I want to show that if a circle has its center in the origin and the radius is $2$, then every point $z$ of that circle has the property that $\left|z^{4}-4z^{2}+3\right|\geq3$.
I know that the circle $C=\left\{ z:\left|z-0\right|=\left|z\right|=2\right\}$ So the norm of $z$ is $2$ but I don't know how to take that and connect it to the inequality. I'm kind of a newbie in this area so any idea would be appreciated.
My answer: $$\left|z^{4}-4z^{2}+3\right|=\left|\left(z^{2}-1\right)\left(z^{2}-3\right)\right|=\left|z^{2}-1\right|\left|z^{2}-3\right|\geq\left|\left|z^{2}\right|-\left|1\right|\right|\left|\left|z^{2}\right|-\left|3\right|\right|=\left|\left|z\right|^{2}-1\right|\left|\left|z\right|^{2}-3\right|=\left|4-1\right|\left|1\right|=3$$
I've used the reverse triangle inequality, and the fact that $\left|z\right|=2\Rightarrow\left|z\right|^{2}=4$
Note that $$z^{4}-4z^{2}+3=(z^2-1)(z^2-3)$$ and recall the reverse triangle inequality: for all $u,v\in\mathbb{C}$, $|u-v|\geq ||u|-|v||$.
Can you take it from here?