought of as a proof technique: to show a function is holomorphic, it's sometimes easier to show its contour integrals vanish than to check Cauchy-Riemann directly.
Key takeaways: Maximum modulus principle: $|f|$ for nonconstant holomorphic $f$ attains its max only on the boundary of a region, never the interior; the mean-value property is the root cause. Morera's theorem: continuous $f$ with $\oint_\gamma f\,dz=0$ for every closed contour $\gamma$ in a domain $\Rightarrow$ $f$ holomorphic there.