If $ \pmb z $ lies on the circle $\pmb |z|=2$ then show that $\frac{|1|}{|z^4-4z^2+3|}\leq\frac13$ .

Question

I tried solving the above problem by substituting $\pmb z^2$ as $\pmb x$ . Which gave me the result $$\frac{|1|}{|(x-1)(x-3)|} \leq \frac13$$ . I am unable to solve further than this .Also here $z$ is a complex number .

Answer 1

If I have read the question correctly you want to show that $|\frac 1 {z^{4}-4z^{3}+3}| \leq \frac 1 3$. This is easy. $|z^{4}-4z^{3}+3|=|(z^{2}-3)(z^{2}-1)|$. Since $|z^{2}-3| \geq |z|^{2}-3 =1$ and $|z^{2}-1| \geq |z|^{2}-1 =3$ we get $|(z^{2}-3)(z^{2}-1)| \geq 3$. Hence $|\frac 1 {z^{4}-4z^{3}+3}| \leq \frac 1 3$.