Many contour integrals cannot be computed exactly, but we often only need to know they are small. The ML inequality gives a quick upper bound from two numbers: how big $f$ gets on the curve, and how long the curve is. It rests on the triangle inequality and the definition of the contour integral.
Key idea
If $|f(z)| \le M$ at every point of $C$ and $C$ has length $L$, then
$$\left|\int_C f(z)\,dz\right| \le M \cdot L.$$
To find $M$ for a fraction, bound the top from above and the bottom from below; the reverse triangle inequality $|z^2 + a^2| \ge |z|^2 - a^2$ is the usual tool.
Why it works
The modulus of a sum is at most the sum of the moduli, and integrals are limits of sums, so $\left|\int_a^b g(t)\,dt\right| \le \int_a^b |g(t)|\,dt$ for complex $g$. Then
$$\left|\int_C f\,dz\right| = \left|\int_a^b f(z(t))z'(t)\,dt\right| \le \int_a^b |f(z(t))|\,|z'(t)|\,dt \le M\int_a^b |z'(t)|\,dt = M \cdot L.$$
Worked examples
Worked example: A full circle
Bound $\left|\displaystyle\oint_{|z| = 3} \frac{dz}{z^2 + 1}\right|$.
- $L = 2\pi \cdot 3 = 6\pi$.
- On $|z| = 3$: $|z^2 + 1| \ge |z|^2 - 1 = 8$, so $\left|\frac{1}{z^2 + 1}\right| \le \frac{1}{8} = M$.
- $M \cdot L = \frac{6\pi}{8} = \frac{3\pi}{4}$. (The true value is $0$, as the residue theorem will show; a bound only promises "no bigger than".)
Answer: $\frac{3\pi}{4}$
Worked example: An arc that vanishes
Show that $\displaystyle\int_{C_R} \frac{dz}{z^2 + 4} \to 0$ as $R \to \infty$, where $C_R$ is the upper half of $|z| = R$.
- $L = \pi R$, and for $R > 2$: $|z^2 + 4| \ge R^2 - 4$.
- $M \cdot L = \frac{\pi R}{R^2 - 4}$, which behaves like $\frac{\pi}{R} \to 0$.
- So the arc contributes nothing in the limit. This is the step that lets residues compute integrals over the real line.
Answer: the bound $\frac{\pi R}{R^2 - 4} \to 0$
Worked example: A segment
Bound $\left|\displaystyle\int_C z^2\,dz\right|$ along the segment from $0$ to $3 + 4i$.
- $L = |3 + 4i| = 5$. On the segment $|z| \le 5$, so $|z^2| \le 25 = M$.
- $M \cdot L = 125$. (The exact value is $\frac{(3 + 4i)^3}{3}$, with modulus $\frac{125}{3}$, comfortably below the bound.)
Answer: $125$
Useful lengths: a segment from $z_1$ to $z_2$ has $L = |z_2 - z_1|$; a full circle of radius $R$ has $L = 2\pi R$; an arc of angle $\varphi$ has $L = R\varphi$. For $M$, find the largest value of the top and the smallest value of the bottom on the curve, using the triangle inequality in the right direction for each.
Common mistake
Bounding the denominator in the wrong direction. To bound $\frac{1}{|z^2 + 1|}$ from above you need $|z^2 + 1|$ from below. The triangle inequality $|z^2 + 1| \le R^2 + 1$ gives $\frac{1}{|z^2 + 1|} \ge \frac{1}{R^2 + 1}$, a lower bound for the fraction, useless here. Use $|z^2 + 1| \ge R^2 - 1$ instead. Check: your $M$ should be at least as large as $|f|$ at the worst point you can find (here $z = \pm iR$, where $|f| = \frac{1}{R^2 - 1}$).