Cauchy's formula and the rigidity of holomorphic functions

1. A boundary integral determines an interior value

For $c\in\mathbb C$ and $R>0$, orient the circle $C:|z-c|=R$ counterclockwise. Suppose $f$ is holomorphic on an open neighbourhood of the closed disk $\overline{D(c,R)}$. For every $a$ with $|a-c|<R$,

$$f(a)=\frac1{2\pi i}\int_C\frac{f(z)}{z-a}\,dz.$$

The denominator is never zero on $C$. The formula is an interior-value theorem; it does not assert this value for an exterior point or a point on the contour.

Open set used in this proof. Let $U$ be the open holomorphy neighbourhood of the closed disk in the hypothesis. All moving circles, including the initial boundary circle, are considered in $U\setminus\{a\}$. In the following proof, “the punctured disk” is interpreted within this surrounding open set with $a$ removed. The circles lie in the closed disk and avoid $a$, so the compact homotopy image has local-primitive disks in $U\setminus\{a\}$.

Proof. Choose $\varepsilon>0$ with $|a-c|+\varepsilon<R$. The outer circle deforms to the small circle $C_\varepsilon:|z-a|=\varepsilon$ without passing through $a$: use centres $(1-t)c+ta$ and radii $(1-t)R+t\varepsilon$. Each moving circle encloses $a$ and stays in the original disk. Apply the grid/local-primitive proof of C02 to $f(z)/(z-a)$ on the punctured disk. Although the moving basepoint is not fixed, its two side tracks are identical and cancel, so closed-loop integrals are unchanged. Hence

$$\int_C\frac{f(z)}{z-a}\,dz=\int_{C_\varepsilon}\frac{f(z)}{z-a}\,dz =2\pi if(a)+\int_{C_\varepsilon}\frac{f(z)-f(a)}{z-a}\,dz.$$

The final term has modulus at most $2\pi\sup_{|z-a|=\varepsilon}|f(z)-f(a)|$, which tends to zero by continuity at $a$. This proves the formula. ∎

This proof does not presume that $(f(z)-f(a))/(z-a)$ has a holomorphic extension at $a$; that fact is a consequence of the regularity obtained below.

Worked example. If $|z|=2$ is positively oriented, then

$$\int_{|z|=2}\frac{e^z}{z-1}\,dz=2\pi i e.$$

For the denominator $z-3$ instead, $e^z/(z-3)$ is holomorphic on a neighbourhood of the closed disk, so Cauchy theorem gives zero. Locate the pole before applying the formula.

2. All higher derivatives and their bounds

For $a$ in the open disk the distance from $a$ to the contour is positive. On a sufficiently small disk around $a$, difference quotients of the kernel $1/(z-a)$ converge uniformly in $z\in C$ to $1/(z-a)^2$. The estimation lemma permits passing that limit under the contour integral. Repeating this argument yields

$$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}\,dz,\qquad n=0,1,2,\ldots.$$

The uniform kernel argument proves the existence of these derivatives; it does not assume $f$ was initially $C^\infty$. The same formula shows they are continuous locally because the denominators stay uniformly separated from zero.

On a circle centred at $a$ of radius $r$, put $M_r=\max_{|z-a|=r}|f(z)|$. Length estimation gives the Cauchy estimates

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

The radius must belong to a disk on which the function is holomorphic up to a neighbourhood of its boundary. Taking a larger radius through a singularity is invalid.

Worked example. For $|z|=2$,

$$\int\frac{e^z}{(z-1)^3}\,dz=\frac{2\pi i}{2!}f''(1)=\pi ie.$$

The factorial is $2!$, not $3!$, because the denominator exponent is one more than the derivative order.

3. Local power series, zero isolation and uniqueness

Fix $a$ and a closed disk $|z-a|\le r$ in the holomorphy region. For $|w-a|<r$, on its boundary expand the kernel using a geometric series:

$$\frac1{z-w}=\frac1{z-a}\sum_{n=0}^{\infty}\left(\frac{w-a}{z-a}\right)^n.$$

For $|w-a|\le\rho<r$ this series converges uniformly on the contour because its ratio has modulus at most $\rho/r<1$. Integrate termwise and use the derivative formula:

$$f(w)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(w-a)^n.$$

Thus every holomorphic function is analytic. Conversely, a power series is differentiable inside its disk of convergence by uniform convergence of the differentiated series on smaller disks, so analytic functions are holomorphic. There is no assumption that real differentiable functions share this property.

If $f(a)=0$ and the first nonzero Taylor coefficient is indexed $m$, then

$$f(z)=(z-a)^mg(z),\qquad g(a)\ne0,$$

with $g$ holomorphic nearby. Continuity keeps $g$ nonzero in a smaller disk, making $a$ an isolated zero, of order $m$. If all Taylor coefficients vanish, $f$ vanishes on a disk about $a$.

Identity theorem. If holomorphic $f,g$ on a domain agree on a set with an accumulation point inside that domain, then they agree everywhere.

Proof. Apply the preceding zero discussion to $h=f-g$. Accumulating zeros at an interior point exclude a first nonzero Taylor coefficient; hence $h$ vanishes on a neighbourhood of that point. Let $A$ be the set of points having a neighbourhood on which $h=0$. This set is nonempty and open. If points of $A$ tend to $b$ in the domain, all derivatives $h^{(n)}(b)$ vanish by continuity of derivatives; Taylor expansion then puts $b$ in $A$. Thus $A$ is also relatively closed. Connectedness makes $A$ the whole domain. ∎

Accumulation only at a missing boundary point does not suffice: $\sin(\pi/z)$ has zeros $z=1/n$ accumulating at zero, but is not identically zero on the punctured plane.

4. Mean values and maximum modulus

Parameterizing the formula at the centre gives the mean-value identity

$$f(a)=\frac1{2\pi}\int_0^{2\pi}f(a+re^{it})\,dt.$$

It averages complex values, not moduli. The triangle inequality implies $|f(a)|\le\max_{|z-a|=r}|f(z)|$.

Strong maximum-modulus principle. If $f$ is holomorphic on a domain and $|f|$ has a local maximum at an interior point, then $f$ is constant.

Proof. Suppose $|f(z)|\le M=|f(a)|$ near $a$. If $M=0$, $f=0$ locally. If $M>0$, multiply $f$ by a unit complex constant so that its value at $a$ is $M>0$. On every sufficiently small circle about $a$, the real part is at most $M$ and its average is $M$. Continuity forces its real part to equal $M$ at every point of the circle; since its modulus is at most $M$, its imaginary part must be zero there. Varying the radius shows the function is constant in a disk. The identity theorem propagates the constant throughout the domain. ∎

Boundary version. If $f$ is continuous on the closure of a bounded domain and holomorphic inside, then the maximum of its modulus on the compact closure is attained on the boundary. If an interior maximum occurs, the strong principle makes $f$ constant on the domain and continuity gives the same value on the boundary. Boundedness and continuity on the closure cannot be omitted without replacing this argument.

5. Liouville, polynomial roots and Morera

Liouville's theorem. A bounded entire function is constant. If $|f|\le M$ on $\mathbb C$, apply the $n=1$ estimate on a disk of arbitrary radius $r$ about any $a$: $|f'(a)|\le M/r$. Sending $r\to\infty$ gives $f'=0$, so the function is constant on the connected plane.

Fundamental theorem of algebra. Every nonconstant complex polynomial has a complex root. If $p$ had no root, $1/p$ would be entire. Its leading term implies $|p(z)|\to\infty$ as $|z|\to\infty$, so $1/p$ is bounded outside a large disk; continuity gives boundedness on the remaining compact disk. Liouville makes $1/p$ constant, contradicting the degree of $p$. Repeated division by linear factors then factors every degree-$n$ polynomial into exactly $n$ linear factors counted with multiplicity.

Morera's theorem. If continuous $f$ on an open set has zero integral around every triangle whose closed interior lies in that set, then $f$ is holomorphic there.

Proof. In a disk contained in the set, define a segment-integral function $F$ from its centre. Triangle cancellation and continuity, exactly as in C02, give $F'=f$. Thus $F$ is holomorphic. The higher-derivative formula already proved implies $F'$ is holomorphic, so $f$ is holomorphic locally at every point. ∎

Morera is a converse to a contour theorem with a continuity hypothesis. A typical application: if holomorphic functions converge uniformly on every compact subset of an open set, the limit is continuous locally, and their triangle integrals converge to its triangle integrals; Morera makes the limit holomorphic. Pointwise convergence alone is not this theorem.

6. Exercises

P01 — Foundation. Evaluate $\int_{|z|=3}z^2/(z-1)\,dz$ counterclockwise.

P02 — Core. Evaluate $\int_{|z|=2}\cos z/z^3\,dz$ counterclockwise.

P03 — Core. If $f$ is holomorphic on a neighbourhood of $|z|\le4$ and $|f|\le7$ there, bound $|f^{(3)}(0)|$.

P04 — Proof. Prove that an entire function satisfying $|f(z)|\le C(1+|z|^m)$, where $m\ge0$ is an integer, is a polynomial of degree at most $m$.

P05 — Diagnosis. Does equality at every point $1/n$ force two entire functions to agree? Does equality at every positive integer do so? Justify each answer.

P06 — Proof. Let $f$ be nonvanishing and holomorphic on a domain. If $|f|$ has an interior local minimum, prove that $f$ is constant. Why is nonvanishing necessary?

P07 — Core. Show that a holomorphic function on $|z|<1$, continuous on $|z|\le1$, with $|f|\le2$ on the boundary satisfies $|f|\le2$ everywhere. If $|f(0)|=2$, what follows?

P08 — Extension. If $f_n$ are holomorphic on a domain and converge uniformly on compact subsets, prove their limit is holomorphic using Morera. Explain how the derivative formula also gives convergence of $f_n'$ on compact subsets.

7. Full solutions

P01. The point $1$ lies inside the circle and the numerator is entire. The formula gives $2\pi i\cdot1^2=2\pi i$.

P02. The exponent $3$ requests the second derivative: the integral is $2\pi i(\cos)''(0)/2!=-\pi i$.

P03. Cauchy estimates with radius four give $|f^{(3)}(0)|\le3!\cdot7/4^3=21/32$. This is a uniform upper bound, not necessarily the sharp bound under additional data.

P04. For every integer $n>m$, the estimate at zero gives $|f^{(n)}(0)|\le n!C(1+R^m)/R^n\to0$ as $R\to\infty$. All Taylor coefficients above degree $m$ vanish. Since the function is entire, its Taylor expansion at zero represents it on every disk, hence everywhere. The zero function is included.

P05. Yes for $1/n$: zero is an interior accumulation point in the common domain $\mathbb C$, so the identity theorem applies. No for positive integers: the functions $0$ and $\sin(\pi z)$ agree at all such integers but are not equal. Infinity is not an interior accumulation point of the complex plane.

P06. Because $f$ is nonzero, $1/f$ is holomorphic. A local minimum of $|f|>0$ is a local maximum of $|1/f|$, so the maximum-modulus principle makes $1/f$, hence $f$, constant. Without nonvanishing, $f(z)=z$ has a local minimum of its modulus at zero and is not constant.

P07. Continuity on the compact closed disk gives an attained maximum. The boundary principle bounds it by two. If $|f(0)|=2$, the maximum occurs in the interior, so the strong principle makes $f$ constant, with constant modulus two.

P08. On each small closed disk in the domain, uniform convergence gives continuity of the limit. For any closed triangle contained in the domain, uniform convergence on its compact boundary and the estimation lemma give $\int f=\lim\int f_n=0$. Morera gives holomorphy.

For derivative convergence, write $U$ for the domain and let $K\subset U$ be a nonempty compact set; the empty case is vacuous. Choose $r>0$ so that the compact thickening

$$K_r=\{z\in\mathbb C:\operatorname{dist}(z,K)\le r\}$$

is contained in $U$. If $U\ne\mathbb C$, take $r<\operatorname{dist}(K,\mathbb C\setminus U)$; if $U=\mathbb C$, any fixed $r>0$ works. Each circle of radius $r$ centred at a point of $K$ lies in $K_r$. Applying the Cauchy derivative estimate to the holomorphic function $f_n-f$ therefore gives

$$\sup_{a\in K}|f_n'(a)-f'(a)|\le\frac1r\sup_{z\in K_r}|f_n(z)-f(z)|\longrightarrow0.$$

Uniform convergence on the single compact set $K_r$ proves uniform derivative convergence on $K$. This argument requires an enclosing compact thickening; it does not call the union of the circles a neighbourhood.

8. Summary

Boundary values determine every interior derivative. Geometric expansion of the Cauchy kernel proves analyticity and uniqueness, after which maximum modulus, Liouville and polynomial-root existence follow. Morera provides a route back from vanishing contour integrals to holomorphy. The next chapter treats maximal convergence disks and annuli, where negative powers record holes and isolated singularities.