Finding the number of zeros of a polynomial in the closed disk

Question

Find the number of zeros of $f(z)=z^6-5z^4+3z^2-1$ in $|z|\leq1$.

My attempts have not gotten far.

I know we can examine the related equation $f(w)=w^3-5w^2+3w-1$ in $|w|\leq1$, letting $w=z^2$.

It is clear that $f(w)=0$ for $|w|=1$ if and only if $w=-1$.

My main problem is that this seems obviously like an application of Rouché's theorem, where we let $g(w)=5w^2$, whereupon we have the relation $|f(w)|< |g(w)|$ for all $|w|=1$, with the sole exception that equality holds when $w=-1$.

I would like to use Rouché to equate the number of zeros of $f$ and $g$, but my understanding of Rouché is that the inequality should be strict with no exceptions.

Answer 1

Hint: On $|z|=1$ : $$|z^6-1|\leqslant|z|^6+1=2\leqslant5|z|^4-3|z|^2\leqslant|-5z^4+3z^2|$$

Answer 2

Alt. hint:   show that the cubic $\,w^3-5w^2+3w-1\,$ has a unique real root which is $\,\gt 1\,$, then use that the product of all three roots is $\,1\,$.

Answer 3

Rouche's theorem.

if $|g(z)| > |h(z)|$ for all $z$ along some closed contour. $g(z) + h(z)$ has as many zeros inside the contour as $g(z)$ has inside the contour.

But what about when $|g(z)| = |h(z)|$? Then there is the possibility that some zeros lie on the contour.

In this case let $g(z) = 5z^4, h(z) = z^6 + 3z^2 - 1$ on the contour $|z| = 1$

$|g(z)| = 5, |h(z)| \le 5$

There will be $4$ zeros in the disk $|z|\le 1$ with some possibly on the contour.