Cauchy's Integral Formula | Mathematics

6.1 Statement

Theorem 6.1 (Cauchy”s Integral Formula). If $f$ is analytic on a connected domain Containing a simple closed positively oriented contour $\gamma$ And $z_0$ is inside $\gamma$ Then

$f(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}\, dz$

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

$\int_\gamma \frac{f(z)}{z - z_0}\, dz = \int_{\gamma_\varepsilon} \frac{f(z)}{z - z_0}\, dz$

On $\gamma_\varepsilon$ : $f(z) = f(z_0) + (z - z_0)f'(\zeta)$ for some $\zeta$ between $z$ and $z_0$ .

$= \int_{\gamma_\varepsilon} \frac{f(z_0)}{z - z_0}\, dz + \int_{\gamma_\varepsilon} f'(\zeta)\, dz = f(z_0) \cdot 2\pi i + 0$

Since $\int_{\gamma_\varepsilon} \frac{dz}{z - z_0} = 2\pi i$ (parameterize $z = z_0 + \varepsilon e^{i\theta}$ ) and $\int_{\gamma_\varepsilon} f'(\zeta)\, dz \to 0$ as $\varepsilon \to 0$ by the ML inequality. $\blacksquare$

6.2 Derivatives

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

$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\, dz$

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

$f^{(n+1)}(z_0) = \lim_{h \to 0} \frac{f^{(n)}(z_0 + h) - f^{(n)}(z_0)}{h} = \lim_{h \to 0} \frac{n!}{2\pi i}\int_\gamma \frac{1}{h}\left[\frac{f(z)}{(z - z_0 - h)^{n+1}} - \frac{f(z)}{(z - z_0)^{n+1}}\right] dz$

$= \frac{(n+1)!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n+2}}\, dz$

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

Corollary 6.3. If $f$ is analytic, then $f$ is infinitely differentiable.

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

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

$|f^{(n)}(z_0)| \leq \frac{n!M}{R^n}$

Proof. From the integral formula: $|f^{(n)}(z_0)| = \frac{n!}{2\pi}\left|\int_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^{n+1}}\, dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}$ . $\blacksquare$

6.4 Liouville’s Theorem

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

Proof. If $|f(z)| \leq M$ for all $z$ Then by Cauchy’s estimates with $R$ arbitrarily large: $|f'(z_0)| \leq \frac{M}{R} \to 0$ as $R \to \infty$ . So $f'(z) = 0$ for all $z$ Meaning $f$ is Constant. $\blacksquare$

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

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

6.5 Fundamental Theorem of Algebra

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

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

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

6.6 Worked Examples: Cauchy’s Integral Formula

Solution

Problem. Evaluate $\int_\gamma \frac{e^z}{z - 1}\, dz$ where $\gamma$ is $|z| = 2$ .

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

$\int_\gamma \frac{e^z}{z - 1}\, dz = 2\pi i \cdot f(1) = 2\pi i \cdot e^1 = 2\pi i e$ . $\blacksquare$

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

By Cauchy’s formula for derivatives with $f(z) = z^2 + 1$ and $z_0 = i$ :

$\int_\gamma \frac{f(z)}{(z - i)^3}\, dz = \frac{2\pi i}{2!}\,f''(i)$ .

$f'(z) = 2z$ , $f''(z) = 2$ . So $f''(i) = 2$ .

$\int_\gamma \frac{z^2 + 1}{(z - i)^3}\, dz = \frac{2\pi i}{2} \cdot 2 = 2\pi i$ . $\blacksquare$

Problem. Evaluate $\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz$ where $\gamma$ is $|z| = 4$ .

Singularities inside $\gamma$ : $z = 0$ and $z = \pi$ .

$\frac{\sin z}{z(z - \pi)} = \frac{1}{\pi}\left(\frac{\sin z}{z - \pi} - \frac{\sin z}{z}\right)$ .

At $z = 0$ : by CIF, $\int_\gamma \frac{\sin z}{z}\, dz = 2\pi i \cdot \sin(0) = 0$ . At $z = \pi$ : by CIF, $\int_\gamma \frac{\sin z}{z - \pi}\, dz = 2\pi i \cdot \sin(\pi) = 0$ .

$\int_\gamma \frac{\sin z}{z(z - \pi)}\, dz = \frac{1}{\pi}(0 - 0) = 0$ .

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

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

$\int_\gamma \frac{e^{2z}}{(z-1)^2(z+1)}\, dz = \frac{1}{4} \cdot 2\pi i \cdot e^{-2} - \frac{1}{4} \cdot 2\pi i \cdot e^2 + \frac{1}{2} \cdot \frac{2\pi i}{1!} \cdot 2e^2$

$= \frac{\pi i e^{-2}}{2} - \frac{\pi i e^2}{2} + 2\pi i e^2 = \frac{\pi i e^{-2}}{2} + \frac{3\pi i e^2}{2}$ .

6.7 Common Pitfalls

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

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Intuition

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.

Cross-References

Cauchy’s Theorem: Cauchy’s theorem provides the foundation for the integral formula by establishing path independence.
Taylor and Laurent Series: The integral formula leads to power series representations of analytic functions.
Liouville’s Theorem: Liouville’s theorem follows from Cauchy’s estimates and characterizes bounded entire functions.
Classical Mechanics
Electromagnetism

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