Cauchy's Integral Formula

Cauchy's theorem says a holomorphic integrand around a closed loop gives zero. Put a single singularity inside the loop and something remarkable happens instead: the integral reads off the function's value at that point. This is Cauchy's integral formula.

If f is holomorphic on and inside a simple closed contour \gamma, and a lies inside \gamma, then f(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z - a}\,dz.

The value at an interior point is an average of the function's values on the boundary, weighted by 1/(z - a). Let us see where it comes from.

Deriving the formula

Step 1 — shrink the contour to a tiny circle. The integrand f(z)/(z - a) is holomorphic everywhere inside \gamma except at z = a. By Cauchy's theorem we may deform \gamma down to a small circle C_\varepsilon of radius \varepsilon about a without changing the integral:

\oint_\gamma \frac{f(z)}{z - a}\,dz = \oint_{C_\varepsilon} \frac{f(z)}{z - a}\,dz.

Step 2 — parametrise the small circle. On C_\varepsilon set z = a + \varepsilon e^{it}, so z - a = \varepsilon e^{it} and dz = i\varepsilon e^{it}\,dt. The \varepsilon e^{it} cancels:

\oint_{C_\varepsilon} \frac{f(z)}{z - a}\,dz = \int_0^{2\pi} \frac{f(a + \varepsilon e^{it})}{\varepsilon e^{it}}\,i\varepsilon e^{it}\,dt = i\int_0^{2\pi} f(a + \varepsilon e^{it})\,dt.

Step 3 — split f. Write f(z) = f(a) + \bigl(f(z) - f(a)\bigr). The constant part comes straight out:

i\int_0^{2\pi} f(a)\,dt + i\int_0^{2\pi} \bigl(f(a + \varepsilon e^{it}) - f(a)\bigr)\,dt = 2\pi i\,f(a) + R_\varepsilon.

Step 4 — kill the remainder with the ML inequality. By the ML inequality, the remainder term over C_\varepsilon is bounded by M\,L with L = 2\pi\varepsilon and M = \max |f(z) - f(a)|. Since f is continuous, that maximum \to 0 as \varepsilon \to 0, so

|R_\varepsilon| \le M \cdot 2\pi\varepsilon \;\longrightarrow\; 0.

Step 5 — read off the formula. The integral does not depend on \varepsilon, so in the limit only the constant term remains:

\oint_\gamma \frac{f(z)}{z - a}\,dz = 2\pi i\,f(a) \quad\Longrightarrow\quad f(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z - a}\,dz.

The boundary knows everything

Rearranged for computation, the formula is a ready-made integral evaluator:

\oint_\gamma \frac{f(z)}{z - a}\,dz = 2\pi i\,f(a).

Read the formula as a statement about information, and it is astonishing. The values of a holomorphic function everywhere inside a contour are completely fixed by its values on the boundary. Nothing in the interior is free; the boundary data pins down every interior point at once. There is no analogue in real calculus — a smooth real function on [0, 1] can do anything in the middle regardless of its endpoints. Holomorphy is a far stiffer condition, and this rigidity is exactly what powers the next result: differentiate this formula and a holomorphic function turns out to be infinitely differentiable.

Move the point inside and out

The circle is the contour \gamma; drag the point a with the sliders. When a sits inside, the integral \oint f(z)/(z - a)\,dz equals 2\pi i\,f(a); push a outside and the integrand becomes holomorphic everywhere inside, so by Cauchy's theorem the integral is zero. The readout shows the sample value for f(z) = z^2.

The formula only fires when a is strictly inside \gamma and f is holomorphic on and within the contour. Reach for it at a point outside \gamma and you do not get f(a) — you get 0, because then the integrand f(z)/(z - a) has no singularity inside the loop and Cauchy's theorem sends the whole integral to zero. So \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z - a}\,dz equals f(a) for interior a but 0 for exterior a — never confuse the two cases.