By factoring $z^4-4z^2+3$ into two quadratic factors an using the triangle inequality, show that if $z$ lies on the circle $|z|=2$ ($z$ is a complex number) then
$$\left|\frac{1}{z^4-4z^2+3}\right| \leq 1/3$$
Approach: I am still trying to figure out how to factor the given polynomial into two factors. Maybe we don't have to and we have to use the fact that $|z^4-4z^2+3| \leq |z^4+4z^2+3|$
I don't know to be honest. How can I approach this?
On
Factor the polynomial as: $(z^2-3)(z^2-1)$. You can think of it as factoring the polynomial $x^2-4x+3$ and taking $x = z^2$. Then you can do:
$$\left|\frac{1}{(z^2-3)(z^2-1)}\right| = \frac{1}{|z^2-3||z^2-1|} \leq \frac{1}{||z^2|-3|||z^2|-1||} \leq $$
$$\frac{1}{(||z|^2-3|)(||z|^2-1|)} \leq \frac{1}{(4-3)(4-1)} = \frac{1}{3}$$
$$ z^4 - 4z^2 + 3 = (z^2 - 3)(z^2 - 1). $$ $$ \lvert z^4 - 4z^2 + 3 \rvert = \lvert z^2 - 3 \rvert \lvert z^2 - 1 \rvert \ge (\lvert z^2 \rvert - 3)(\lvert z^2 \rvert - 1) = ({\lvert z \rvert}^2 - 3)({\lvert z \rvert}^2 - 1) = (4-3)(4-1) = 3 $$ which proves the result after rearranging.