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

Question

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...

Answer 1

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 $$ |z^2 -2| \ge |z|^2 - 2 = 2^2 - 2 = 2 \\ \Longrightarrow |z^4 - 4z^2 +3| \ge | z^2 -2|^2 -1 \ge 2^2 - 1 = 3 \, . $$

Answer 2

HINT

Recall that $$\left|\frac1w \right|=|w|^{-1}$$

and then refer to Martin R answer.