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Cauchy's Integral Formula

Source lecture(s): PHY622 Lec4

Intuition

The value of an analytic function inside a contour is determined entirely by its values on the boundary.

Formal Definition

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

Mathematical Formulation

Repeated differentiation gives \(f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz\).

Derivation

Apply the residue theorem to \(f(z)/(z-z_0)\); the only pole inside \(C\) is simple at \(z_0\) with residue \(f(z_0)\).

Worked Example

\(f(z)=z^2\) gives \(f(0)=\frac{1}{2\pi i}\oint z^2/z\,dz = \frac{1}{2\pi i}\oint z\,dz = 0\) (since \(z\) is analytic).

Common Mistakes

  • Applying when \(f\) is not analytic inside \(C\).
  • Confusing the pole order in the \(n\)-th derivative formula.

Quiz

Q1. What does Cauchy's integral formula tell you about derivatives?

Answer

They are also given by contour integrals.