Rouche's Theorem application for $z^6-5z^4+3z^2-1$ in $|z|\leq 1$

Question

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

Taking $g(z)=1$ would be the obvious choice, but it's not the right one. The next choice would be $z^6-1$ because we know the roots of unity, that doesn't work either. Am I missing something obvious? $g(z)=-5z^4$ would work, but then I'd have to expand this and I'm pretty sure the point of the theorem is to avoid Mathematica!

Any insight is appreciated.

Answer 1

If you choose $g(z) = -5z^4+ 3z^2$ and $f(z) = z^6-1$, then $|f(z)| \le 2$ and $|g(z)| \ge 2$ on $|z| = 1$. Moreover, on $|z|=1$ we have $|g(z)| = 2$ only at $\pm 1$ and $|f(1)| = |f(-1)| = 0 < 2$. Thus, Rouché's theorem applies and you need only count the roots of $z^2(-5z^2+3)$ within $|z|=1$.

Answer 2

If the application of Rouche seems inconclusive, one can try to enhance the root separation by some Dandelin-Graeffe root-squaring iterations. Set $w=z^2$, $p(w)=w^3-5w^2+3w-1$ and compute \begin{align} p_1(w^2)&=-p(-w)·p(w)=w^2(w^2+3)^2-(5w^2+1)^2 \\&=w^6-19w^4-w^2-1 \\ p_1(w)&=w^3-19w^2-w-1 \end{align} which already has a largely sufficiently dominant term $-19w^2$. As $p_1(w)$ has 2 roots inside the unit circle, so does $p(w)$, which gives 4 roots of $f(z)$ inside the unit circle.