Cauchy's theorem says integrals vanish when nothing bad is inside. When something bad is inside, the same theorem still helps: it lets us slide and shrink the contour freely, as long as we never drag it across a point where $f$ fails to be analytic. Then any contour around a point can be replaced by a tiny circle around it, where the integral is easy.
Key idea
If $C_1$ can be deformed continuously into $C_2$ through a region where $f$ is analytic, then $\oint_{C_1} f\,dz = \oint_{C_2} f\,dz$. In particular, for any simple closed contour $C$ traced counterclockwise with $a$ inside,
$$\oint_C \frac{dz}{(z - a)^n} = \begin{cases} 2\pi i, & n = 1, \\ 0, & n = 2, 3, \ldots \end{cases}$$
and if $a$ is outside $C$, the integral is $0$ for every $n$.
Why it works
Take a big contour $C$ and a small circle $c$ around $a$, both counterclockwise. Cut the ring between them with two close parallel slits; the boundary of the slit ring is a single closed contour (along $C$, in along one slit, around $c$ clockwise, out along the other slit), and $f$ is analytic inside it. By Cauchy's theorem its integral is $0$. The two slits cancel (same path, opposite directions), leaving $\oint_C f\,dz - \oint_c f\,dz = 0$. For $(z - a)^{-n}$ on the small circle, the circle computation gives $2\pi i$ for $n = 1$ and $0$ otherwise.
Worked examples
Worked example: A square instead of a circle
Compute $\displaystyle\oint_C \frac{5\,dz}{z - i}$, where $C$ is the square with corners $\pm 2 \pm 2i$, counterclockwise.
- The only bad point is $i$, inside the square.
- Shrink the square to a small circle around $i$ without crossing $i$: the integral of $\frac{1}{z - i}$ is $2\pi i$.
- Times $5$: $10\pi i$.
Answer: $10\pi i$
Worked example: A square power
Compute $\displaystyle\oint_{|z| = 4} \frac{dz}{(z + 1)^2}$.
- $-1$ is inside, but the power is $2$.
- $\frac{1}{(z + 1)^2}$ has the antiderivative $-\frac{1}{z + 1}$ away from $-1$, so the integral is $0$.
Answer: $0$
Worked example: Partial fractions
Compute $\displaystyle\oint_{|z| = 2} \frac{dz}{z(z - 3)}$ and $\displaystyle\oint_{|z| = 4} \frac{dz}{z(z - 3)}$.
- $\frac{1}{z(z - 3)} = \frac{1}{3}\left(\frac{1}{z - 3} - \frac{1}{z}\right)$.
- On $|z| = 2$ only $0$ is inside: $\frac{1}{3}(0 - 2\pi i) = -\frac{2\pi i}{3}$.
- On $|z| = 4$ both are inside: $\frac{1}{3}(2\pi i - 2\pi i) = 0$.
Answer: $-\frac{2\pi i}{3}$ and $0$
Common mistake
Expecting $\oint \frac{dz}{(z - a)^2}$ to be $2\pi i$ like the case $n = 1$. Only the power $-1$ survives a trip around $a$. Every other power has an antiderivative $\frac{(z - a)^{1 - n}}{1 - n}$ that is single-valued away from $a$, so its closed integrals are $0$. Check by parametrizing: $\int_0^{2\pi} ir^{1 - n}e^{i(1 - n)t}\,dt$ is $0$ unless $n = 1$.