ML Inequality and Complex Integrals I have a question about the ML-Inequalty and how it applies on complex integrals.
Getting this page from Wikipedia http://en.wikipedia.org/wiki/Estimation_lemma  the denominator in the f(z) is $(z² + 1)²$. Using the absolute value with the triangular inequality we'll get $(z² +1)² \ge |z²| - 1 = r² - 1 $ where $ |z| = r $.
So, I'll get after some algebra this:
$
| 1 /(z² + 1)² | \le 1/(r² -1)²
$
So, here is the question : Do I always need to use the triangle inequality to get the highest upper bound for the $f(z)$ function ? 
Just another thought : 
If I have the function $f(z) = 1/(z² - 1)$ , why can't I use the :
$ M  = 1/(a² -1) $
Because from the triangular inequality I should get : 
$
|z²-1| \le a² + 1
$
and
$
|z²-1| \ge a² - 1
$
After some algebra : 
$ 1/|z² +1| \le 1/(a² + 1) \le 1/(a²-1) $
In some texts I saw people using the middle term.
Thanks!

For your first question, you don't "always have to use the triangle inequality". Any bound that is good enough will do, but the triangle inequality is a powerful tool and often useful.
For your particular question, it is correct that
$$|z^2-1| \le a^2+1$$
if $|z| \ge a (\ge 1)$ but this implies that
$$
\frac{1}{|z^2-1|} \ge \frac{1}{a^2+1}.
$$
Note the change in the direction of the inequality! Hence your estimate goes in the wrong direction. You want to bound $|f|$ from above, not from below.
On the other hand, a sharp estimate is rarely needed. It's often just as good to use a weaker estimate, such as
$$
\frac{1}{z^2-1} \le \frac{C}{a^2}
$$
for some constant $C$, or in "big-O" notation $\le O(a^{-2})$. In fact, it's a (fairly) easy exercise to show that if $p$ is a polynomial of degree, then $|p(z)| \le C|z|^n$ for some constant $C$ is $|z|$ is large enough, and this is just the kind of estimate you need for most basic applications of the residue theorem.
