When $f$ has an antiderivative, closed integrals vanish. Cauchy's theorem says much more: analyticity alone is enough, with no antiderivative in sight, as long as $f$ is analytic inside the contour too. The proof joins Green's theorem with the Cauchy–Riemann equations, and the result is the backbone of the rest of the course.
Idea clave
Cauchy's theorem. If $f$ is analytic on and inside a simple closed contour $C$, then
$$\oint_C f(z)\,dz = 0.$$
"Inside" matters: one point inside $C$ where $f$ is not analytic (such as $0$ for $\frac{1}{z}$) can make the integral nonzero.
Por qué funciona
Write $f = u + iv$ and $dz = dx + i\,dy$. Then $f\,dz = (u\,dx - v\,dy) + i(v\,dx + u\,dy)$. Apply Green's theorem, $\oint P\,dx + Q\,dy = \iint (Q_x - P_y)\,dA$, to each part (assuming continuous partials):
$$\oint_C f\,dz = \iint_D (-v_x - u_y)\,dA + i\iint_D (u_x - v_y)\,dA.$$
Both integrands are $0$ by the Cauchy–Riemann equations $u_y = -v_x$ and $u_x = v_y$. (Goursat later showed the theorem holds without assuming continuous partials.)
Worked examples
Ejemplo resuelto: An entire integrand
Compute $\displaystyle\oint_{|z - 1| = 3} e^{z^2}\cos z\,dz$.
- $e^{z^2}\cos z$ is entire: analytic on and inside every contour.
- By Cauchy's theorem, the integral is $0$. No parametrization needed (and none would be easy).
Respuesta: $0$
Ejemplo resuelto: A singular point outside
Compute $\displaystyle\oint_{|z| = 1} \frac{e^z}{z - 3}\,dz$.
- The only point where the integrand fails to be analytic is $z = 3$.
- $|3| = 3 > 1$, so it lies outside the circle; the integrand is analytic on and inside $C$.
- The integral is $0$.
Respuesta: $0$
Ejemplo resuelto: Counting the bad points
Which points where $f(z) = \dfrac{1}{(z^2 + 4)(z - 3)}$ is not analytic lie inside $|z - i| = 2$?
- $f$ fails at $2i$, $-2i$ and $3$.
- Distances from $i$: $|2i - i| = 1$, $|-2i - i| = 3$, $|3 - i| = \sqrt{10}$. Only $2i$ is within $2$.
- So Cauchy's theorem does not apply here; the integral is not automatically $0$ (later: it is $2\pi i$ times a residue at $2i$).
Respuesta: only $2i$
- List the points where $f$ is not analytic (zeros of denominators, cuts of logarithms, anywhere $\overline{z}$ or $|z|$ appears).
- Check which of them lie inside or on $C$.
- If none does, the integral is $0$. If some do, the residue methods later in the course compute it.
Error común
Applying the theorem with a bad point inside. $\frac{1}{z}$ is analytic on the circle $|z| = 1$, yet $\oint_{|z| = 1}\frac{dz}{z} = 2\pi i$, because $0$ is inside. The hypothesis is "analytic on and inside $C$". Before writing $0$, list the points where $f$ is not analytic and check each one: inside or outside? Also check that $f$ is analytic at all; $\oint_{|z| = 2}\overline{z}\,dz = 8\pi i$, since $\overline{z}$ is not analytic anywhere.