Cauchy integral theorem and path independence

Cauchy's Integral Theorem (Cauchy-Goursat). Let $f$ be analytic at every point inside and on a simple closed contour $C$, where the enclosed region has no holes (simply connected). Then

where: $C$ is a simple closed contour traversed once, and "analytic inside and on $C$" means $f$ is differentiable at every point of the closed region bounded by $C$ — not just on the boundary curve itself. The result is a direct consequence of the Cauchy-Riemann equations, which is why it holds only for analytic functions and fails the moment a singularity sits inside $C$.

Path independence. If $f$ is analytic on a simply connected domain $D$ and $z_1, z_2 \in D$, then $\int_C f(z)\,dz$ takes the same value along every path $C$ in $D$ from $z_1$ to $z_2$ — it depends only on the endpoints. Two such paths $C_1$ and $C_2$, joined end to end with $C_2$ reversed, form a closed contour on which the integral vanishes by the theorem, so the two open-path integrals must agree.

Antiderivative and evaluation. Path-independence lets us define $F(z) = \int_{z_0}^{z} f(\zeta)\,d\zeta$ unambiguously (any path in $D$), with $F'(z) = f(z)$. Evaluation then mirrors ordinary calculus:

where: $F$ is any antiderivative of $f$ on $D$, and $z_1, z_2$ are the path's endpoints — the shape of the path in between is irrelevant once $f$ is analytic throughout $D$.

Deformation of contours. If $f$ is analytic in the region between two simple closed contours $C_1$ and $C_2$ (with $C_2$ inside $C_1$, both traversed counter-clockwise — the classic picture is a "hole" excised around a singularity), then $\oint_{C_1} f(z)\,dz = \oint_{C_2} f(z)\,dz$. A large or awkward contour can therefore be replaced by a small circle hugging each enclosed singularity, which is exactly the figure below.

Problem. Evaluate $\displaystyle\oint_C e^z\,dz$ for $C: |z|=2$, and separately evaluate $\displaystyle\int_0^{\pi i} e^z\,dz$ along any path.

Step 1. Check analyticity. $e^z$ is entire — analytic at every point of $\mathbb{C}$ — so it is analytic inside and on any simple closed contour, in particular $C: |z|=2$.

Step 2. Apply the theorem to the closed contour. Since the hypothesis of Cauchy's Integral Theorem is satisfied, $\oint_C e^z\,dz = 0$ immediately, with no parametrization needed.

Step 3. Apply path-independence to the open path. $e^z$ has the antiderivative $F(z) = e^z$ (itself), valid everywhere, so $\int_0^{\pi i} e^z\,dz = F(\pi i) - F(0) = e^{\pi i} - e^0$.

Step 4. Evaluate. Using $e^{\pi i} = -1$, $\int_0^{\pi i} e^z\,dz = -1 - 1 = \mathbf{-2}$. Both answers required zero parametrization: the closed integral is $0$ by the theorem, and the open integral is $-2$ by the antiderivative shortcut.

Competitive stretch

Problem. Given $\displaystyle\oint_{C_2}\frac{1}{z}\,dz = 2\pi i$ for $C_2: |z|=1$ (found earlier by direct parametrization), evaluate $\displaystyle\oint_{C_1}\frac{1}{z}\,dz$ for $C_1: |z|=5$.

Step 1. Check whether the theorem applies to $C_1$. $f(z)=1/z$ fails to be analytic at $z=0$, which lies inside $C_1$, so Cauchy's Integral Theorem does not apply — this is not a case that gives zero.

Step 2. Use deformation instead. $1/z$ is analytic in the annular region between $C_1: |z|=5$ and $C_2: |z|=1$ (both enclose only the singularity $z=0$, same orientation), so the deformation principle gives $\oint_{C_1}\frac{1}{z}\,dz = \oint_{C_2}\frac{1}{z}\,dz$.

Step 3. Substitute the known value. $\oint_{C_2}\frac{1}{z}\,dz = 2\pi i$, so $\displaystyle\oint_{C_1}\frac{1}{z}\,dz = \mathbf{2\pi i}$, reusing the earlier direct computation instead of re-parametrizing the larger circle.

• Applying the theorem when a singularity lies inside $C$. Concluding $\oint_C f\,dz = 0$ without checking that $f$ is analytic at every interior point invalidates the whole argument — a single interior singularity is fatal to the hypothesis.
• Confusing "analytic on $C$" with "analytic inside and on $C$." The theorem needs the whole closed region bounded by $C$, not merely the boundary curve.
• Forgetting simple-connectedness. Path-independence and the antiderivative argument need a domain with no holes; domains with holes are exactly where nonzero closed-contour integrals, like $1/z$ around the origin, arise.
• Mismatched orientation in deformation. Both contours must be traversed the same way (typically counter-clockwise) for $\oint_{C_1} f\,dz = \oint_{C_2} f\,dz$ to hold as stated.

Ask the tutor: if the Green's-theorem step or the deformation argument still feels hand-wavy, ask the AI tutor to walk through an extra worked contour before moving on to Cauchy's Integral Formula.