Finding number of roots of a polynomial in the unit disk I would like to know how to find the number of (complex) roots of the polynomal $f(z) = z^4+3z^2+z+1$ inside the unit disk. The usual way to solve such a problem, via Rouché's theorem does not work, at least not in an "obvious way".
Any ideas?
Thanks!
edit: here is a rough idea I had: For any $\epsilon >0$, let $f_{\epsilon}(z) = z^4+3z^2+z+1-\epsilon$. By Rouché's theorem, for each such $\epsilon$, $f_{\epsilon}$ has exactly 2 roots inside the unit disc. Hence, by continuity, it follows that $f$ has 2 roots on the closed unit disc, so it remains to determine what happens on the boundary. Is this reasoning correct? what can be said about the boundary?

This one is slightly tricky, but you can apply Rouché directly.
Let $g(z) = 3z^2 + 1$. Note that $|g(z)| \geq 2$ for $|z| = 1$ with equality only for $z = \pm i$ (because $g$ maps the unit circle onto the circle with radius $3$ centered at $1$).
On the other hand for all $|z| = 1$ we have the estimate $h(z) = |f(z) - g(z)| = |z(z^3 + 1)| \leq 2$ and for $z = \pm i$ we have $h(\pm i) = \sqrt{2} < 2 \leq |g(\pm i)|$. Therefore $|f(z) - g(z)| < |g(z)|$ for all $|z| = 1$ and thus Rouché can be applied.
