Recipe

Evaluate a contour integral with the derivative formula

Use when: the integrand has a single point $z_0$ raised to a power in the denominator, $\dfrac{f(z)}{(z-z_0)^{k}}$ , with $f$ analytic and $z_0$ strictly inside a simple closed contour $C$

Goal: the integral's exact value, obtained from a derivative of $f$ at one point instead of any integration

Taught in The derivative formula

Quick recipe

Confirm the formula applies
Read the exponent and solve for $n$
Compute $f^{(n)}(z_0)$
Solve for the integral using $\dfrac{n!}{2\pi i}\oint_C\dfrac{f(z)}{(z-z_0)^{n+1}}\,dz=f^{(n)}(z_0)$

1
Check three things. First, $C$ is a simple closed contour. Second, $f$ is analytic on $C$ and everywhere inside it. Third, $z_0$ (the value that makes the denominator zero) sits strictly inside $C$ , not on it and not outside.

Example

$\oint_{|z|=2}\dfrac{e^z}{(z-1)^2}\,dz$ : The circle $|z|=2$ is simple and closed. The function $f(z)=e^z$ is entire, so it is analytic inside and on the circle. Also $z_0=1$ satisfies $|1|=1<2$ , so it is strictly inside.

Why: the formula's derivation needs $f$ to have no singularities inside $C$ , and it needs $z_0$ to be enclosed. Skip this check and you may apply the formula where it does not hold.
2
The kernel's exponent in $(z-z_0)^{n+1}$ is always one more than the derivative order $n$ . Set the actual exponent equal to $n+1$ and solve for $n$ .

Example

$(z-1)^2$ has exponent $2$ , so $n+1=2$ and $n=1$ . This integral recovers $f'(z_0)$ , not $f(z_0)$ .
- the exponent is $1$ — $n=0$ , and Step 3 is just finding $f(z_0)$ , with no differentiation.
- the exponent is $2$ or more — $n\ge1$ , and Step 3 needs you to differentiate $f$ that many times.
Why: this off-by-one is the most common place the computation goes wrong. So it gets its own step instead of hiding inside the next one.
3
Differentiate $f$ a total of $n$ times with ordinary rules. Then evaluate the result at $z_0$ .

Example

$f(z)=e^z$ , $n=1$ : $f'(z)=e^z$ , so $f'(1)=e$ .

Why: this is the one piece of data the formula needs. Everything about the contour $C$ has already dropped out.
4
Rearrange the formula to isolate the integral. Then put in $n$ and $f^{(n)}(z_0)$ from the last two steps.

Example

$\oint_{|z|=2}\dfrac{e^z}{(z-1)^2}\,dz=\dfrac{2\pi i}{1!}f'(1)=\dfrac{2\pi i}{1}\cdot e=2\pi i\,e$ .

Why: this is the formula from the technical explanation above, solved for the integral instead of for $f^{(n)}(z_0)$ .

Common mistakes

Reading the exponent in the denominator as $n$ instead of $n+1$ . This shifts the derivative order and the factorial by one.
Forgetting to check that $z_0$ lies inside $C$ , not on it and not outside it.
Using the formula when $f$ has a singularity inside $C$ other than $z_0$ .
Finding $f^{(n)}$ at the wrong point. This happens when $(z-z_0)$ is not written in that exact form, for example treating $z^3$ as centered somewhere other than $z_0=0$ .