Recipe

Evaluate a contour integral using an antiderivative

Use when: you're asked for $\int_C f(z)\,dz$ and you can write down a candidate $F$ with $F'(z)=f(z)$

Goal: the contour integral's value as $F(z_1)-F(z_0)$ , in one step, with no parametrization or integration along $C$ required

Taught in Antiderivatives and path independence

Quick recipe

Propose a candidate antiderivative
Differentiate the candidate and compare it to $f$
Check that $F$ is analytic on a domain containing all of $C$
Evaluate $F$ at the endpoints and subtract

1
Guess $F$ the way you would for a real antiderivative. Run the usual differentiation rules backward, with $z$ in place of $x$ .

Example

Take $f(z)=z$ . The real rule $\int x\,dx=x^2/2$ carries over. So try $F(z)=z^2/2$ .

Why: with no candidate $F$ to test, the rest of the recipe has nothing to check.
2
Compute $F'(z)$ with the ordinary rules. Confirm it matches $f(z)$ at every point, not just at one value of $z$ .

Example

$F(z)=z^2/2$ gives $F'(z)=z=f(z)$ . So the candidate checks out.

Why: $F$ earns the name "antiderivative of $f$ " only once $F'(z)=f(z)$ is verified. Matching the pattern of the real rule is not enough.
3
Find the largest domain where $F$ is analytic. Confirm that it holds every point of $C$ , from start to end. Watch for a branch cut or a singularity of $F$ that crosses the contour.

Example

$\operatorname{Log}z$ is analytic on $\mathbb{C}$ minus its branch cut. So it is a valid antiderivative of $1/z$ on any domain that avoids the cut. The unit circle $|z|=1$ , traced all the way around, crosses any branch cut through the origin. So the domain check fails for that contour.
- $F$ is analytic on a domain containing all of $C$ — go on to Step 4; the shortcut applies.
- $F$ has a branch cut or singularity that $C$ crosses or winds around — the shortcut does not apply to this contour; parametrize $C$ and integrate from the definition instead.
Why: the theorem behind this shortcut, $\int_C f(z)\,dz=F(z_1)-F(z_0)$ , was proved for a contour that lies fully in a domain where $F$ is analytic. Outside that domain, the chain-rule argument fails.
4
Compute $F(z_1)-F(z_0)$ . Here $z_0$ and $z_1$ are the start and end points of $C$ , exactly as given.

Example

Let $C$ run from $z_0=0$ to $z_1=1+i$ . Then $F(1+i)-F(0)=\dfrac{(1+i)^2}{2}-0=\dfrac{1+2i-1}{2}=\dfrac{2i}{2}=i$ .

Why: this is the identity Step 3 just showed applies here. The route $C$ took between the endpoints has already dropped out.

Common mistakes

Treating "I can write a formula that looks like an antiderivative" as enough. You must differentiate it to confirm $F'=f$ .
Skipping the domain check in Step 3. Then you apply the endpoint formula to a contour that winds around a singularity or crosses a branch cut. It quietly gives a wrong answer instead of failing loudly.
Forgetting that a passed domain check holds only for that domain. $\operatorname{Log}z$ is a valid antiderivative of $1/z$ near most points. It is never valid on a domain that holds a full loop around the origin.
Reporting $F(z_0)-F(z_1)$ instead of $F(z_1)-F(z_0)$ . This quietly flips the sign.