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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?


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?

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up vote 2 down vote accepted

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.

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