On the previous page we saw that contour
integrals of powers around the unit circle almost always vanished — only
1/z, which is not holomorphic at the origin, survived. That
was a hint of a sweeping law. Cauchy's integral theorem says the vanishing was
no coincidence:
-
If f is holomorphic on and inside a simple closed contour
\gamma (a region with no holes), then
\displaystyle\oint_\gamma f(z)\,dz = 0.
No singularity inside the loop means no net contribution. Let us prove it by turning the complex
integral into two real ones and unleashing Green's theorem.
Deriving it from Green's theorem
Write f = u + iv and
dz = dx + i\,dy. We expand the product and collect real and
imaginary line integrals, then convert each with
Green's theorem.
Step 1 — expand the integrand. Multiply
(u + iv)(dx + i\,dy):
(u + iv)(dx + i\,dy) = (u\,dx - v\,dy) + i\,(v\,dx + u\,dy).
Step 2 — split into two real circulations. So the contour integral is a real
line integral plus i times another:
\oint_\gamma f\,dz = \oint_\gamma (u\,dx - v\,dy) + i\oint_\gamma (v\,dx + u\,dy).
Step 3 — apply Green's theorem to each. Green's theorem turns a circulation
\oint (P\,dx + Q\,dy) into the double integral
\iint_D (Q_x - P_y)\,dA over the enclosed region
D. For the first integral
P = u,\ Q = -v; for the second
P = v,\ Q = u:
\oint_\gamma f\,dz = \iint_D (-v_x - u_y)\,dA + i\iint_D (u_x - v_y)\,dA.
Step 4 — invoke Cauchy–Riemann. Because f is
holomorphic, the Cauchy–Riemann
equations u_x = v_y and
u_y = -v_x hold throughout D. Read off each
integrand:
-v_x - u_y = -v_x - (-v_x) = 0, \qquad u_x - v_y = u_x - u_x = 0.
Step 5 — conclude. Both double integrals are integrals of zero, so
\oint_\gamma f(z)\,dz = 0.
The Cauchy–Riemann equations are precisely the conditions that make both Green's-theorem
integrands vanish. Holomorphy and a vanishing loop integral are two faces of the same fact.
Path-independence
A vanishing loop integral has an immediate, powerful consequence. Take two points
z_0 and z_1 and two paths
\gamma_1, \gamma_2 from one to the other.
Traverse \gamma_1 forwards and \gamma_2
backwards: together they form a closed loop, so by Cauchy's theorem the total is zero. Reversing
a path negates its integral, hence
\int_{\gamma_1} f\,dz - \int_{\gamma_2} f\,dz = 0 \quad\Longrightarrow\quad \int_{\gamma_1} f\,dz = \int_{\gamma_2} f\,dz.
For a holomorphic f, the value of
\int f\,dz between two points does not depend on which
route you take — only on where you start and where you end. This is exactly the behaviour of a
gradient field in real vector calculus, and it lets us define an antiderivative
F with F' = f and write
\int_{z_0}^{z_1} f\,dz = F(z_1) - F(z_0), a complex fundamental
theorem of calculus. The freedom to deform a contour without changing its integral is
the engine behind Cauchy's integral formula and the whole residue calculus to come.
Bend the path, keep the answer
The two coloured curves both run from z_0 to
z_1. Bend the upper path with the bow slider: provided
f is holomorphic in the region they enclose, both give the
same integral, and the closed loop formed by going out along one and back along the
other integrates to zero — Cauchy's theorem in a picture.
The theorem needs f to be holomorphic everywhere inside the
loop, not merely on the loop itself. The "simply connected / no holes" condition on the enclosed
region is essential and easy to forget. If even a single singularity — say a pole — is
trapped inside the contour, then \oint_\gamma f\,dz is
generally not zero. That is exactly what happened with
1/z around the unit circle, where the integral came out to
2\pi i. Far from being an exception, that nonzero value is the whole
point of what comes next: measuring precisely how much an enclosed singularity contributes is the
job of residue theory.