How do I find the upper bound of a real number when $z$ has been restricted From what I understand we manipulate the real number in question so that it takes the same format as a triangle inequality (whichever one it may be) and then we can use this to break up multiplications as moduli are multiplicative but I want to see this in action so that I can have a visual understanding of the process.
A question such as : 
Give an upper bound for the real number 
$|\frac{1}{z^2+1}|$ for any $z$ such that  $|z|=2$
or a more difficult example with polynomials on the numerator and denominator would be greately appreciated. 
I have a particular question I want to solve but I don't want to post it because I want to use your advice to solve it myself.
Appreciated.

You could proceed like this. 
Because $\left| \frac{1}{z^2+1} \right| = \frac{1}{|z^2+1|}$, it follows that if $L$ is a lower bound for $|z^2+1|$ then $M = \frac{1}{L}$ is an upper bound for $\left| \frac{1}{z^2+1} \right|$. 
So now we must find a lower bound for $|z^2+1|$. 
The triangle inequality has the form $|a+b| \le |a| + |b|$ which can be rewritten $|a| \ge |a+b| - |b|$. So the idea is to set up some triangle inequality where $|a| = |z^2+1|$. 
Here's a way to do this:
$$|z^2| = |z^2 + 1 - 1| \le |z^2 + 1| + |-1| = |z^2+1| + 1
$$
and therefore, using that $|z|=2$, we obtain
$$|z^2 + 1| \ge |z^2| - 1 = |z|^2 - 1 = 2^2 - 1 = 3
$$
and so
$$\left| \frac{1}{z^2+1} \right| \le \frac{1}{3}
$$
Added after the comment of @BrevanEllefsen: Furthermore, upon substituting $z=2i$ this inequality becomes an equality, and therefore $\frac{1}{3}$ is the least upper bound of $\left|\frac{1}{z^2+1}\right|$ for $|z|=2$, in fact $\frac{1}{3}$ is the maximum value.
