Cauchy Integral Formula & Derivative Formula

Derivation

Cauchy Integral Formula & Derivative Formula

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Statement

If \( f \) is analytic on an open set containing a positively oriented simple closed contour \( C \) and its interior, then for every point \( z_0 \) strictly inside \( C \) the boundary values of \( f \) determine \( f(z_0) \) and every derivative of \( f \) at \( z_0 \) exactly: \[ f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}\,dz, \qquad f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz . \] As corollaries: every bounded entire function is constant (Liouville), and every nonconstant complex polynomial has a root (fundamental theorem of algebra).

Why it matters

This is the "holographic" heart of complex analysis: the values of an analytic function on a closed curve fix its value — and all its derivatives — at every interior point. Nothing like this is true for merely smooth real functions. It converts local complex differentiability (a single derivative!) into infinite differentiability, Taylor expandability, and extreme rigidity. Almost every contour-integral technique used in physics — residues, dispersion relations, Green's functions, partition-function asymptotics — descends from this formula.

Physically it is the complex-analysis twin of a boundary-value problem in field theory: just as the electrostatic potential inside a charge-free region is determined by its boundary data (its real and imaginary parts are harmonic), an analytic "field" in the plane is fully reconstructed from its boundary. The rotational symmetry of the reconstruction kernel is what produces the mean-value property and, through it, Liouville's theorem.

Assumptions

Assumption.\( f \) is analytic on an open set containing \( C \) and its whole interior. If dropped — say \( f \) has a pole or essential singularity inside \( C \) — the integral picks up residue contributions and no longer returns \( f(z_0) \); the residue theorem replaces this formula. \( C \) is a simple closed, positively oriented, rectifiable contour. Reversing orientation flips the sign; a contour winding \( k \) times multiplies the result by the winding number \( k \); a non-rectifiable curve makes the integral itself undefined. \( z_0 \) lies strictly inside \( C \). If \( z_0 \) is outside, the integrand is analytic throughout the interior and the integral is \( 0 \) by Cauchy–Goursat; if \( z_0 \) sits on \( C \), the integral diverges and only a principal value (giving \( \pi i f(z_0) \) at a smooth boundary point) survives. The region between \( C \) and a small circle about \( z_0 \) admits a crosscut making it simply connected. Dropping this (e.g. working on a multiply connected domain without cuts) invalidates the contour-deformation step; one must then track winding numbers around each hole. Continuity of \( f \) at \( z_0 \) (automatic from analyticity). It is the only property used in the shrinking-circle limit; without it the error term in Step 6 need not vanish.

Derivation

\[ I \equiv \oint_C \frac{f(z)}{z-z_0}\,dz \]

Define the object to be evaluated; the integrand is analytic everywhere inside \( C \) except the single point \( z_0 \). A

\[ \oint_C \frac{f(z)}{z-z_0}\,dz \;=\; \oint_{C_\varepsilon} \frac{f(z)}{z-z_0}\,dz, \qquad C_\varepsilon:\ |z-z_0|=\varepsilon \]

Contour deformation: join \( C \) and the small circle \( C_\varepsilon \) by a crosscut; the integrand is analytic on the cut annular region, so Cauchy–Goursat (assumed prior result) gives zero around its boundary, and the two crosscut traversals cancel. Valid for every \( \varepsilon \) smaller than the distance from \( z_0 \) to \( C \). C

\[ z = z_0 + \varepsilon e^{i\theta}, \qquad dz = i\varepsilon e^{i\theta}\,d\theta, \qquad \theta \in [0, 2\pi] \]

Parametrize the small circle; positive orientation means \( \theta \) increasing. B

\[ \oint_{C_\varepsilon} \frac{f(z)}{z-z_0}\,dz = \int_0^{2\pi} \frac{f\!\left(z_0+\varepsilon e^{i\theta}\right)}{\varepsilon e^{i\theta}}\, i\varepsilon e^{i\theta}\,d\theta = i\int_0^{2\pi} f\!\left(z_0+\varepsilon e^{i\theta}\right) d\theta \]

Substitute; the factors \( \varepsilon e^{i\theta} \) cancel identically — this exact cancellation is why the simple pole is special. A

\[ I = i\int_0^{2\pi} f(z_0)\,d\theta + i\int_0^{2\pi} \left[ f\!\left(z_0+\varepsilon e^{i\theta}\right) - f(z_0) \right] d\theta = 2\pi i\, f(z_0) + E(\varepsilon) \]

Add and subtract the constant \( f(z_0) \); the first integral is elementary. A

\[ |E(\varepsilon)| \le 2\pi \sup_{|z-z_0|=\varepsilon} \left| f(z)-f(z_0) \right| \xrightarrow[\varepsilon \to 0]{} 0 \quad\Rightarrow\quad E(\varepsilon)=0 \ \text{for all admissible } \varepsilon \]

ML estimate plus continuity of \( f \) at \( z_0 \). Crucially, Step 2 showed \( I \) is independent of \( \varepsilon \); a constant whose bound tends to zero is exactly zero, so no approximation survives. C

\[ f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}\,dz \]

Combine Steps 5–6 and divide by \( 2\pi i \). This is the Cauchy integral formula. A

\[ \frac{f(z_0+h)-f(z_0)}{h} = \frac{1}{2\pi i}\oint_C f(z)\,\frac{1}{h}\left[ \frac{1}{z-z_0-h} - \frac{1}{z-z_0} \right] dz = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0-h)(z-z_0)}\,dz \]

Write the difference quotient using the formula at both \( z_0+h \) and \( z_0 \) (both interior for \( |h| \) small); the bracket telescopes algebraically: \( \frac{1}{z-z_0-h}-\frac{1}{z-z_0} = \frac{h}{(z-z_0-h)(z-z_0)} \). A

\[ \left| \oint_C f(z)\left[ \frac{1}{(z-z_0-h)(z-z_0)} - \frac{1}{(z-z_0)^2} \right] dz \right| \le \frac{|h| \, M \, L}{(d/2)\,d^{\,2}} \xrightarrow[h\to 0]{} 0 \]

Legality of the limit under the integral: with \( d = \operatorname{dist}(z_0, C) > 0 \), \( M=\max_C |f| \), \( L=\operatorname{length}(C) \), and \( |h|<d/2 \), the bracket equals \( \frac{h}{(z-z_0-h)(z-z_0)^2} \) and is uniformly \( O(|h|) \) on \( C \). The ML bound forces the difference quotient to converge to the integral of \( f(z)/(z-z_0)^2 \). C

\[ f'(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^2}\,dz \]

The limit exists — so \( f' \) is itself given by a contour integral of the same type. B

\[ f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz \]

Induction on \( n \): assume the formula at order \( n \); the identical difference-quotient telescoping and uniform ML bound of Steps 8–9 applied to the kernel \( (z-z_0)^{-(n+1)} \) produce the factor \( (n+1) \) from \( \frac{d}{dz_0}(z-z_0)^{-(n+1)} = (n+1)(z-z_0)^{-(n+2)} \), building up \( n! \). Each pass proves the previous integral is differentiable — so \( f \) is infinitely differentiable. C

\[ \left| f^{(n)}(z_0) \right| \le \frac{n!}{2\pi} \cdot \frac{M_R}{R^{n+1}} \cdot 2\pi R = \frac{n!\, M_R}{R^n}, \qquad M_R = \max_{|z-z_0|=R} |f(z)| \]

Cauchy estimates: take \( C \) to be the circle of radius \( R \) about \( z_0 \) and apply the ML bound to Step 11. B

\[ f \ \text{entire},\ |f| \le M \ \Rightarrow\ |f'(z_0)| \le \frac{M}{R} \xrightarrow[R\to\infty]{} 0 \ \Rightarrow\ f' \equiv 0 \ \Rightarrow\ f \ \text{constant} \]

Liouville's theorem: an entire function admits every radius \( R \) in Step 12 with \( n=1 \); a bound independent of \( R \) forces the derivative to vanish at every point. C

\[ p(z) \neq 0 \ \forall z \ \Rightarrow\ \frac{1}{p} \ \text{entire and bounded} \ \Rightarrow\ \frac{1}{p} \ \text{constant} \ \Rightarrow\ \text{contradiction} \]

Fundamental theorem of algebra: if a nonconstant polynomial \( p \) had no root, \( 1/p \) would be entire; since \( |p(z)| \to \infty \) as \( |z| \to \infty \) (leading term dominates), \( 1/p \) is bounded, hence constant by Step 13 — contradicting \( p \) nonconstant. So \( p \) has a root, and by polynomial division exactly \( \deg p \) roots with multiplicity. C

Result

\[ f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz, \qquad n = 0, 1, 2, \dots \]

Reading. The value of an analytic function — and every one of its derivatives — at any interior point is a weighted average of its values on the surrounding contour, with weight \( n!/\bigl(2\pi i\,(z-z_0)^{n+1}\bigr) \). The case \( n=0 \) is the Cauchy integral formula; the interior is a hologram of the boundary. Immediate consequences: analytic once means analytic infinitely often; bounded entire functions are constant (Liouville); every nonconstant polynomial has a complex root.

Units check. Assign \( f \) units \( [F] \) and \( z \) units \( [L] \). The right side carries \( [F]\cdot[L] / [L]^{n+1} = [F]/[L]^n \) (the \( dz \) supplies one power of \( [L] \); \( n! \) and \( 2\pi i \) are pure numbers) — exactly the units of an \( n \)-th derivative \( d^n f/dz^n \). For \( n=0 \) both sides carry \( [F] \). Consistent.

Limiting cases

\( n = 0 \): the derivative formula reduces to the Cauchy integral formula \( f(z_0) = \frac{1}{2\pi i}\oint_C f(z)/(z-z_0)\,dz \).
\( f \equiv 1 \): recovers the winding integral \( \oint_C dz/(z-z_0) = 2\pi i \) — the entire content is the geometry of encirclement.
\( z_0 \) outside \( C \): the integrand is analytic throughout the interior, and the integral collapses to \( 0 \) by Cauchy–Goursat.
\( C \) a circle of radius \( r \) centred on \( z_0 \): the formula becomes the mean value property \( f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0 + re^{i\theta})\,d\theta \).
\( R \to \infty \) with \( f \) entire and bounded: Cauchy's estimate \( |f'| \le M/R \to 0 \) gives Liouville's theorem as the infinite-radius limit.

Breaks when

\( f \) fails to be analytic somewhere inside \( C \). A pole or essential singularity of \( f \) (e.g. \( f(z) = e^{1/z} \) with \( 0 \) inside \( C \)) contributes extra residue terms; the formula silently returns the wrong answer if applied anyway. Functions that are real-smooth but not holomorphic — \( f(z) = \bar z \) or \( |z|^2 \), which violate the Cauchy–Riemann equations — satisfy no such formula at all: \( \oint_{|z|=1} \bar z/(z)\,dz = \oint_{|z|=1} \bar z\, z^{-1} dz = 2\pi i \cdot 0 \) fails to reproduce \( \bar z \) at interior points.
Branch cuts thread the contour. For \( \sqrt{z} \) or \( \log z \) with the branch point \( 0 \) inside \( C \), no single-valued analytic branch exists on the full interior; the deformation argument of Step 2 is illegal and the integral depends on where the cut is drawn.
\( z_0 \) lies on the contour. The integral diverges; only a principal value exists, giving \( \pi i f(z_0) \) (half the interior value) at a smooth boundary point — the source of the \( \tfrac{1}{2} \) in Sokhotski–Plemelj and in dispersion theory.
Multiply connected domains or self-intersecting contours. Each hole and each extra loop multiplies contributions by winding numbers; blind use of the simple formula miscounts by integer factors.

Failure modes

The lost factorial. Writing \( f^{(n)}(z_0) = \frac{1}{2\pi i}\oint f/(z-z_0)^{n+1} dz \) without the \( n! \) — a factor of \( 2 \) wrong already at the second derivative.
Off-by-one in the pole order. A pole of order \( n+1 \) extracts the \( n \)-th derivative; students routinely pair \( (z-z_0)^n \) with \( f^{(n)} \) and are one derivative off.
Orientation blindness. A clockwise contour gives \( -2\pi i f(z_0) \); parametrizing \( \theta: 2\pi \to 0 \) without noticing flips every answer's sign.
Averaging about the wrong centre. Using the mean value property \( f(z_0) = \frac{1}{2\pi}\int f\,d\theta \) on a circle that does not have \( z_0 \) at its centre — the simple average only holds at the centre; off-centre one needs the Poisson kernel.
Applying the formula with a hidden singularity. Evaluating \( \oint_{|z|=2} \frac{\tan z}{z-1}\,dz \) as \( 2\pi i \tan 1 \), forgetting that \( \tan z \) has its own poles at \( \pm\pi/2 \approx \pm 1.57 \) inside \( |z| = 2 \).
Boundary-point evaluation. Plugging \( z_0 \in C \) into the formula and reporting \( 2\pi i f(z_0) \) instead of recognizing the principal-value half: \( \pi i f(z_0) \).

Discussion

The formula is a statement of rigidity: analytic functions have no local freedom. Knowing \( f \) on any closed curve pins down \( f \) and all derivatives inside; knowing \( f \) on any tiny disc pins it down on the whole connected domain (analytic continuation). Contrast the real-variable world, where a \( C^\infty \) bump function can be identically zero outside an interval yet nonzero inside — complex analyticity forbids any such stealth. This rigidity is exactly what makes complex methods so powerful and so dangerous: one analytic expression valid in a strip is automatically valid wherever it can be continued, which is how physicists get away with rotating contours, resumming series, and defining \( \zeta(-1) = -\tfrac{1}{12} \).

On the fields thread: the real and imaginary parts of an analytic function are harmonic, \( \nabla^2 u = \nabla^2 v = 0 \), so the Cauchy formula is the complex-analytic packaging of a two-dimensional electrostatics boundary-value problem. "Boundary values determine the interior" is the Dirichlet principle; the kernel \( \frac{1}{2\pi i (z - z_0)} \) plays the role of the Green's function of the disc, and taking real parts of the circle case yields the Poisson integral formula. The mean value property (\( f \) at the centre equals its average over the circle) is Gauss's mean-value theorem for potentials in a charge-free region — the same statement that underlies Earnshaw's theorem: no maxima or minima of a potential in empty space, hence no stable electrostatic trapping.

On the symmetry thread: the mean value property is a direct child of rotational symmetry. The kernel restricted to a centred circle is \( \frac{1}{2\pi}\,d\theta \) — the unique rotation-invariant probability measure on the circle — so the value at the symmetric point is the symmetric average. Liouville's theorem then reads as a scaling statement: the estimate \( |f^{(n)}| \le n! M_R / R^n \) says derivative information decays under dilation, and a function invariant under arbitrarily large dilations of its bound must be scale-free, i.e. constant. In physics language, causality plus boundedness pushes analyticity into half-planes, and the same contour machinery yields the Kramers–Kronig relations: the real and imaginary parts of a causal response function \( \chi(\omega) \) are each other's Hilbert transforms — absorption determines dispersion because \( \chi \) is analytic in the upper half \( \omega \)-plane.

The strongest form of the result is due to Goursat: only existence of the complex derivative is assumed — no continuity of \( f' \) — yet Steps 8–11 manufacture derivatives of all orders and, with the geometric expansion of \( (z-w)^{-1} \), a Taylor series converging on the largest disc avoiding singularities. So holomorphic \( \Rightarrow \) analytic, a genuine theorem rather than a definition. Morera's theorem closes the loop (vanishing contour integrals \( \Rightarrow \) analyticity), giving a robust criterion stable under locally uniform limits — the reason boundedness-plus-convergence arguments (Montel, Vitali) work in complex analysis and power tools like the Riemann mapping theorem. The Cauchy estimates also quantify the radius of convergence: \( \limsup |f^{(n)}(z_0)/n!|^{1/n} = 1/R_{\text{sing}} \), so the nearest complex singularity dictates real Taylor convergence — why the innocuous real function \( 1/(1+x^2) \) has radius exactly \( 1 \): its poles sit at \( \pm i \).

Common misconceptions. (i) The formula does not require knowing \( f \) is "nice" inside — analyticity alone forces infinite smoothness; there is no separate hypothesis to check beyond analyticity on and inside \( C \). (ii) It is not an approximation: the shrinking-circle error in Step 6 is exactly zero, not merely small. (iii) Liouville does not say bounded analytic functions on a disc are constant — boundedness must hold on the entire plane; \( \frac{1}{1-z} \) is bounded on \( |z| \le \tfrac{1}{2} \) and far from constant.

Worked examples

Example 1 — Cauchy integral formula. Evaluate \( \displaystyle I = \oint_{|z|=2} \frac{e^z}{z-1}\,dz \) (positively oriented).

\[ f(z) = e^z, \qquad z_0 = 1, \qquad |z_0| = 1 < 2 \]

Identify the analytic numerator and the pole location: \( e^z \) is entire, and \( z_0 = 1 \) lies strictly inside the circle \( |z| = 2 \). All hypotheses hold. A

\[ I = 2\pi i\, f(z_0) = 2\pi i\, e^1 \]

Apply the \( n=0 \) formula with symbols first: \( I = 2\pi i f(z_0) \), then insert \( z_0 = 1 \). B

\[ I = 2\pi (2.71828\ldots)\, i = 17.079\, i \]

Numbers last: \( e \approx 2.71828 \), \( 2\pi \approx 6.28319 \). The answer is purely imaginary because \( f(z_0) \) is real and the \( 2\pi i \) supplies the phase. A

\[ \oint_{|z|=2} \frac{e^z}{z-1}\,dz = 2\pi i\, e \approx 17.08\, i \quad (\text{dimensionless}) \]

Reading. One pass around any contour enclosing \( z = 1 \) — the radius-2 circle, a square, anything — returns \( 2\pi i \) times the value of \( e^z \) at the enclosed point. Deforming the contour changes nothing so long as \( z = 1 \) stays inside.

Units check. \( z \) and \( f \) are dimensionless here; \( [F]\,[L]/[L] = [F] \) — a pure number, as obtained.

Example 2 — derivative formula with an ML sanity bound. Evaluate \( \displaystyle J = \oint_{|z|=1} \frac{e^{2z}}{z^3}\,dz \) and check it against the crude ML estimate.

\[ f(z) = e^{2z}, \qquad z_0 = 0, \qquad n+1 = 3 \ \Rightarrow\ n = 2 \]

Match the integrand to the kernel \( f(z)/(z-z_0)^{n+1} \): a third-order pole at the origin extracts the second derivative. A

\[ J = \frac{2\pi i}{n!}\, f^{(n)}(z_0) = \frac{2\pi i}{2!}\, f''(0), \qquad f''(z) = 4e^{2z} \ \Rightarrow\ f''(0) = 4 \]

Rearrange the boxed result for the integral, differentiate symbolically twice (each derivative brings down a factor \( 2 \)), then evaluate at \( z_0 = 0 \). B

\[ J = \frac{2\pi i}{2}\cdot 4 = 4\pi i \approx 12.566\, i \]

Insert numbers last: \( 4\pi \approx 12.566 \). A

\[ |J| \le \frac{\max_{|z|=1} |e^{2z}|}{1^3}\cdot 2\pi (1) = 2\pi e^2 \approx 46.4 \quad\text{and}\quad 12.57 < 46.4 \ \checkmark \]

Consistency check via the ML inequality: \( |e^{2z}| = e^{2\operatorname{Re}z} \le e^2 \) on the unit circle, length \( L = 2\pi \). The exact answer respects the bound, as it must. B

\[ \oint_{|z|=1} \frac{e^{2z}}{z^3}\,dz = 4\pi i \approx 12.57\, i \quad (\text{dimensionless}) \]

Reading. The triple pole acts as a second-derivative probe: the contour integral measures \( f''(0)/2! \) — the coefficient of \( z^2 \) in the Taylor series of \( e^{2z} \), namely \( 4/2 = 2 \), times \( 2\pi i \).

Units check. With dimensionless \( z \), \( [F]\,[L]/[L]^3 = [F]/[L]^2 \), the units of a second derivative; all quantities here are pure numbers.

Problems

Evaluate \( \displaystyle \oint_{|z|=2} \frac{z^2}{z-i}\,dz \) (positive orientation).
Solution
Here \( f(z) = z^2 \) is entire and \( z_0 = i \) lies inside \( |z| = 2 \) since \( |i| = 1 < 2 \). The Cauchy integral formula gives \( \oint_C \frac{z^2}{z-i}\,dz = 2\pi i\, f(i) = 2\pi i\, (i)^2 = 2\pi i \,(-1) = -2\pi i \approx -6.283\, i \). Note the answer is independent of the radius — any contour enclosing \( i \) gives the same value.
Evaluate \( \displaystyle \oint_{|z|=1} \frac{e^{z}}{z^2}\,dz \).
Solution
Second-order pole at \( z_0 = 0 \): match to the derivative formula with \( n+1 = 2 \), so \( n = 1 \) and \( f(z) = e^z \). Then \( \oint_C \frac{e^z}{z^2}\,dz = \frac{2\pi i}{1!}\, f'(0) = 2\pi i\, e^0 = 2\pi i \approx 6.283\, i \). Equivalently, the integral reads off the coefficient of \( z^1 \) in \( e^z = 1 + z + z^2/2 + \cdots \), which is \( 1 \), times \( 2\pi i \).
Evaluate \( \displaystyle \oint_{|z|=3} \frac{dz}{(z-1)(z-2)} \), then redo it for the contour \( |z| = 1.5 \).
Solution
Partial fractions: \( \frac{1}{(z-1)(z-2)} = \frac{1}{z-2} - \frac{1}{z-1} \). On \( |z| = 3 \) both points \( z = 1 \) and \( z = 2 \) are interior, and each term is a Cauchy kernel with \( f \equiv 1 \): \( \oint \frac{dz}{z-2} = 2\pi i \) and \( \oint \frac{dz}{z-1} = 2\pi i \). Hence the total is \( 2\pi i - 2\pi i = 0 \). On \( |z| = 1.5 \) only \( z = 1 \) is inside; the \( \frac{1}{z-2} \) piece is analytic there and integrates to zero by Cauchy–Goursat, leaving \( -\oint \frac{dz}{z-1} = -2\pi i \approx -6.283\, i \). The exterior zero for the large contour is a general phenomenon: for \( |z| = R \to \infty \) the ML bound scales as \( (2\pi R)\, R^{-2} \to 0 \), and the integral is \( R \)-independent once both poles are enclosed.
A function \( f \) is analytic on \( |z| \le 5 \) and satisfies \( |f(z)| \le 10 \) on the circle \( |z| = 5 \). Find the best bound on \( |f'''(0)| \) provided by the Cauchy estimates, and evaluate it numerically.
Solution
Cauchy's estimate with \( n = 3 \), \( R = 5 \), \( M_R = 10 \): \( |f'''(0)| \le \frac{n!\, M_R}{R^n} = \frac{3! \times 10}{5^3} = \frac{60}{125} = 0.48 \). (By the maximum modulus principle \( |f| \le 10 \) on the boundary already controls the interior, so no better information is hiding inside.) The bound is sharp in the scaling sense: \( f(z) = 10\,(z/5)^3 \) attains \( |f'''(0)| = 10 \cdot 3!/5^3 = 0.48 \) exactly.
Suppose \( f \) is entire and \( |f(z)| \le 2 + |z|^{3/2} \) for all \( z \). Prove that \( f \) is a polynomial of degree at most \( 1 \), and quantify the argument by bounding \( |f''(z_0)| \) for arbitrary \( z_0 \) using the circle of radius \( R = 10^6 \).
Solution
Apply Cauchy's estimate for \( n = 2 \) on the circle \( |z - z_0| = R \): every point on it has \( |z| \le |z_0| + R \), so \( M_R \le 2 + (|z_0| + R)^{3/2} \) and \( |f''(z_0)| \le \frac{2!\,\bigl[2 + (|z_0| + R)^{3/2}\bigr]}{R^2} \). As \( R \to \infty \) the numerator grows like \( R^{3/2} \) while the denominator grows like \( R^2 \), so the bound behaves as \( 2 R^{-1/2} \to 0 \). Since \( f''(z_0) \) is a fixed number \( \le \) every one of these bounds, \( f''(z_0) = 0 \) for every \( z_0 \); hence \( f'' \equiv 0 \), \( f' \) is constant, and \( f(z) = a z + b \) is a polynomial of degree \( \le 1 \). (Consistency: for \( f = az + b \) the growth bound forces nothing further — degree-1 growth sits below \( |z|^{3/2} \).) Numerically, for \( |z_0| \le 1 \) and \( R = 10^6 \): \( M_R \le 2 + (1 + 10^6)^{3/2} \approx 1.0000015 \times 10^9 \), so \( |f''(z_0)| \le \frac{2 \times 1.000 \times 10^9}{10^{12}} \approx 2.0 \times 10^{-3} \) — already tiny, and shrinking like \( 2/\sqrt{R} \) as the circle grows. This is the same mechanism as Liouville's theorem (the case of exponent \( 0 \)): sub-\( |z|^n \) growth kills the \( n \)-th derivative.