Proving that a polynomial has no roots on the unit circle

Question

I want to prove that if $|z|=1 $ then $z^8-3z^2+1 \neq 0$. I tried to prove the reciprocal by taking norms in $z^8-3z^2+1= 0$ and then solving for $ |z|$ but it does not work. I also assume $| z|=1 $ and then trying to see that $| z^8-3z^2+1 |> 0 $ but it did not work neither.

Any ideas on this?

Answer 1

If $z^8-3z^2+1 = 0$ then $$ 3 |z|^2 = |3 z^2| = |z^8 + 1| \le |z|^8 + 1 $$ and that is not possible if $|z| = 1$.

Answer 2

Try the reverse triange inequality: $$|z^8 - 3z^2 + 1| \ge |3z^2| - |z^8| - 1.$$

Answer 3

$$|z|=1 \implies |z^8|=1 $$

Therefore, $|3z^2-1|=|z^8|=1$

Note that $|3z^2-1| \ge ||3z^2|-1| =2$

Which is not consistent with $|3z^2-1|=1$