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Cauchy's Integral Formula

Theorem 6.1 (Cauchy’s Integral Formula). If ff is analytic on a connected domain Containing a simple closed positively oriented contour γ\gamma And z0z_0 is inside γ\gamma Then

f(z0)=12πi∫γf(z)z−z0 dzf(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}\, dz

Proof. Let γε\gamma_\varepsilon be a small circle of radius ε\varepsilon around z0z_0. Since f(z)z−z0\frac{f(z)}{z - z_0} is analytic on the region between γ\gamma and γε\gamma_\varepsilon

∫γf(z)z−z0 dz=∫γεf(z)z−z0 dz\int_\gamma \frac{f(z)}{z - z_0}\, dz = \int_{\gamma_\varepsilon} \frac{f(z)}{z - z_0}\, dz

On γε\gamma_\varepsilon: f(z)=f(z0)+(z−z0)f′(ζ)f(z) = f(z_0) + (z - z_0)f'(\zeta) for some ζ\zeta between zz and z0z_0.

=∫γεf(z0)z−z0 dz+∫γεf′(ζ) dz=f(z0)⋅2πi+0= \int_{\gamma_\varepsilon} \frac{f(z_0)}{z - z_0}\, dz + \int_{\gamma_\varepsilon} f'(\zeta)\, dz = f(z_0) \cdot 2\pi i + 0

Since ∫γεdzz−z0=2πi\int_{\gamma_\varepsilon} \frac{dz}{z - z_0} = 2\pi i (parameterize z=z0+εeiθz = z_0 + \varepsilon e^{i\theta}) and ∫γεf′(ζ) dz→0\int_{\gamma_\varepsilon} f'(\zeta)\, dz \to 0 as ε→0\varepsilon \to 0 by the ML inequality. ■\blacksquare

Theorem 6.2 (Cauchy’s Integral Formula for Derivatives). Under the same conditions,

f(n)(z0)=n!2πi∫γf(z)(z−z0)n+1 dzf^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz

Proof. We proceed by induction. The base case n=0n = 0 is Theorem 6.1. For the inductive step, Assume the formula holds for nn. Using the difference quotient:

f(n+1)(z0)=lim⁡h→0f(n)(z0+h)−f(n)(z0)h=lim⁡h→0n!2πi∫γ1h[f(z)(z−z0−h)n+1−f(z)(z−z0)n+1]dzf^{(n+1)}(z_0) = \lim_{h \to 0} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} = \lim_{h \to 0} \frac{n!}{2\pi i}\int_\gamma \frac{1}{h}\left[\frac{f(z)}{(z - z_0 - h)^{n+1}} - \frac{f(z)}{(z - z_0)^{n+1}}\right] dz

=(n+1)!2πi∫γf(z)(z−z0)n+2 dz= \frac{(n+1)!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz

Where we justified passing the limit inside the integral by uniform convergence of the integrand On compact subsets. ■\blacksquare

6.3 Consequences of Cauchy’s Integral Formula

Section titled “6.3 Consequences of Cauchy’s Integral Formula”

Corollary 6.3. If ff is analytic, then ff is infinitely differentiable.

This is remarkable: a single complex derivative implies the existence of all derivatives.

Corollary 6.4 (Cauchy’s Estimates). If ff is analytic on and inside a circle ∣z−z0∣=R|z - z_0| = R And ∣f(z)∣≤M|f(z)| \leq M on the circle, then

∣f(n)(z0)∣≤n!MRn|f^{(n)}(z_0)| \leq \frac{n!M}{R^n}

Proof. From the integral formula: ∣f(n)(z0)∣=n!2π∣∫∣z−z0∣=Rf(z)(z−z0)n+1 dz∣≤n!2π⋅MRn+1⋅2πR=n!MRn|f^{(n)}(z_0)| = \frac{n!}{2\pi}\left|\int_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^{n+1}}\, dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}. ■\blacksquare

Theorem 6.5 (Liouville’s Theorem). Every bounded entire function is constant.

Proof. If ∣f(z)∣≤M|f(z)| \leq M for all zz Then by Cauchy’s estimates with RR arbitrarily large: ∣f′(z0)∣≤MR→0|f'(z_0)| \leq \frac{M}{R} \to 0 as R→∞R \to \infty. So f′(z)=0f'(z) = 0 for all zz Meaning ff is Constant. ■\blacksquare

Corollary 6.6. If ff is entire and ∣f(z)∣≥M|f(z)| \geq M for all zz (bounded away from zero), then ff is constant.

Proof. 1/f1/f is entire and bounded by 1/M1/M So constant by Liouville. ■\blacksquare

Theorem 6.7 (Fundamental Theorem of Algebra). Every non-constant polynomial p(z)∈C[z]p(z) \in \mathbb{C}[z] has a root in C\mathbb{C}.

Proof. Suppose p(z)p(z) has no root. Then f(z)=1/p(z)f(z) = 1/p(z) is entire. Since ∣p(z)∣→∞|p(z)| \to \infty as ∣z∣→∞|z| \to \infty, f(z)→0f(z) \to 0 So ff is bounded. By Liouville’s theorem, ff is constant, so pp Is constant, a contradiction. ■\blacksquare

Corollary 6.8. Every polynomial of degree n≥1n \geq 1 has exactly nn roots in C\mathbb{C} Counting multiplicities.

6.6 Worked Examples: Cauchy’s Integral Formula

Section titled “6.6 Worked Examples: Cauchy’s Integral Formula”
Solution

Problem. Evaluate ∫γezz−1 dz\int_\gamma \frac{e^z}{z - 1}\, dz where γ\gamma is ∣z∣=2|z| = 2.

Solution. The function ezz−1\frac{e^z}{z - 1} has a singularity at z=1z = 1 Which lies inside γ\gamma. By Cauchy’s integral formula with f(z)=ezf(z) = e^z and z0=1z_0 = 1:

∫γezz−1 dz=2πi⋅f(1)=2πi⋅e1=2πie\int_\gamma \frac{e^z}{z - 1}\, dz = 2\pi i \cdot f(1) = 2\pi i \cdot e^1 = 2\pi i e. ■\blacksquare

Problem. Evaluate ∫γz2+1(z−i)3 dz\int_\gamma \frac{z^2 + 1}{(z - i)^3}\, dz where γ\gamma is ∣z∣=2|z| = 2.

By Cauchy’s formula for derivatives with f(z)=z2+1f(z) = z^2 + 1 and z0=iz_0 = i:

∫γf(z)(z−i)3 dz=2πi2! f′′(i)\int_\gamma \frac{f(z)}{(z - i)^3}\, dz = \frac{2\pi i}{2!}\,f''(i).

f′(z)=2zf'(z) = 2z, f′′(z)=2f''(z) = 2. So f′′(i)=2f''(i) = 2.

∫γz2+1(z−i)3 dz=2πi2⋅2=2πi\int_\gamma \frac{z^2 + 1}{(z - i)^3}\, dz = \frac{2\pi i}{2} \cdot 2 = 2\pi i. ■\blacksquare

Problem. Evaluate ∫γsin⁡zz(z−π) dz\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz where γ\gamma is ∣z∣=4|z| = 4.

Singularities inside γ\gamma: z=0z = 0 and z=πz = \pi.

sin⁡zz(z−π)=1π(sin⁡zz−π−sin⁡zz)\frac{\sin z}{z(z - \pi)} = \frac{1}{\pi}\left(\frac{\sin z}{z - \pi} - \frac{\sin z}{z}\right).

At z=0z = 0: by CIF, ∫γsin⁡zz dz=2πi⋅sin⁡(0)=0\int_\gamma \frac{\sin z}{z}\, dz = 2\pi i \cdot \sin(0) = 0. At z=πz = \pi: by CIF, ∫γsin⁡zz−π dz=2πi⋅sin⁡(π)=0\int_\gamma \frac{\sin z}{z - \pi}\, dz = 2\pi i \cdot \sin(\pi) = 0.

∫γsin⁡zz(z−π) dz=1π(0−0)=0\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz = \frac{1}{\pi}(0 - 0) = 0.

Problem. Evaluate ∫γe2z(z−1)2(z+1) dz\int_\gamma \frac{e^{2z}}{(z - 1)^2(z + 1)}\, dz where γ\gamma is ∣z∣=3|z| = 3.

By partial fractions: 1(z−1)2(z+1)=1/4z+1−1/4z−1+1/2(z−1)2\frac{1}{(z-1)^2(z+1)} = \frac{1/4}{z+1} - \frac{1/4}{z-1} + \frac{1/2}{(z-1)^2}.

∫γe2z(z−1)2(z+1) dz=14⋅2πi⋅e−2−14⋅2πi⋅e2+12⋅2πi1!⋅2e2\int_\gamma \frac{e^{2z}}{(z-1)^2(z+1)}\, dz = \frac{1}{4} \cdot 2\pi i \cdot e^{-2} - \frac{1}{4} \cdot 2\pi i \cdot e^2 + \frac{1}{2} \cdot \frac{2\pi i}{1!} \cdot 2e^2

=πie−22−πie22+2πie2=πie−22+3πie22= \frac{\pi i e^{-2}}{2} - \frac{\pi i e^2}{2} + 2\pi i e^2 = \frac{\pi i e^{-2}}{2} + \frac{3\pi i e^2}{2}.

  • Forgetting the n!n! in the derivative formula: f(n)(z0)=n!2πi∫γf(z)(z−z0)n+1 dzf^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz. The n!n! is easy to omit.
  • Applying CIF when z0z_0 lies on the contour. The theorem requires z0z_0 strictly inside γ\gamma.
  • Confusing the orientation: the contour must be positively oriented (counterclockwise).
  • Neglecting to check that ff is analytic on and inside γ\gamma, not just on γ\gamma.

Cauchy’s integral formula says the value of an analytic function inside a contour is completely determined by its values on the contour itself. This is extraordinary: the interior is enslaved to the boundary. Imagine knowing the temperature along the edge of a drum skin and being able to compute the temperature at any interior point. The formula works because analytic functions cannot have local extrema, so information propagates inward from the boundary. The derivative formula extends this: not just the function but all its derivatives are recoverable from boundary data, making complex analysis a boundary theory.