advanced29 min studyLesson 363 in Mathematics

Cauchy's integral formula and derivative estimates

Recover interior derivatives from surrounding values and use contour bounds to prove analytic rigidity.

On this page

What you’ll understand

  • Apply the Cauchy formula with the correct factorial and orientation.
  • Derive derivative bounds from a circle maximum.
  • Explain how integral formulas imply local power series and Liouville's theorem.

A holomorphic function's values around a small surrounding circle determine its value at the center and every derivative there. Cauchy's integral formula is the mechanism behind that remarkable rigidity. It converts local information into contour integrals and turns a bound on surrounding values into a bound on interior variation. The formula is not valid merely because the numerator looks familiar: check holomorphicity throughout a neighborhood of the closed disk and keep the contour orientation, center, and derivative order explicit.

A surrounding circle reconstructs the interior

Suppose f is holomorphic on an open neighborhood of a closed disk centered at a with radius R. With counterclockwise boundary orientation, its value at any interior point is the integral of f divided by the displacement from that point, scaled by 1/(2πi). The kernel has a singularity at the evaluation point, but the original f is holomorphic there. One proof removes a tiny circle around that point, applies Cauchy's theorem to the intervening region, and evaluates the limiting small-circle contribution.

Differentiating with respect to the interior evaluation point yields formulas for all higher derivatives. The contour remains a positive distance from a compact interior neighborhood, so the kernel derivatives are uniformly controlled and differentiation under the integral is justified. This establishes that a holomorphic function has derivatives of every order, rather than assuming such smoothness at the outset. Reversing orientation reverses the integral; the standard displayed formula always uses positive orientation.

f(m)(a)=m!2πi∮∣ζ−a∣=Rf(ζ)(ζ−a)m+1 dζ,m=0,1,2,…f^{(m)}(a)=\frac{m!}{2\pi i}\oint_{|\zeta-a|=R}\frac{f(\zeta)}{(\zeta-a)^{m+1}}\,d\zeta,\qquad m=0,1,2,\ldots
The circle is traversed counterclockwise and f is holomorphic on an open neighborhood of the closed disk. A denominator of power m+1 corresponds to derivative order m and includes the factorial m!.↔ Scroll wide equations horizontally.

Bounds turn an identity into a theorem

Let M bound the modulus of f on the circle. The contour length is 2πR and the denominator has constant modulus Rm+1R^{m+1}. Applying the length bound to Cauchy's formula gives the derivative estimate below. Notice that larger available circles improve derivative control only when their boundary maxima do not grow too quickly. For an entire function that is globally bounded by one constant, the first-derivative estimate tends to zero as R tends to infinity at every center.

That proves Liouville's theorem: a bounded entire function is constant. Being bounded on a single disk is not enough, and being bounded only along the real axis is not enough. For example, a complex exponential can remain bounded or oscillatory along selected directions while growing along others. The estimate measures a whole surrounding circle, so it explicitly prevents replacing a complex-domain hypothesis with a convenient real slice.

MR=max⁡∣ζ−a∣=R∣f(ζ)∣⟹∣f(m)(a)∣≤m!MRRmM_R=\max_{|\zeta-a|=R}|f(\zeta)|\quad\Longrightarrow\quad |f^{(m)}(a)|\le\frac{m!M_R}{R^m}
The estimate follows from the same closed-disk holomorphicity hypothesis as the integral formula. For a bounded entire function, a common bound M_R≤M for every R forces the first derivative to vanish.↔ Scroll wide equations horizontally.

Local series are controlled by a nearby circle

For points inside a smaller concentric circle, expand the Cauchy kernel as a geometric series. Its convergence is uniform along the fixed integration circle, so term-by-term integration is legitimate. The coefficients become the derivatives at the center divided by factorials. This proves a local Taylor representation and explains why holomorphicity is equivalent to local analyticity. A smooth real function need not equal its Taylor series, so that equivalence is a special consequence of complex differentiability.

A practical calculation separates exact evaluation from an estimate. If the numerator's derivative at the center is known, the formula gives an exact contour integral. If only a modulus bound is available, it gives an inequality and cannot identify the actual derivative. The largest usable circle is constrained by the function's domain and singularities, not by the desire for a small bound. This distinction will reappear when Taylor series stop at nearby singularities and require analytic continuation beyond their original disks.

Match the pole power to the derivative order

Evaluate ∮∣z∣=2e2z/z4 dz\oint_{|z|=2} e^{2z}/z^4\,dz with counterclockwise orientation. Check the numerator's analytic domain and the factorial explicitly.

  1. The numerator f(z)=e2zf(z)=e^{2z} is entire, so it is holomorphic on a neighborhood of the closed radius-2 disk. The denominator power is 4, corresponding to derivative order m=3.

  2. The third derivative is f(3)(z)=8e2zf^{(3)}(z)=8e^{2z}, so f(3)(0)=8f^{(3)}(0)=8. Cauchy's derivative formula says the integral is 2πi f(3)(0)/3!2\pi i\, f^{(3)}(0)/3!.

  3. Divide by 3!=63!=6 before simplifying: 2πi⋅8/6=8πi/32\pi i\cdot8/6=8\pi i/3. As a check, the coefficient of z³ in the numerator is 23/3!=4/32^3/3!=4/3, and multiplying that coefficient by 2πi gives the same integral.

The result

The integral is 8πi/38\pi i/3. The singularity belongs to the kernel, while the numerator satisfies the holomorphicity hypothesis needed for the formula.

Explore the model

Interactive lab

Interactive lab · function

Compare a real slice with a complex derivative claim

Pause and Reset; set Coefficient a=1 and Position x=1. Predict slope 2 for the real slice of z², then inspect the tangent readout. In a fresh comparison Reset and set a=1,x=0 to predict slope 0. For z² the complex derivative agrees with that real slice because holomorphicity is known separately. This quadratic plot neither integrates a complex contour nor proves Cauchy's bound from one sampled line.

Compare a real slice with a complex derivative claimThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA real-slice derivative can agree with the complex derivative, but the surrounding-domain argument supplies the analytic guarantee.

Check your understanding

If f is holomorphic near the closed disk |z−a|≤R and |f|≤M on its boundary, which bound is correct?

Try it yourself

A function is holomorphic on a neighborhood of ∣z−1∣≤3|z-1|\le3 and has modulus at most 5 on its boundary. Bound ∣f′′(1)∣|f''(1)| and the modulus of ∮∣z−1∣=3f(z)/(z−1)3 dz\oint_{|z-1|=3}f(z)/(z-1)^3\,dz.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the second-derivative estimate, then remember that the corresponding contour integral is 2πi times the derivative divided by 2!.

Compare with the solution

Cauchy's estimate gives ∣f′′(1)∣≤2!⋅5/32=10/9|f''(1)|\le2!\cdot5/3^2=10/9. The contour integral equals πif′′(1)\pi i f''(1), so its modulus is at most 10π/910\pi/9. Directly, its length bound is (6π)(5/27)=10π/9(6\pi)(5/27)=10\pi/9, confirming the factorial bookkeeping. Neither inequality determines the actual derivative or integral.

Take it with you

  • Cauchy's formula reconstructs every interior derivative from a surrounding contour.
  • The derivative estimate is m! times the boundary maximum divided by radius to power m.
  • Whole-domain control, not one real slice, gives analytic rigidity.

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