Q:

Prove that $\left|\frac{1}{z^4-4z^2+3}\right|\leq \frac{1}{3}$, if $|z|=2$

Prove that $$\left|\dfrac{1}{z^4-4z^2+3}\right|\leq \dfrac{1}{3},\quad\text{ if}\quad |z|=2$$

I know that I can complete squares on the denominator ($(z^2-2)^2-1$). I know the triangular inequality and some other few things... but I can't work so well with absolute value on $\mathbb{C}$, if anyone could help me...

A:

For $|z|=2$ the triangle inequality gives
$$
 |z^2 -3|  \ge |z^2| - 3 = |z|^2 - 3 = 1 \, ,\\
 |z^2 -1|  \ge |z^2| - 1 = |z|^2 - 1 = 3 \, ,
$$
and therefore 
$$
 |z^4 - 4z^2 +3| = |(z^2-3)(z^2-1)| = |z^2-3 | \cdot | z^2-1 |  \ge 1 \cdot 3 = 3 \, .
$$
Alternatively: You already found that
$$
 z^4 - 4z^2 +3 = (z^2 -2)^2 -1
$$
and for $|z| = 2$ the triangle inequality gives
$$
 |z^2 -2| \ge |z|^2 - 2 = 2^2 - 2 = 2 \\
\Longrightarrow
 |z^4 - 4z^2 +3| \ge | z^2 -2|^2 -1 \ge 2^2 - 1 = 3 \, .
$$

